Math Homework Answers - Recent questions and answers in Geometry Answers
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Powered by Question2AnswerAnswered: Form the intersection for the following sets. R = {10, 15, 20} S = {20, 25}
https://www.mathhomeworkanswers.org/74500/form-the-intersection-for-the-following-sets-r-10-15-20-s-20-25?show=239805#a239805
<p><span style="background-color:rgb(255, 255, 255); color:rgb(2, 10, 27); font-family:proximanova,helvetica,arial,sans-serif; font-size:16px">For the intersection of R and S, we need to find what number set R and set S has in common. Since 20 is in R and 20 is in S then </span>
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<span style="background-color:rgb(255, 255, 255); color:rgb(2, 10, 27); font-family:proximanova,helvetica,arial,sans-serif; font-size:16px">R ∩ S=20. </span></p>Geometry Answershttps://www.mathhomeworkanswers.org/74500/form-the-intersection-for-the-following-sets-r-10-15-20-s-20-25?show=239805#a239805Thu, 23 Mar 2017 14:07:02 +0000Answered: area of square problem
https://www.mathhomeworkanswers.org/239729/area-of-square-problem?show=239734#a239734
<p style="text-align:justify">You are using squared graph paper, right? When you have drawn a 2x2 square you just count how many squares of graph paper are enclosed. If the side of the square is 2 then the area enclosed is 4 squares.</p>
<p style="text-align:justify">But how do you draw a square with area 2? What you do is draw the diagonals of each of the 4 enclosed squares. For the top left and bottom right squares you draw the diagonal from the bottom left corner to the top right corner; and for the other two squares you draw the other diagonal. This gives you a square inside the bigger square tilted by 45 degrees.</p>
<p style="text-align:justify">You have also divided the area of 4 squares into 8 triangles. So 8 triangles have a total area of 4 little squares. Each triangle has an area of ½. The tilted square contains 4 of these triangles. So if 8 triangles is equivalent to 4, then 4 triangles is equivalent to 2. That's how you know the area of your tilted square is 2. So its sides have length √2, the length of each diagonal. No Pythagoras!</p>
<p style="text-align:justify">Looking at squares geometrically probably also helps you to answer your other question, when you start with the tilted square. Because the area of the tilted square is 2, its side length is √2. But the side is the diagonal of a unit square. This means to get the length of the side of the square knowing the length of its diagonal, you just divide the length of the diagonal by √2.</p>Geometry Answershttps://www.mathhomeworkanswers.org/239729/area-of-square-problem?show=239734#a239734Wed, 22 Mar 2017 07:44:31 +0000Answered: Suppose you know the length of the diagonal of a square. How can you find the side length of the square? Explain.
https://www.mathhomeworkanswers.org/239726/suppose-length-diagonal-square-find-length-square-explain?show=239730#a239730
d^2 = a^2 + a^2<br />
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d^2 = 2a^2<br />
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a^2 = d^2/2<br />
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a= sgrt(d^2/2) = d/sgrt2 = (d*sgrt2)/2Geometry Answershttps://www.mathhomeworkanswers.org/239726/suppose-length-diagonal-square-find-length-square-explain?show=239730#a239730Tue, 21 Mar 2017 22:00:21 +0000Answered: round 4.123 rounded to the nearest hundredth place
https://www.mathhomeworkanswers.org/21368/round-4-123-rounded-to-the-nearest-hundredth-place?show=239721#a239721
<p><span style="color:#EE82EE"><span style="font-family:georgia,serif"><span style="font-size:26px">4.12<img alt="cool" src="https://www.mathhomeworkanswers.org/qa-plugin/ckeditor4/plugins/smiley/images/shades_smile.gif" style="height:20px; width:20px" title="cool">+</span></span></span></p>Geometry Answershttps://www.mathhomeworkanswers.org/21368/round-4-123-rounded-to-the-nearest-hundredth-place?show=239721#a239721Tue, 21 Mar 2017 16:04:58 +0000Answered: If a trapezoid is isosceles, then each pair of base angles is_________?
https://www.mathhomeworkanswers.org/239565/trapezoid-isosceles-then-each-pair-base-angles-is_________?show=239574#a239574
CongruentGeometry Answershttps://www.mathhomeworkanswers.org/239565/trapezoid-isosceles-then-each-pair-base-angles-is_________?show=239574#a239574Sat, 18 Mar 2017 09:21:15 +0000Answered: the sum of three terms of gp is 13 and sum of their squares is 91 what are the three terms
https://www.mathhomeworkanswers.org/92535/the-three-terms-and-sum-their-squares-what-are-the-three-terms?show=239281#a239281
<p style="text-align:justify">Three terms can be represnted by a, ar, ar^2 where a is the first term and r the common ratio.
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a(1+r+r^2)=13 so a^2(1+r+r^2)^2=169 by squaring.
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a^2+a^r^2+a^2r^4=91=a^2(1+r^2+r^4)
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Divide the two equations: 169/91=(1+r+r^2)^2/(1+r^2+r^4).
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(r^3-1)=(r-1)(1+r+r^2) and (r^6-1)=(r^2-1)(1+r^2+r^4);
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1+r+r^2=(r^3-1)/(r-1); 1+r^2+r^4=(r^6-1)/(r^2-1); (1+r+r^2)^2=(r^3-1)^2/(r-1)^2.
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So 169/91=13/7=(r^3-1)^2(r^2-1)/((r-1)^2(r^6-1))=(r^3-1)(r+1)/((r^3+1)(r-1))=(r^3-1)/((r^2-r+1)(r-1))=(r^2+r+1)/(r^2-r+1).
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Therefore cross-multiplying: 13(r^2-r+1)=7(r^2+r+1); 6r^2-20r+6=0; 3r^2-10r+3=0.
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This factorises: (3r-1)(r-3)=0, giving us two complementary zeroes: r=3 and r=1/3.
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So we can find a: a(1+r+r^2)=13; 13a=13, so a=1 when we substitute r=3.
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The three terms are 1, 3, 9 because 1+3+9=13 and 1+9+81=91.
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If we use r=1/3 we get 9, 3, 1 as the terms.</p>Geometry Answershttps://www.mathhomeworkanswers.org/92535/the-three-terms-and-sum-their-squares-what-are-the-three-terms?show=239281#a239281Thu, 09 Mar 2017 17:33:17 +0000Answered: if ae= 3x+3 and ec = 5x-15 find ac
https://www.mathhomeworkanswers.org/13570/if-ae-3x-3-and-ec-5x-15-find-ac?show=239255#a239255
Actually JkHenry, If you're trying to find AC, you have to plug in "x." First, set 3x+3 and 5x-15 equal. 5x-15=3x+3. Then, add 15 to both sides and you'll have 5x=3x+18. Now, subtract 3x from both sides, getting this result; 2x=18. Now there's not much to go. Simplify 2x=18 to x=9. Now plug in 9 as "x" to the equations and you should get the same answer for both of them. 5*9=45. 45-15= 30. 3*9=27. 27+3=30. :DGeometry Answershttps://www.mathhomeworkanswers.org/13570/if-ae-3x-3-and-ec-5x-15-find-ac?show=239255#a239255Wed, 08 Mar 2017 14:26:48 +0000Answered: Given trapezoid KLMN with KL | LM is 15m long, ∠M=115° and the diagonal LN makes an angle of 35° with LM. Find the area of the trapezoid in square meters.
https://www.mathhomeworkanswers.org/239050/given-trapezoid-diagonal-makes-angle-trapezoid-square-meters?show=239062#a239062
<p style="text-align:justify"><img alt="" src="http://www.mathhomeworkanswers.org/?qa=blob&qa_blobid=6025634570078042836" style="height:249px; width:400px"></p>
<p style="text-align:justify">LP is the perpendicular from L to NM produced.</p>
<p style="text-align:justify">The area of the trapezoid is the sum of the areas of the triangles NKL and NLM. These triangles have the same height given by LP. A few extra angles are shown. Using the sine rule in triangle NLM:</p>
<p style="text-align:justify">15/sin30=15/(1/2)=30=NM/sin35, so NM=30sin35. LP/LM=sin65, so LP=15sin65, the common height of the triangles.</p>
<p style="text-align:justify">Area of triangle NKL=½15*15sin65; area of triangle NLM is ½30sin35*15sin65.</p>
<p style="text-align:justify">Therefore the area of KLMN is ½15sin65(15+30sin35)=½225sin65(1+2sin35)=<a href="denied:tel:218.9229" rel="nofollow">218.9229</a> sq m approx.</p>
<p style="text-align:justify"> </p>
<p style="text-align:justify"> </p>Geometry Answershttps://www.mathhomeworkanswers.org/239050/given-trapezoid-diagonal-makes-angle-trapezoid-square-meters?show=239062#a239062Wed, 01 Mar 2017 20:05:57 +0000Answered: geometry proof
https://www.mathhomeworkanswers.org/127562/geometry-proof?show=238926#a238926
<p><img alt="" src="http://www.mathhomeworkanswers.org/?qa=blob&qa_blobid=12176512439869882289" style="height:159px; width:350px"></p>
<p style="text-align:justify">ABCD is a parallelogram.</p>
<p style="text-align:justify">In triangles ABD and BCD, BD, the diagonal of ABCD, is common to both triangles.</p>
<p style="text-align:justify">Angles BDA and DBC are equal (alternate angles between parallel lines)</p>
<p style="text-align:justify">Angles ABD and BDC are equal for the same reason.</p>
<p style="text-align:justify">The triangles are congruent (ASA) therefore BC=AD and AB=DC, the opposite sides of ABCD are equal (congruent).</p>Geometry Answershttps://www.mathhomeworkanswers.org/127562/geometry-proof?show=238926#a238926Sun, 26 Feb 2017 12:40:01 +0000Answered: How to draw 125, 100, 80, 70degree angles without using compass?
https://www.mathhomeworkanswers.org/37508/how-to-draw-125-100-80-70degree-angles-without-using-compass?show=238855#a238855
<p><img alt="" src="http://www.mathhomeworkanswers.org/?qa=blob&qa_blobid=12209012609095319752" style="height:248px; width:400px"></p>
<p style="text-align:justify">The picture illustrates how to form the required angles. Graph paper is best to use, along with a straight edge or ruler. A protractor is not needed. The angles at P, Q and R are right angles, and you can make these using a set square if you are not using graph paper. The line measurements can be any units as long as the horizontal and vertical measurements use the same unit.</p>
<p style="text-align:justify">Some of the angles are marked. If you position the points A, B and D accurately you will find the following angles:</p>
<p style="text-align:justify">AOQ=125; the angle of 100 is shown on BO produced and AOB=AOQ-BOP=125-45=80; DOQ=AOR=55, and AOD=AOQ-DOQ=125-55=70. So the angles you need are AOQ=125 (shown), AOB=80 and supplementary angle 100 (shown) on BO produced, AOD=70. (AOQ is an approximation with 0.01% error.)</p>Geometry Answershttps://www.mathhomeworkanswers.org/37508/how-to-draw-125-100-80-70degree-angles-without-using-compass?show=238855#a238855Fri, 24 Feb 2017 12:43:35 +0000Answered: How to make a 125 degree angle without compass
https://www.mathhomeworkanswers.org/238853/how-to-make-a-125-degree-angle-without-compass?show=238854#a238854
<p>Use graph paper as shown:</p>
<p><img alt="" src="http://www.mathhomeworkanswers.org/?qa=blob&qa_blobid=9161651350196885881" style="height:267px; width:350px"></p>
<p style="text-align:justify">This construction uses graph paper, but no protractor or compasses are needed, just a straight edge or ruler. The angle is very close to 125 degrees (99.99% accurate), and is probably as accurate as you're going to get by hand. Instead of graph paper you can use a set square and ruler to give you the right angle and measurements.</p>Geometry Answershttps://www.mathhomeworkanswers.org/238853/how-to-make-a-125-degree-angle-without-compass?show=238854#a238854Fri, 24 Feb 2017 11:38:21 +0000Answered: How many parallelograms are in an octagonal prism?
https://www.mathhomeworkanswers.org/49490/how-many-parallelograms-are-in-an-octagonal-prism?show=238819#a238819
it is 8 parallelogramsGeometry Answershttps://www.mathhomeworkanswers.org/49490/how-many-parallelograms-are-in-an-octagonal-prism?show=238819#a238819Thu, 23 Feb 2017 22:46:30 +0000Answered: how to find the height of a trapezoid with only bases and sides
https://www.mathhomeworkanswers.org/238730/how-to-find-the-height-of-trapezoid-with-only-bases-and-sides?show=238735#a238735
<p>In any trapezoid, if the sides are equal, then the trapezoid is symetrical, as shown below.</p>
<p><img alt="" src="http://www.mathhomeworkanswers.org/?qa=blob&qa_blobid=7136858989081293882" style="height:334px; width:600px"></p>
<p>By a simple application of Pythagoras' theorem, we can show that</p>
<p><span style="text-decoration: underline;"><strong>Trapezoid height = 8 </strong></span></p>Geometry Answershttps://www.mathhomeworkanswers.org/238730/how-to-find-the-height-of-trapezoid-with-only-bases-and-sides?show=238735#a238735Thu, 23 Feb 2017 07:58:10 +0000Answered: in a triangle abc, if ab is the greatest side, then prove that angle c > 60
https://www.mathhomeworkanswers.org/144330/in-triangle-abc-if-the-greatest-side-then-prove-that-angle-60?show=238316#a238316
nobody knows except RodGeometry Answershttps://www.mathhomeworkanswers.org/144330/in-triangle-abc-if-the-greatest-side-then-prove-that-angle-60?show=238316#a238316Fri, 17 Feb 2017 18:35:15 +0000Answered: Equilateral triangle ABC, P be point in it such that BP^2+ CP^2 = AP^2. Prove angle BPC = 150
https://www.mathhomeworkanswers.org/98387/equilateral-triangle-abc-point-such-that-prove-angle-bpc-150?show=238315#a238315
150 is your answer.................................................................Geometry Answershttps://www.mathhomeworkanswers.org/98387/equilateral-triangle-abc-point-such-that-prove-angle-bpc-150?show=238315#a238315Fri, 17 Feb 2017 18:34:34 +0000Answered: Find the length and volume of the wire
https://www.mathhomeworkanswers.org/238178/find-the-length-and-volume-of-the-wire?show=238181#a238181
<p style="text-align:justify">The copper wire will form a tight spiral around the cylinder, but we can approximate to this by treating the wire as a number of segments. Each segment has a length equal to the circumference of the cylinder, which is 49π cm (about 3 times its diameter). The segments are placed side by side so we can work out how many there are by dividing the diameter of the wire into the length of the cylinder: 18/0.6=30.</p>
<p style="text-align:justify">So we now know that we need 30 segments of wire, where each segment is 49π cm long. The volume of the segment of wire is length times cross-sectional area, π(0.3)^2=0.09π sq cm, so the volume of a segment is 49π*0.09π cc=43.5250cc. There are 30 of these side by side so the total volume is 30*43.5250=1305.75cc. The length of wire is 30*49π=4618.14cm.</p>
<p style="text-align:justify">This is a fair approximation, because each segment will actually be slightly longer so as to join up with the next segment to form a continuous length, but because the cylinder's dimensions are very much greater than the wire's as a cylinder, the difference won't be significant. So 4618cm and 1306cc should be accurate enough for the length and volume of wire.</p>Geometry Answershttps://www.mathhomeworkanswers.org/238178/find-the-length-and-volume-of-the-wire?show=238181#a238181Thu, 16 Feb 2017 11:48:56 +0000Answered: Find the equation of line which contains the point (-2,3) and is parallel to the line 3x+5y=17. ( write the numerical coefficient of each term to complete the required equation )
https://www.mathhomeworkanswers.org/238016/contains-parallel-numerical-coefficient-required-equation?show=238040#a238040
<p style="text-align:justify">Parallel lines have the same slope, so the slope is -3/5, because 5y=17-3x, y=-3x/5+17/5.
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The equation of the line is y=-3x/5+a where we have to find a by plugging in the given point:
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3=6/5+a, so a=9/5 and y=-3x/5+9/5, which can be written 3x+5y=9 to match its parallel counterpart.</p>Geometry Answershttps://www.mathhomeworkanswers.org/238016/contains-parallel-numerical-coefficient-required-equation?show=238040#a238040Wed, 15 Feb 2017 13:53:43 +0000Answered: A small airplane at an altitude of 5000 feet is flying east at 300 feet per second, and you are watching it with a small telescope as it passes directly overhead.
https://www.mathhomeworkanswers.org/238028/airplane-altitude-flying-watching-telescope-directly-overhead?show=238039#a238039
<p style="text-align: justify;">In t seconds the plane moves 300t feet further east. The plane was initially at an altitude of 5000 feet. What we have is a right triangle with the long leg=5000 and the short leg=300t. As the plane moves the telescope is tilted further and further away from the vertical. The slope (tangent) of this angle is 300t/5000=3t/50.</p>
<p style="text-align: justify;">(A) Now we can put t=5, 10 and 20: 0.3, 0.6, 1.2. The slopes relative to the ground are the inverses of these:</p>
<p style="text-align: justify;">10/3, 5/3, 5/6. The corresponding angles are 73.30, 59.04, 39.81 degrees. The angles from the vertical are these angles subtracted from 90 degrees: 16.70, 30.96, 50.19 degrees.</p>
<p style="text-align: justify;">(B) The slope (from the ground, or angle of tilt of the telescope) is 50/3t or arctan(50/3t) (same as tan^-1(50/3t)).</p>Geometry Answershttps://www.mathhomeworkanswers.org/238028/airplane-altitude-flying-watching-telescope-directly-overhead?show=238039#a238039Wed, 15 Feb 2017 13:46:35 +0000Answered: area of regular hexagon inscribed in a circle of radius 6 cm
https://www.mathhomeworkanswers.org/41449/area-of-regular-hexagon-inscribed-in-a-circle-of-radius-6-cm?show=237922#a237922
<p style="text-align: justify;">A hexagon can be broken down into 6 equilateral triangles of side r, where r is the radius of the circumscribed circle. The height of each triangle is r√3/2 and half the base is r/2, making the area of each triangle r^2√3/4. Since there are six of these the total area is 6r^2√3/4=3r^2√3/2. If r=6, area=54√3=93.53 sq cm approx.</p>
<p style="text-align: justify;">(The height of the triangle can be found using Pythagoras: √(r^2-(r/2)^2)=√(3r^2/4)=r√3/2.)</p>
<p style="text-align: justify;"> </p>Geometry Answershttps://www.mathhomeworkanswers.org/41449/area-of-regular-hexagon-inscribed-in-a-circle-of-radius-6-cm?show=237922#a237922Tue, 14 Feb 2017 12:14:04 +0000Answered: a formula for finding diagonal of square
https://www.mathhomeworkanswers.org/117831/a-formula-for-finding-diagonal-of-square?show=237867#a237867
<p><strong>Sorry don't know how to solve.....................</strong></p>Geometry Answershttps://www.mathhomeworkanswers.org/117831/a-formula-for-finding-diagonal-of-square?show=237867#a237867Mon, 13 Feb 2017 19:40:47 +0000Answered: what is the formula a pentagon?
https://www.mathhomeworkanswers.org/127625/what-is-the-formula-a-pentagon?show=237866#a237866
<p>I think you answered your own question there. You said <span style="font-family:open sans,sans-serif; font-size:18px">Area=5/2 SE. so that's your answer</span></p>Geometry Answershttps://www.mathhomeworkanswers.org/127625/what-is-the-formula-a-pentagon?show=237866#a237866Mon, 13 Feb 2017 19:40:08 +0000Answered: prove that C,D,H,E are concyclic
https://www.mathhomeworkanswers.org/237771/prove-that-c-d-h-e-are-concyclic?show=237787#a237787
<p><img alt="" src="http://www.mathhomeworkanswers.org/?qa=blob&qa_blobid=8111204587555463665" style="height:447px; width:400px"></p>
<p>The picture shows the secants from point P.</p>
<p style="text-align:justify">The quadrilateral CDHE is required to be proved to be a cyclic quadrilateral. That means that CDH+CEH=180=DCE+EHD.</p>
<p style="text-align:justify">Join CB and AD, and join AH and BE. From this construction we get two pairs of similar triangles: APD and BPC, and APH and BPE, because of the common angle at P, and equal angles PCB=DAP, PAH=BEP (angles in the same segment).</p>
<p style="text-align:justify">We can therefore write:</p>
<p style="text-align:justify">PD/PB=PA/PC=DA/BC (triangles PCB and DAP) and PB/PH=PE/PA=BE/HA (triangles PAH and BEP).</p>
<p style="text-align:justify">From this we get: PB.PA=PC.PD=PE.PH.</p>
<p style="text-align:justify">So PC.PD=PE.PH and therefore PC/PE=PH/PD. Therefore, triangles PCE and PHD are also similar because P is the included common angle. This means PDH=PEC. But CDH=180-PDH (supplementary angles on a straight line), so CDH=180-PEC (PEC is the same angle as CEH). These are opposite angles of the quadrilateral CDHE, and this is a definitive property of cyclic quadrilaterals, so CDHE is cyclic. The other two angles must also be supplementary because the angles of a quadrilateral add up to 360 degrees.</p>Geometry Answershttps://www.mathhomeworkanswers.org/237771/prove-that-c-d-h-e-are-concyclic?show=237787#a237787Sun, 12 Feb 2017 15:47:35 +0000Answered: how do i calculate the lenght and breadth of a rectangle if the area = 210^m
https://www.mathhomeworkanswers.org/236844/how-calculate-the-lenght-and-breadth-rectangle-the-area-210?show=236851#a236851
<p style="text-align: justify;">If L=length and B=breadth LB=210. Its perimeter is 2(L+B). Area is assumed to be 210 sq m.</p>
<p style="text-align: justify;">L=210/B. So there are infinitely many solutions, but if L and B are integer metres, we need to break 210 into factors:</p>
<p style="text-align: justify;">(1,210), (2,105), (3,70), (5,42), (6,35), (7,30), (10,21), (14,15)</p>
<p style="text-align: justify;">The perimeters corresponding to these pairs of length and breadth are:</p>
<p style="text-align: justify;">422, 214, 146, 94, 82, 74, 62, 58.</p>Geometry Answershttps://www.mathhomeworkanswers.org/236844/how-calculate-the-lenght-and-breadth-rectangle-the-area-210?show=236851#a236851Tue, 31 Jan 2017 22:33:39 +0000Answered: need help in grade 8 question
https://www.mathhomeworkanswers.org/236557/need-help-in-grade-8-question?show=236559#a236559
<p style="text-align: justify;">This requires imagination. Get some ideas by looking around, or in magazines, or online. See how shapes are used, how colours are used. The letter O is a circle, the letter C can be a crescent (one circle overlapping another). If you decide on a logo you have the opportunity to make letters from geometrical shapes. The letter A or V is a type of triangle, for example. You could even use the letters in your own name, and you don't have to have all the letters the same size or in a line. Make up a fictitious company name or name of a store. You can place them on top of one another. You can use shades of the same colour to suggest shadow and three dimensions. You can cut out shapes from cardboard and experiment by arranging them in different ways. Let your imagination run wild!</p>Geometry Answershttps://www.mathhomeworkanswers.org/236557/need-help-in-grade-8-question?show=236559#a236559Sat, 28 Jan 2017 01:50:50 +0000Answered: need help in reading and writing in math
https://www.mathhomeworkanswers.org/236540/need-help-in-reading-and-writing-in-math?show=236558#a236558
<ol>
<li style="text-align:justify">After the first cut and pile the total thickness will be 0.02mm. On the second cut we double this to 0.04mm; then 0.08, 0.16, 0.32, 0.64, 1.28, 2.56, 5.12, 10.24. So the final thickness is 10.24mm=1.024cm.</li>
<li style="text-align:justify">Day 1, 50 people, Day 2, 78, Day 3, 106; the increase is 28 per day. We have to get from 50 to 200 people, so that's an increase of 150 from Day 1. 150/28=5 and then some. So after 6 days we have 6*28=168 extra people, making 168+50=218.</li>
<li style="text-align:justify">One way of working this out is to pick a triangle of a particular shape and size and write down how many of that size there are; then pick a different sized triangle and do the same for that. The answer is likely to be a multiple of 5. I can see more than 20 triangles. How many can you see?</li>
<li style="text-align:justify">0.4*0.4*0.4*0.4=0.0256. So the height after 4 bounces is 0.0256*2.3=0.05888m=5.888cm. Two more bounces takes us to just under 1cm.</li>
<li style="text-align:justify">Add all the areas together. Take the area of each lake and divide by the total then multiply by 100 to get the percentage.</li>
<li style="text-align:justify">Time how fast you breathe in normally. Let's suppose this is 4 seconds. Remember that you breathe slower when asleep. If you're very energetic you may breathe faster on average. There are 60*60*24=86,400 seconds in a day, so that's 21,600 breaths a day. Multiply by 365.25 to find how many breaths in a year=7,889,400 breaths a year. Now multiply by your age. You can round the figures of course. It's only a fun exercise.</li>
<li style="text-align:justify">a) 1/10 b) 7/10 c) 5/10 d) 7/10 The probability is simply the ratio of the marbles you are counting to the total number of marbles, 10.</li>
</ol>
<p style="text-align:justify">You may have to write in your words what the answers are. I hope this helps.</p>Geometry Answershttps://www.mathhomeworkanswers.org/236540/need-help-in-reading-and-writing-in-math?show=236558#a236558Sat, 28 Jan 2017 01:35:01 +0000Answered: Prove that <ARQ=<BQR
https://www.mathhomeworkanswers.org/236511/prove-that-arq-bqr?show=236512#a236512
<p style="text-align:justify"><BAR+<QCD=180 (opposite angles in a cyclic quad), let z=<QCD, <BAR=180-z;
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<br>
Also, <ABQ+<RDC=180 (ditto), let y=<ABQ, <RDC=180-y.</p>
<p style="text-align:justify">Let x=<APR=<RPD (given because of bisected angle BPC).
<br>
<br>
<ARQ+<QRD=180 (angles on a straight line), let w=<ARQ, w+180-z+x=180 (triangle APR), so w=z-x.</p>
<p style="text-align:justify"><QCP=180-z (angles on straight line).
<br>
<br>
<BQR+<CQR=180 (ditto), let v=<BQR, <CQR=180-v=x+180-z (external angle of triangle PQC), so v=z-x. But w=z-x so v=w and <ARQ=<BQR (i) proved.
<br>
<br>
And since <BQR+<RQC=180, <ARQ+<RQC=180 (ii) proved.</p>Geometry Answershttps://www.mathhomeworkanswers.org/236511/prove-that-arq-bqr?show=236512#a236512Fri, 27 Jan 2017 11:25:48 +0000what is math?
https://www.mathhomeworkanswers.org/236269/what-is-math
<p>I need asnwer for this question <span style="font-size:116%"><em>I</em><em>f</em><em>f</em><span style="font-family:mathjax_main; font-size:120%">(</span><em>x</em><span style="font-family:mathjax_main; font-size:120%">)</span><span style="font-family:mathjax_main; font-size:120%">=</span><em>a</em><span style="font-family:mathjax_main; font-size:120%">,</span><em>t</em><em>h</em><em>e</em><em>n</em><em>F</em><span style="font-family:mathjax_main; font-size:120%">(</span><em>x</em><span style="font-family:mathjax_main; font-size:120%">)</span><span style="font-family:mathjax_main; font-size:120%">=</span><em>a</em><em>x</em><span style="font-family:mathjax_main; font-size:120%">+</span><em>C</em></span>
<br>
<span style="font-size:116%"><em>I</em><em>f</em><em>f</em><span style="font-family:mathjax_main; font-size:120%">(</span><em>x</em><span style="font-family:mathjax_main; font-size:120%">)</span><span style="font-family:mathjax_main; font-size:120%">=</span><em>x</em><em>a</em><span style="font-family:mathjax_main; font-size:120%">,</span><em>t</em><em>h</em><em>e</em><em>n</em><em>F</em><span style="font-family:mathjax_main; font-size:120%">(</span><em>x</em><span style="font-family:mathjax_main; font-size:120%">)</span><span style="font-family:mathjax_main; font-size:120%">=</span><em>x</em><em>a</em><span style="font-family:mathjax_main; font-size:60%">+</span><span style="font-family:mathjax_main; font-size:60%">1</span><em>a</em><span style="font-family:mathjax_main; font-size:84.9%">+</span><span style="font-family:mathjax_main; font-size:84.9%">1</span><span style="font-family:mathjax_main; font-size:120%">+</span><em>C</em><span style="font-family:mathjax_main; font-size:120%">(</span><em>u</em><em>n</em><em>l</em><em>e</em><em>s</em><em>s</em><em>a</em><span style="font-family:mathjax_main; font-size:120%">=</span><span style="font-family:mathjax_main; font-size:120%">−</span><span style="font-family:mathjax_main; font-size:120%">1</span><span style="font-family:mathjax_main; font-size:120%">)</span></span>
<br>
<span style="font-size:116%"><em>I</em><em>f</em><em>f</em><span style="font-family:mathjax_main; font-size:120%">(</span><em>x</em><span style="font-family:mathjax_main; font-size:120%">)</span><span style="font-family:mathjax_main; font-size:120%">=</span><span style="font-family:mathjax_main; font-size:84.9%">1</span><em>x</em><span style="font-family:mathjax_main; font-size:120%">,</span><em>t</em><em>h</em><em>e</em><em>n</em><em>F</em><span style="font-family:mathjax_main; font-size:120%">(</span><em>x</em><span style="font-family:mathjax_main; font-size:120%">)</span><span style="font-family:mathjax_main; font-size:120%">=</span><span style="font-family:mathjax_main; font-size:120%">ln</span><span style="font-family:mathjax_main; font-size:120%">(</span><em>x</em><span style="font-family:mathjax_main; font-size:120%">)</span><span style="font-family:mathjax_main; font-size:120%">+</span><em>C</em></span>
<br>
<span style="font-size:116%"><em>I</em><em>f</em><em>f</em><span style="font-family:mathjax_main; font-size:120%">(</span><em>x</em><span style="font-family:mathjax_main; font-size:120%">)</span><span style="font-family:mathjax_main; font-size:120%">=</span><em>e</em><em>x</em><span style="font-family:mathjax_main; font-size:120%">,</span><em>t</em><em>h</em><em>e</em><em>n</em><em>F</em><span style="font-family:mathjax_main; font-size:120%">(</span><em>x</em><span style="font-family:mathjax_main; font-size:120%">)</span><span style="font-family:mathjax_main; font-size:120%">=</span><em>e</em><em>x</em><span style="font-family:mathjax_main; font-size:120%">+</span><em>C</em></span>
<br>
<span style="font-size:116%"><em>I</em><em>f</em><em>f</em><span style="font-family:mathjax_main; font-size:120%">(</span><em>x</em><span style="font-family:mathjax_main; font-size:120%">)</span><span style="font-family:mathjax_main; font-size:120%">=</span><span style="font-family:mathjax_main; font-size:120%">cos</span><span style="font-family:mathjax_main; font-size:120%">(</span><em>x</em><span style="font-family:mathjax_main; font-size:120%">)</span><span style="font-family:mathjax_main; font-size:120%">,</span><em>t</em><em>h</em><em>e</em><em>n</em><em>F</em><span style="font-family:mathjax_main; font-size:120%">(</span><em>x</em><span style="font-family:mathjax_main; font-size:120%">)</span><span style="font-family:mathjax_main; font-size:120%">=</span><span style="font-family:mathjax_main; font-size:120%">sin</span><span style="font-family:mathjax_main; font-size:120%">(</span><em>x</em><span style="font-family:mathjax_main; font-size:120%">)</span><span style="font-family:mathjax_main; font-size:120%">+</span><em>C</em></span></p>Geometry Answershttps://www.mathhomeworkanswers.org/236269/what-is-mathMon, 23 Jan 2017 15:44:53 +0000Answered: What is the mean ratio
https://www.mathhomeworkanswers.org/127754/what-is-the-mean-ratio?show=236213#a236213
<p>Do you mean the Golden Mean?</p>
<p style="text-align:justify">Consider a line split into two segments a and b, so the line length is a+b. If the longer segment is a and b/a=a/(a+b). The ratio of a to b is the Golden Ratio or Golden Mean.</p>
<p style="text-align:justify">b(a+b)=a^2, a^2-ab-b^2=0, a^2-ab+b*2/4-b^2=b^2/4, (a-b/2)^2=5b^2/4, a-b/2=±b√5/2, a=b/2±b√5/2.</p>
<p style="text-align:justify">So 2a/b=1±√5=3.236 or -1.236, making a/b=1.61803 approx (we can reject the negative solution because a and b are both positive). So the Golden Ratio or Golden Mean is 1.61803. The reciprocal of this is 0.61803. Therefore we have that the difference of the Golden Mean and its reciprocal is exactly 1, that is, G-1/G=1.</p>Geometry Answershttps://www.mathhomeworkanswers.org/127754/what-is-the-mean-ratio?show=236213#a236213Sun, 22 Jan 2017 17:49:56 +0000Answered: Geometry problem with pythagoras theorem
https://www.mathhomeworkanswers.org/235895/geometry-problem-with-pythagoras-theorem?show=236008#a236008
<p style="text-align:justify">In the picture we need to find out the angle, x<90, that D makes with a1/a2. We can use the cosine rule.</p>
<p style="text-align:justify">C^2=D^2+a2^2+2(a2)(D)cosx; B^2=D^2+a1^2-2(a1)(D)cosx.</p>
<p style="text-align:justify">[Note that if x=90 (Pythagorean assumption), C^2=D^2+a2^2 and B^2=D^2+a1^2 and C^2-B^2=a2^2-a1^2. (C-B)(C+B)=(a2+a1)(a2-a1).</p>
<p style="text-align:justify">77.6(2D+292.6)=40870(40870-2a1); 155.2D+22705.76=1670356900-81740a1.</p>
<p style="text-align:justify">155.2D=1670334194.24-81740a1. From this it's clear that D and a1 cannot be found uniquely. However, because the right-hand side must be positive a1<<a href="denied:tel:1670334194.24" rel="nofollow">1670334194.24</a>/81740=20434.72 approx. Any value in excess of this would make D negative. This means that a2>20435.28. a2>a1.]</p>
<p style="text-align:justify">If x≠90, C^2-B^2=a2^2-a1^2+2(a1+a2)Dcosx.</p>
<p style="text-align:justify">C-B=77.6, C=B+77.6; a1+a2=40870.</p>
<p style="text-align:justify">(C-B)(C+B)=40870(a2-a1)+2*40870Dcosx.</p>
<p style="text-align:justify">77.6(2D+292.6)=40870(40870-2a1)+81740Dcosx.</p>
<p style="text-align:justify">155.2D+22705.76=1670356900-81740a1+81740Dcosx.</p>
<p style="text-align:justify">D(155.2-81740cosx)=1670334194.24-81740a1. We have three variables, but none can be found uniquely. But we can confirm that a1<20434.72 and a2>20435.28, provided cosx<155.2/81740=0.0018987 and x>89.89 degrees. For x<89.89, the left-hand and right-hand sides must both be negative. If x=84 degrees, for example, (cosx is about 0.1), -8018.8D=1670334194.24-81740a1. So a1>20434.72 and a2<20435.28. So the magnitude of x affects the relationship between a1 and a2.</p>
<p style="text-align:justify">Perimeter = 40870+292.6+2D=41162.6+2D.</p>
<p style="text-align:justify">B+C>a1+a2 in order to produce a triangle. 2D+292.6>40870, D>20288.7, so B>20396.2 and C>20473.8.</p>
<p style="text-align:justify">EXAMPLE SOLUTION</p>
<ol>
<li style="text-align:justify">Let D=20400, and x=90, then a1=(1670334194.24-155.2*20400)/81740=20395.99 and a2=20474.01 (approx). B=20507.5, C=20585.1.</li>
<li style="text-align:justify">Let D=20400, and cosx=0.00175 (x=89.90 approx), then a1=(1670334194.24-12.155)/81740=20434.72, a2=20435.28, B=20507.5, C=20585.1.</li>
</ol>
<p style="text-align:justify">The examples show that 40870^2 is such a large number that it is barely affected when anything is subtracted from it so a1 and a2 are both close to 40870/2=20435.</p>Geometry Answershttps://www.mathhomeworkanswers.org/235895/geometry-problem-with-pythagoras-theorem?show=236008#a236008Tue, 17 Jan 2017 22:12:58 +0000Answered: The length of diagonals of a rhombus are 12 and 18. Find the perimeter in simplest radical form.?
https://www.mathhomeworkanswers.org/127911/length-diagonals-rhombus-find-perimeter-simplest-radical?show=235976#a235976
<p><span style="background-color:rgb(255, 255, 255); color:rgb(51, 51, 51); font-family:sans-serif,arial,verdana,trebuchet ms; font-size:13px"><a href="http://brainly.com" rel="nofollow">http://brainly.com</a></span></p>Geometry Answershttps://www.mathhomeworkanswers.org/127911/length-diagonals-rhombus-find-perimeter-simplest-radical?show=235976#a235976Tue, 17 Jan 2017 18:44:01 +0000Answered: adjeent angles in a parallelogram are 5x° and3x°. the smallest angles in the parallelogram are each
https://www.mathhomeworkanswers.org/127934/adjeent-parallelogram-and3x%C2%B0-smallest-angles-parallelogram?show=235967#a235967
<p><span style="background-color:rgb(255, 255, 255); color:rgb(51, 51, 51); font-family:sans-serif,arial,verdana,trebuchet ms; font-size:13px"><a href="http://brainly.com" rel="nofollow">http://brainly.com</a></span></p>Geometry Answershttps://www.mathhomeworkanswers.org/127934/adjeent-parallelogram-and3x%C2%B0-smallest-angles-parallelogram?show=235967#a235967Tue, 17 Jan 2017 18:38:55 +0000Answered: what is the area of a pentagon with the given apothem A=10in,S=14.5in
https://www.mathhomeworkanswers.org/128053/what-is-the-area-of-pentagon-with-the-given-apothem-10in-14-5in?show=235953#a235953
<p><span style="background-color:rgb(255, 255, 255); color:rgb(51, 51, 51); font-family:sans-serif,arial,verdana,trebuchet ms; font-size:13px"><a href="http://brainly.com" rel="nofollow">http://brainly.com</a></span></p>Geometry Answershttps://www.mathhomeworkanswers.org/128053/what-is-the-area-of-pentagon-with-the-given-apothem-10in-14-5in?show=235953#a235953Tue, 17 Jan 2017 17:38:03 +0000Answered: find the Locus L, of points equidistant from B and C, also the Locus L2 of points 4cm from A.
https://www.mathhomeworkanswers.org/128090/find-locus-points-equidistant-from-also-the-locus-points-from?show=235943#a235943
<p><span style="background-color:rgb(255, 255, 255); color:rgb(51, 51, 51); font-family:sans-serif,arial,verdana,trebuchet ms; font-size:13px"><a href="http://brainly.com" rel="nofollow">http://brainly.com</a></span></p>Geometry Answershttps://www.mathhomeworkanswers.org/128090/find-locus-points-equidistant-from-also-the-locus-points-from?show=235943#a235943Tue, 17 Jan 2017 17:22:08 +0000Answered: help with geometric proof involving triangles
https://www.mathhomeworkanswers.org/128172/help-with-geometric-proof-involving-triangles?show=235929#a235929
<p><span style="background-color:rgb(255, 255, 255); color:rgb(51, 51, 51); font-family:sans-serif,arial,verdana,trebuchet ms; font-size:13px"><a href="http://brainly.com" rel="nofollow">http://brainly.com</a></span></p>Geometry Answershttps://www.mathhomeworkanswers.org/128172/help-with-geometric-proof-involving-triangles?show=235929#a235929Tue, 17 Jan 2017 17:02:20 +0000Answered: finding angle of a triangle
https://www.mathhomeworkanswers.org/235914/finding-angle-of-a-triangle?show=235917#a235917
<p style="text-align:justify">Assume BC is the diameter of the Ferris wheel centre O. AOB is an isosceles triangle where AO=BO (radii). So angles ABO and BAO are equal. They must add up to 180-120=60, so each must be 30 degrees. <strong>ABO=30 degrees</strong>. The diameter of the wheel and the position of point E are not relevant in this solution. I guess there are other parts to the question where they would be important.</p>Geometry Answershttps://www.mathhomeworkanswers.org/235914/finding-angle-of-a-triangle?show=235917#a235917Tue, 17 Jan 2017 15:28:54 +0000Answered: Quadrilateral WXYZ has right angles at angle W and angle Y and an acute angle at angle X. Altitudes are dropped from X and Z to diagonal WY, meeting WY at O and P. Prove that OW = PY.
https://www.mathhomeworkanswers.org/235558/quadrilateral-angles-altitudes-dropped-diagonal-meeting?show=235915#a235915
<p><img alt="" src="http://www.mathhomeworkanswers.org/?qa=blob&qa_blobid=5530022634127196183" style="height:505px; width:400px"></p>
<p><span style="font-size:12px">Consider the diagonal WY split into segments: PY, OP and OW.</span></p>
<p><span style="font-family:arial,helvetica,sans-serif"><span style="font-size:12px">Let x=PY, y=OP, z=OW.</span></span></p>
<p style="text-align:justify"><span style="font-family:arial,helvetica,sans-serif"><span style="font-size:12px">The<span style="color:rgb(69, 69, 69)"> following right-angled triangles are similar: OWX, PWZ; OXY, PYZ, because we have complementary angles in right-angles triangles and we have complementary angles as a result of split right angles.</span></span></span></p>
<p style="text-align:justify"><span style="font-family:arial,helvetica,sans-serif"><span style="font-size:12px"><span style="color:rgb(69, 69, 69)">We can write the following side ratios: OW/OX=PZ/PW; OX/OY=PY/PZ. OW=OX.PZ/PW, and PY=OX.PZ/OY, so OW/PY=OY/PW, which means that if OW=PY then OY=PW. OY=x+y; PW=y+z. So z/x=(x+y)/(y+z). Cross-multiply: z(y+z)=x(x+y); yz+z^2-x^2-xy=yz+(z-x)(z+x)-xy=y(z-x)+(z+x)(z-x)=(x+y+z)(z-x)=0. Therefore x+y+z=0, which isn't true so z-x=0 and z=x, i.e., OW=PY (and OY=PW).</span></span></span></p>
<p style="text-align: justify;"><span style="font-size:12px">Incidentally, the other diagonal XZ can be considered as the diameter of a circle that circumscribes the quadrilateral. The right angles at W and Y are simply created by the semicircles for each half of the quadrilateral.</span></p>Geometry Answershttps://www.mathhomeworkanswers.org/235558/quadrilateral-angles-altitudes-dropped-diagonal-meeting?show=235915#a235915Tue, 17 Jan 2017 14:05:41 +0000Answered: please check the locus answer
https://www.mathhomeworkanswers.org/235700/please-check-the-locus-answer?show=235789#a235789
<p style="text-align:justify">Yes the bisector of angle C is correct for (b) and the perpendicular bisector of BC is correct for (c).</p>
<p><img alt="" src="http://www.mathhomeworkanswers.org/?qa=blob&qa_blobid=5427410028092092088" style="height:282px; width:400px"></p>
<p>This is how you draw them (see picture).</p>
<p style="text-align:justify">Point of compasses on C and any radius smaller than BC or CD but greater than half their length, draw an arc to cut BC and CD. The blue arc is all you need for the angle bisector but I've shown the whole circle, because we can use it for part (c). Also, using the same radius draw a circle with centre B.</p>
<p style="text-align:justify">Now we need to draw the green arcs. Put the point of the compasses on the point on CD where the blue arc cuts. Use a radius a bit larger than before and make one of the green arcs (or you could draw the whole circle). Do the same on BC where the blue arc cuts so that the two arcs cross. Now you can draw the red angle bisector. That's part (b). </p>
<p style="text-align:justify">The blue arc was part of the circle centre C. You already drew the circle centre B, so now you can draw the red perpendicular bisector where the two circles intersect. That's part (c).</p>
<p style="text-align:justify">Because the question is about a park, we can expect that we only need the red lines that are inside the park area.</p>
<p style="text-align:justify">Note that the position of A makes no difference to (b) and (c) because we are only interested in the side BC and the angle C, none of which involve A. However, to draw to scale in (a) you probably have to assume that angles D and A are right angles.</p>Geometry Answershttps://www.mathhomeworkanswers.org/235700/please-check-the-locus-answer?show=235789#a235789Fri, 13 Jan 2017 14:51:42 +0000Answered: F=k q1 q2/r2
https://www.mathhomeworkanswers.org/128212/f-k%C2%A0q1%C2%A0q2-r2?show=235763#a235763
<p><span style="background-color:rgb(255, 255, 255); color:rgb(51, 51, 51); font-family:sans-serif,arial,verdana,trebuchet ms; font-size:13px"><a href="http://brainly.com" rel="nofollow">http://brainly.com</a></span></p>Geometry Answershttps://www.mathhomeworkanswers.org/128212/f-k%C2%A0q1%C2%A0q2-r2?show=235763#a235763Fri, 13 Jan 2017 13:08:23 +0000Answered: what is the surface area of acylinder 7cm on top and 6cm on bottom using 3.14 as pi
https://www.mathhomeworkanswers.org/128213/what-the-surface-area-acylinder-7cm-top-and-6cm-bottom-using?show=235762#a235762
<p><span style="background-color:rgb(255, 255, 255); color:rgb(51, 51, 51); font-family:sans-serif,arial,verdana,trebuchet ms; font-size:13px"><a href="http://brainly.com" rel="nofollow">http://brainly.com</a></span></p>Geometry Answershttps://www.mathhomeworkanswers.org/128213/what-the-surface-area-acylinder-7cm-top-and-6cm-bottom-using?show=235762#a235762Fri, 13 Jan 2017 13:07:45 +0000Answered: need help in loci geometry question
https://www.mathhomeworkanswers.org/235636/need-help-in-loci-geometry-question?show=235654#a235654
<a href="http://brainly.com" rel="nofollow">http://brainly.com</a>Geometry Answershttps://www.mathhomeworkanswers.org/235636/need-help-in-loci-geometry-question?show=235654#a235654Wed, 11 Jan 2017 16:38:57 +0000Answered: derive the lateral area of a frustum of a cone
https://www.mathhomeworkanswers.org/128501/derive-the-lateral-area-of-a-frustum-of-a-cone?show=235598#a235598
<a href="http://www.brainly.com" rel="nofollow">http://www.brainly.com</a>Geometry Answershttps://www.mathhomeworkanswers.org/128501/derive-the-lateral-area-of-a-frustum-of-a-cone?show=235598#a235598Tue, 10 Jan 2017 18:08:59 +0000Answered: four interior angles af an heptagon
https://www.mathhomeworkanswers.org/128510/four-interior-angles-af-an-heptagon?show=235595#a235595
<a href="http://www.brainly.com" rel="nofollow">http://www.brainly.com</a>Geometry Answershttps://www.mathhomeworkanswers.org/128510/four-interior-angles-af-an-heptagon?show=235595#a235595Tue, 10 Jan 2017 18:07:42 +0000Answered: what is an spect ratio of 1 to 1
https://www.mathhomeworkanswers.org/128543/what-is-an-spect-ratio-of-1-to-1?show=235585#a235585
Aspect ratio of 1:1 is a square. Zero would be a point.Geometry Answershttps://www.mathhomeworkanswers.org/128543/what-is-an-spect-ratio-of-1-to-1?show=235585#a235585Tue, 10 Jan 2017 15:56:25 +0000For triangle FGH, which of the following is an expression for y in terms of x?
https://www.mathhomeworkanswers.org/234640/for-triangle-fgh-which-the-following-expression-for-terms-of
The triangle has legs of 4 meters and x meters. The hypotenuse is of y meters.<br />
<br />
I thought the answer was x+4, but the answer key says it is sq. rt.(x^2 + 16)<br />
<br />
Can you please explain why I am wrong and why the answer key is correct.<br />
<br />
Thanks so much.Geometry Answershttps://www.mathhomeworkanswers.org/234640/for-triangle-fgh-which-the-following-expression-for-terms-ofThu, 29 Dec 2016 12:57:54 +0000ayşe 72 m'lik bir ipi 2tam 1bölu 4'tur parçaya ayrılıyor buna göre ayşe kaç eş parça elde itmiş
https://www.mathhomeworkanswers.org/234447/ay%C5%9Fe-mlik-1b%C3%B6lu-par%C3%A7aya-ayr%C4%B1l%C4%B1yor-buna-g%C3%B6re-ay%C5%9Fe-par%C3%A7a-itmi%C5%9F
ben kendime çok guveniyorumGeometry Answershttps://www.mathhomeworkanswers.org/234447/ay%C5%9Fe-mlik-1b%C3%B6lu-par%C3%A7aya-ayr%C4%B1l%C4%B1yor-buna-g%C3%B6re-ay%C5%9Fe-par%C3%A7a-itmi%C5%9FFri, 23 Dec 2016 13:26:12 +0000Geometry Problem
https://www.mathhomeworkanswers.org/233766/geometry-problem
Triangle ABC has the following measurements: AB = 12cm, BC = 8cm, AC = 14 cm. Find the height BE on the side AC, and the area of the triangle.Geometry Answershttps://www.mathhomeworkanswers.org/233766/geometry-problemTue, 13 Dec 2016 20:34:12 +0000Polygon ABCDE rotates 45° clockwise about point F to form polygon FGHIJ, shown in the figure. Find the measure of each angle of polygon ABCDE.
https://www.mathhomeworkanswers.org/233681/polygon-rotates-clockwise-polygon-figure-measure-polygon
options are<br />
<br />
a) 90<br />
<br />
b)117<br />
<br />
c)127Geometry Answershttps://www.mathhomeworkanswers.org/233681/polygon-rotates-clockwise-polygon-figure-measure-polygonMon, 12 Dec 2016 18:08:38 +0000Given: XY=XZ; YO bisects XYZ: ZO bisects XZY
https://www.mathhomeworkanswers.org/233593/given-xy-xz-yo-bisects-xyz-zo-bisects-xzy
<p>Given: XY is congruent to XZ; YO bisects angle XYZ; ZO bisects angle XZY
<br>
<br>
Prove: YO is congruent ZO </p>
<p>Diagram: <a rel="nofollow" href="http://www.mathhomeworkanswers.org/?qa=blob&qa_blobid=7365655606075730709">http://www.mathhomeworkanswers.org/?qa=blob&qa_blobid=7365655606075730709</a></p>Geometry Answershttps://www.mathhomeworkanswers.org/233593/given-xy-xz-yo-bisects-xyz-zo-bisects-xzySun, 11 Dec 2016 00:03:37 +0000if ab = 6 feet, ad= 2 feet, and m<dae = 65, find bc
https://www.mathhomeworkanswers.org/233323/if-ab-6-feet-ad-2-feet-and-m-dae-65-find-bc
<p><span style="font-family:arial,helvetica,sans-serif">it has a picture with it, but it can be solved without the picture</span></p>Geometry Answershttps://www.mathhomeworkanswers.org/233323/if-ab-6-feet-ad-2-feet-and-m-dae-65-find-bcTue, 06 Dec 2016 16:28:19 +0000What solid figure has 10 edges
https://www.mathhomeworkanswers.org/232857/what-solid-figure-has-10-edges
Math problemGeometry Answershttps://www.mathhomeworkanswers.org/232857/what-solid-figure-has-10-edgesWed, 30 Nov 2016 01:59:30 +0000