(3a+2)to the 3rd power

Question

find the product

Answer 1

This problem can be rewritten as (3a+2)(3a+2)(3a+2) then we multiply by distribution and combine like terms

3a*3a = 9a^2

3a*2 = 6a

2*3a = 6a

2*2 = 4

so far we have 9a^2 + 12a + 4 (3a+2)

9a^2*3a = 27a^3

12a*3a = 36a^2

4*3a = 12a

9a^2*2 = 18a^2

12a*2 = 24a

4*2 =  8

So we have 27a^3 + 36a^2 + 18a^2 + 12a + 24a + 8

Simplify this to 27a^3 + 54a^2 + 36a + 8

Answer 2

2.3*1.87+3*3.12

Answer 3

                       i I Kinda Get It Im Still Fuzzy ON The (3a+2) To The 3rd Power.....!.....Thing..................'''''''''''',,,,,,,,,,,,,,,,..........''''''

 

Answer 4

i will like to fine. y is seven more than nine times x

Answer 5

hi i would like to solve the equation 3/4m=7-1/4m (note that the slashes represents fraction bars ..thanks

Answer 6

(7 x10^5) x (2 x 10^2)

Answer 7

(3a + 2)^3 =  (3a+2)(3a+2)(3a+2)

multiply( 3a+2)(3a +2)  suggest multiplying in VERTICAL format

(9a^2+12a+4)

multiply the above by (3a+2) again  being careful to keep your like terms lined up

27^3+54a^2+36a+8

Answer 8

Simplify using positive exponents: (3t 5) -2

Answer 9

5 to the third power -12 divided by 3

Answer 10

163/4 hr work 2 day + 123/4  hr for 2 days how many hour did you work

Answer 11

Answers to choose from: -16x36y16, 2x36y16, 16013y8, 16x36y16.

Answer 12

UNDERSTANDING EXPONENANTS

Answer 13

q.27a^3+54a^2+36a+8

ans.=27a^2[a+2]+36a+8+64-64

=27a^2[a+2]+36[a+2]-64

=[27a^2+36]+[a+2]-64

Answer 14

r(x) = x to second power minus 1 over x to 4 power minus 16  and sketch the graph and dot - cross analysis complete the analysis function

Answer 15

2+8/2*3-1+3*3*3

Answer 16

explain why 4.6,4 3/5 and 23/5 are all the same

Answer 17

2a(a – 5) + 4(a – 5)

Answer 18

You have to multiply the first two terms using FOIL (First, Outer, Inner, Last). Once you have the result (trinomial / 3 term answer), take the first part of the trinomial and multiply it by the remaining term (3a + 2), then do the same for the second and third part of the trinomial. Once you have those answers, line them up from greatest power to least power and combine like terms.

Hope this helps some and doesn't make it more confusing for you.

Answer 19

1/2 + 2/3= 7/6

Answer 20

7)3(x+x+1+2)

Answer 21

x-y=2

xy=15

Answer 22

my answer is 125a

Answer 23

8 over 10 = what over 100

Answer 24

what is the earth

Answer 25

7 - 4 + 5 * 0 + 1

Answer 26

27a cubed plus 8

Answer 27

9a^2+15a+6

Answer 28

9a^2+12a+4

Answer 29

27a^3 + 54a^2 + 36a +8 just use the shortcut. 1. Cube the first term. 2. Thrice the product of the square of the 1st term and the last term. 3. Thrice the product of the 1st term and the square of the last term. 4. Cube the last term.

Answer 30

What is 2 over 7 plus 3 over 7

Answer 31

2-1^2+ (2-1)^3 evaluate

Answer 32

12 . . ____________________________________________

Answer 33

How many times does 2 go in to 2

Answer 34

8886

Answer 35

deffinition of fourth-power-free integer

Answer 36

(3a+2)3

=(3a)3 + (3*(3a)2*2) + (3*3a*(2)2 + (2)3

= 27a3+54a2+36a+8

Answer 37

VII-ATIS,

Kahit ano sagot

di Ko po alam

ay ambot

Yolanda Scandal

Answer 38

6844

Answer 39

فرض مراقبة عدد 2 في السلامية 7اساسي

Answer 40

y2-7-2 and y2+4=

Answer 41

5x-6x+x

Answer 42

Simplify using the correct order of operations 5(2-4)+6÷3

Answer 43

First I did what was in the parentheses.
Since he did not give the value of the variable which is a.

So I plugged in 3. Now 3 times 3 equals 9 + 2 =11.Now I did 11 too the power of 3.11 times 3 =33 times 3 = 99

SO MY ANSWER IS 99

Answer 44

Let 3a=A, and 2=B, so (3a+2)³=(A+B)³ ··· Eq.1    Use standard polynomial form for (A+B)³:

(A+B)³=A³+3A²B+3AB²+B³ ··· Eq.2    Plug A=3a, and B=2 into Eq.2.   We have:

(3a+2)³=(3a)³+3(3a)²(2)+3(3a)(2)²+(2)³

            =27a³+54a²+36a+8 ··· Eq.3

CK: Plug a=1 (or any number) into Eq.3.  LHS=(3x1+2)³=5³=125

RHS=(27x1)³+54x(1)²+36x(1)+8=27+54+36+8=125   Thus, LHS=RHS    CKD

The answer is: (3a+2)³=27a³+54a²+36a+8   

Answer 45

3a+8

Answer 46

why does parentesis affect the outcome of a problem with exponets?