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4 years ago

2/5(a+2b)^2 (if a=3 and b=-4)

Mathematics

1 answer:

kondor19780726 [428]4 years ago

4 0

The answer is 10. Hope this helps.

Multiply the numbers

2/5 •(3-8)^2

Subtract (3-8)

2/5• (-5)^2

A negative base raised to an even power equals a positive

2/5• (5)^2

Reduce the expression with 5.

2•5

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Brainliest to right answer! 3/7r + ⅝s when r = 14 and s = 8.

yanalaym [24]

Answer:

3/7r + 5/8s = 11

Step-by-step explanation:

r = 14

s = 8

3/7r + 5/8s = ?

3/7 (14) + 5/8 (8) = ?

3 x 2 + 5 = ?

6 + 5 = 11

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3 years ago

How do you round 5.04

kodGreya [7K]

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3 years ago

12a^3b^2 +18a²b^2 – 12ab^2 Factor completely

AleksAgata [21]

The factorization of 12a^3b^2 +18a²b^2 – 12ab^2 is $6 a b^{2}(a+2)(2 a-1)$

Solution:

Given, expression is $12 a^{3} b^{2}+18 a^{2} b^{2}-12 a b^{2}$

We have to factorize the given expression completely.

Now, take the expression

$12 a^{3} b^{2}+18 a^{2} b^{2}-12 a b^{2}$

Taking $b^2$ as common term,

$b^{2}\left(12 a^{3}+18 a^{2}-12 a\right)$

Taking "a" as common term,

$b^{2}\left(a\left(12 a^{2}+18 a-12\right)\right)$

Taking "6" as common term,

$b^{2}\left(a\left(6\left(2 a^{2}+3 a-2\right)\right)\right)$

Splitting "3a" as "4a - a" we get,

$b^{2}\left(a\left(6\left(2 a^{2}+4 a-a-2\right)\right)\right)$

$\begin{array}{l}{b^{2}(a(6(2 a(a+2)-1(a+2))))} \\\\ {b^{2}(a(6((a+2) \times(2 a-1))))} \\\\ {6 a b^{2}(a+2)(2 a-1)}\end{array}$

Hence, the factored form of given expression is $6 a b^{2}(a+2)(2 a-1)$

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3 years ago

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eimsori [14]

Uh I did not learn this yet but I think The answer is 2,4

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3 years ago

Solve the system by elimination. First, clear the denominators.

NARA [144]

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3 years ago