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3 years ago

I NEED HELP W 36 pls pls pls pls help help!

Mathematics

1 answer:

kodGreya [7K]3 years ago

4 0

Answer:

Step-by-step explanation:

When you substitute the inside of f(g(2)), you get g(x)=2-4. Then you substitute that answer of -2 and plug it into f(x)=2x+5. Since f(-2)=2(-2)+5 equals to 1, that should be your answer.

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VladimirAG [237]

Answer: First option.

Step-by-step explanation:

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zysi [14]

Answer:

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c} = \frac{9}{2}a^5b^3c$

Step-by-step explanation:

Given

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c}$

Required

Simplify

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c}$

Cancel out 18

$\frac{27a^9 * b^5 * 4c^2 }{a^4 * 12b^2 * 2c}$

Divide 4 and 2

$\frac{27a^9 * b^5 * 2c^2 }{a^4 * 12b^2 *c}$

Divide 27 and 12 by 3

$\frac{9a^9 * b^5 * 2c^2 }{a^4 * 4b^2 *c}$

Apply law of indices

$\frac{9a^{9-4} * b^{5-2} * 2c^{2-1} }{4}$

$\frac{9a^5 * b^3 * 2c }{4}$

Divide 2 and 4

$\frac{9a^5 * b^3 * c}{2}$

$\frac{9a^5b^3c}{2}$

Rewrite as:

$\frac{9}{2}a^5b^3c$

Hence:

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c} = \frac{9}{2}a^5b^3c$

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