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3 years ago

(x^2−1)(3x−4)(3x+4) what is the product

Mathematics

1 answer:

Andrei [34K]3 years ago

6 0

Answer:

9x^4-25x^2+16

Step-by-step explanation:

(x^2−1)(3x−4)(3x+4)=(*)

(x^2−1)((3x)^2−4^2) =

(x^2 - 1)(9x^2-16)=

x^2*9x^2 - x^2*16 - 1*9x^2 +16=

9x^4-16x^2-9x^2+16=

9x^4-25x^2+16

(*) (A-B)(A+B) =A^2 - B^2

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4 years ago

Write the name for the decimal value of point M on the number line.

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<h3>What is a number line?</h3>

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3 years ago

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If we ignore the constant in the denominator, then
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3 years ago

Read 2 more answers

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

aev [14]

Answer: (B) -625

<u>Step-by-step explanation:</u>

Given the sequence {-500, -100, -20, -4, -0.8, ... }, we know that that the first term (a) is -500 and the ratio (r) is $\dfrac{-100}{-500}=\dfrac{1}{5}$

Input those values into the Sum formula:

$S_n=\dfrac{a(1-r^n)}{1-r}\\\\S_8=\dfrac{-500\bigg(1-\dfrac{1}{5}^n\bigg)}{1-\dfrac{1}{5}}\\\\.\quad=\dfrac{-500\bigg(1-\dfrac{1}{390,625}\bigg)}{\dfrac{4}{5}}\\\\.\quad=\dfrac{-500\bigg(\dfrac{390,624}{390,625}\bigg)}{\dfrac{4}{5}}\\\\.\quad=\dfrac{5(-500)(390,624)}{4(390,625)}\\\\.\quad=-624.998\\\\.\quad=-625\quad \text{(rounded to 2 decimal places)}$

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4 years ago