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3 years ago

5

Hey can you please help me posted picture of question :)

1 answer:

IRINA_888 [86]3 years ago

4 0

If you observe carefully, you can see that the graph is crossing x-axis at the points 1 and 9. This means x = 1 and x = 9 are the roots of the polynomials and x - 1 and x - 9 are thus the factors of the polynomial.

We can express the polynomial as the product of its factors. So the above polynomial can be expressed as (x - 1)(x - 9)

Thus, the answer to the question is option D

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Which expressions have a value of 16/81? Check all that apply.

lana [24]

Which expression have a value of 16/81? check all that apply. (2/3)^4, (16/3)^4, (4/81)^2, and (4/9)^2

Answer:

First and last option is correct.

$(\frac{2}{3})^{4}=\frac{16}{81}$

$(\frac{4}{9})^{2}=\frac{16}{81}$

Step-by-step explanation:

Given:

There are four options.

(2/3)^4, (16/3)^4, (4/81)^2, and (4/9)^2

We need to check all given options for value of 16/81.

Solution:

Using rule.

$(\frac{a}{b})^{n}=\frac{a^{n}}{b^{n}}$

Solve for option $(\frac{2}{3})^4$ .

$(\frac{2}{3})^{4}=\frac{2^4}{3^4}=\frac{2\times 2\times 2\times 2}{3\times 3\times 3\times 3}=\frac{16}{81}$

Solve for option $(\frac{16}{3})^4$ .

$(\frac{16}{3})^{4}=\frac{16^4}{3^4}=\frac{16\times 16\times 16\times 16}{3\times 3\times 3\times 3}=\frac{65536}{81}$

Solve for option $(\frac{4}{81})^2$ .

$(\frac{4}{81})^{2}=\frac{4^2}{81^2}=\frac{4\times 4}{81\times 81}=\frac{16}{6561}$

Solve for option $(\frac{4}{9})^2$ .

$(\frac{4}{9})^{2}=\frac{4^2}{9^2}=\frac{4\times 4}{9\times 9}=\frac{16}{81}$

Therefore, expression $(\frac{2}{3})^4$ and $(\frac{4}{9})^2$ have a value of $\frac{16}{81}$ .

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Answer:

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93-12=81

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