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2 years ago

Which expression below gives the average rate of change of the function g(x)= -x^2 -4x on the interval 6 ≤ x ≤ 8?

1 answer:

6 0

The average rate of change of a function, g(x), over an interval [a, b] is given by

$Average\ rate\ of\ change= \frac{g(b)-g(a)}{b-a}$

Thus, the average rate of change of $g(x)=-x^2-4x$ over the interval $6\leq x\leq8$ is given by:

$Average\ rate\ of\ change= \frac{[-(8)^2-4(8)]-[-(6)^2-4(6)]}{8-6} \\ \\ = \frac{(-64-32)-(-36-24)}{2} = \frac{-96-(-60)}{2} = \frac{-96+60}{2} = \frac{-36}{2} \\ \\ -18$

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which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

2 years ago

Myra saved $15 a month for 18 months. She bought a book for $46.50 and a tennis racquet for $129.95. How much does she have left

Elis [28]

Answer:

Step-by-step explanation:

15*18=270.

270-46.50=223.5

223.5-129.95=93.55.

A. $93.55

7 0

2 years ago

Michael is making a big batch of granola bars to bring to his soccer game. The oats in the granola bars contain 568.7 grams of c

givi [52]

Your answer would be the last one for sure !

6 0

2 years ago

Read 2 more answers

The general form of the equation of a circle is 7x2 + 7y2 − 28x + 42y − 35 = 0. The equation of this circle in standard form is

Ede4ka [16]

Answer:

The standard form of a circle is (x-h)^2 + (y-k)^2 = r^2 with (h,k) being the center of the circle and r being the radius. In this case the circle's equation in standard form is (x-2)^2 + (y+3)^2 = 18. Knowing this it's easy to see that the center of the circle (h,k) is (2,-3). Finally the radius is $\sqrt{18}$ or in simplified terms, 3 $\sqrt{2}$