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3 years ago

What is the average rate of change for this quadratic function for the interval from x = 0 to x = 2?

Mathematics

1 answer:

xxTIMURxx [149]3 years ago

8 0

Answer is b hope this helps

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To solve the equation below, you would?

xxMikexx [17]

Step-by-step explanation:

step 1. x - 70 = 135

step 2. x = 205. add 70 to each side

8 0

3 years ago

Which of these expressions is equivalent to 6x – 10x + 20?

erastova [34]

Answer:

4(5-x)

Step-by-step explanation:

6x-10x+20 4(5-x)

-4x+20 20-4x

and those two are both equivalent

4 0

2 years ago

Please help with the questions in the image

algol13

First integral:

Use the rational exponent to represent roots. You have

$\displaystyle \int\sqrt[8]{x^9}\;dx = \int x^{\frac{9}{8}}\;dx$

And from here you can use the rule

$\displaystyle \int x^n\;dx=\dfrac{x^{n+1}}{n+1}+C$

to derive

$\displaystyle \int\sqrt[8]{x^9}\;dx = \dfrac{x^{\frac{17}{8}}}{\frac{17}{8}}=\dfrac{8x^{\frac{17}{8}}}{17}$

Second integral:

Simply split the fraction:

$\dfrac{3+\sqrt{x}+x}{x}=\dfrac{3}{x}+\dfrac{\sqrt{x}}{x}+\dfrac{x}{x}=\dfrac{3}{x}+\dfrac{1}{\sqrt{x}}+1$

So, the integral of the sum becomes the sum of three immediate integrals:

$\displaystyle \int \dfrac{3}{x}\;dx = 3\log(|x|)+C$

$\displaystyle \int \dfrac{1}{\sqrt{x}}\;dx = \int x^{-\frac{1}{2}}\;dx = 2\sqrt{x}+C$

$\displaystyle \int 1\;dx = x+C$

So, the answer is the sum of the three pieces:

$3\log(|x|) + 2\sqrt{x} + x+C$

Third integral:

Again, you can split the integral of the sum in the sum of the integrals. The antiderivative of the cosine is the sine, because $\sin'(x)=\cos(x)$ . So, you have

$\displaystyle \int \left( \cos(x)+\dfrac{1}{7}x\right)\;dx = \int \cos(x)\;dx + \dfrac{1}{7}\int x\;dx = \sin(x)+\frac{1}{14}x^2+C$

7 0

3 years ago

What is -42.15+12.45

JulsSmile [24]

-42.15+12.45=-30...............................

3 0

3 years ago

Read 2 more answers

Consider A = (-9,4) and B (11,17). = Point P₂ partitions the segment from B to A in a 3:5 ratio. Find the coordinates of point P

Bogdan [553]

An easy way to do this is to parameterize the directed line segment from B to A by the function

$\ell_{B\to A}(t) = (1-t) B + t A = (1 - t) (11, 17) + t (-9, 4)$

with 0 ≤ t ≤ 1.

The point P₂ splits up AB so that BP₂ = 3/8 AB and AP₂ = 5/8 AB. Then we reach the point P₂ when t = 3/8, so its coordinates are

$P_2 = \ell_{B\to A}}\left(\dfrac38\right) = \dfrac58 (11,17) + \dfrac38 (-9,4) = \boxed{\left(\dfrac72, \dfrac{97}8\right)}$

3 0

2 years ago