Answer:
What's the question?
Step-by-step explanation:
Answer:
Answer is 
Step-by-step explanation:
To find the interval of x. Use our equations to equal each other.



Integrate.
![\frac{-x^3}{3}+x^2\\(\frac{-2^3}{3}+2^2)-[\frac{-0^3}{3}+0^2]\\-\frac{8}{3} +4-0\\-\frac{8}{3}+\frac{12}{3} =4/3](https://tex.z-dn.net/?f=%5Cfrac%7B-x%5E3%7D%7B3%7D%2Bx%5E2%5C%5C%28%5Cfrac%7B-2%5E3%7D%7B3%7D%2B2%5E2%29-%5B%5Cfrac%7B-0%5E3%7D%7B3%7D%2B0%5E2%5D%5C%5C-%5Cfrac%7B8%7D%7B3%7D%20%2B4-0%5C%5C-%5Cfrac%7B8%7D%7B3%7D%2B%5Cfrac%7B12%7D%7B3%7D%20%20%3D4%2F3)
Using Desmos I have Graphs of both of the equations you have provided. The problem asks us to find the shaded region between those curves/equations.
Proof Check your interval of x.
Answer:
1/2 percent
Step-by-step explanation:
If you look at the box, the middle number is 12 or since its the middle "1/2"
so the fraction would be 1/2, hope this helped.
Given :
- The length of a rectangle is 4m more than the width.
- The area of the rectangle is 45m²
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To Find :
- The length and width of the rectangle.
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Solution :
We know that,

So,
Let's assume the length of the rectangle as x and the width will be (x – 4).
⠀
Now, Substituting the given values in the formula :







⠀
Since, The length can't be negative, so the length will be 9 which is positive.
⠀


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Answer: y = -2x -2
Step-by-step explanation:
Making y the subject of the formula
Subtract 4 from both sides
⇒ y = -2 ( x - 1 ) - 4
y = -2x + 2 -4
y = -2x -2