You cannot factor out a coefficient in this expression, because 9 and 8 do not have a common factor besides 1.
The first step is to write this equation into general form. The
general form of an equation is:
ax^2 + bx + c = 0
To make this equation to general form, you have to simplify
the equation first.
2/3(x-4) (x+5) = 1
2/3 (x^2 + 5x – 4x – 20) = 1
2/3(x^2 + x -20) = 1
2/3x^2 + 2/3x – 40/3 = 1
2/3x^2 + 2/3x – 40/3 – 1 = 0
2/3x^2 +2/3x – 43/3 = 0
Therefore, a = 2/3 ; b = 2/3 ; c = -43/3
Answer:
The slope is 1/2
Step-by-step explanation:
The rate of y over the rate of x so it would be y/x. You put the equation y2-y1/x2-x1
Which would look like -3-3/-4-8= -6/-12=6/12 and then you simplify it to 1/2
The answer to this is 14. To find the hypotenuse from the short do 2S, and for Short to Long it is S Root 3.
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Answer:
Therefore the mass of the of the oil is 409.59 kg.
Step-by-step explanation:
Let us consider a circular disk. The inner radius of the disk be r and the outer diameter of the disk be (r+Δr).
The area of the disk
=The area of the outer circle - The area of the inner circle
= ![\pi (r+\triangle r)^2- \pi r^2](https://tex.z-dn.net/?f=%5Cpi%20%28r%2B%5Ctriangle%20r%29%5E2-%20%5Cpi%20r%5E2)
![=\pi [r^2+2r\triangle r+(\triangle r)^2-r^2]](https://tex.z-dn.net/?f=%3D%5Cpi%20%5Br%5E2%2B2r%5Ctriangle%20r%2B%28%5Ctriangle%20r%29%5E2-r%5E2%5D)
![=\pi [2r\triangle r+(\triangle r)^2]](https://tex.z-dn.net/?f=%3D%5Cpi%20%5B2r%5Ctriangle%20r%2B%28%5Ctriangle%20r%29%5E2%5D)
Since (Δr)² is very small, So it is ignorable.
∴![A=2\pi r\triangle r](https://tex.z-dn.net/?f=A%3D2%5Cpi%20r%5Ctriangle%20r)
The density ![\delta (r)= \frac{40}{1+r^2}](https://tex.z-dn.net/?f=%5Cdelta%20%28r%29%3D%20%5Cfrac%7B40%7D%7B1%2Br%5E2%7D)
We know,
Mass= Area× density
![=(2r \pi \triangle r)(\frac{40}{1+r^2}})](https://tex.z-dn.net/?f=%3D%282r%20%5Cpi%20%5Ctriangle%20r%29%28%5Cfrac%7B40%7D%7B1%2Br%5E2%7D%7D%29)
Total mass ![M=\sum_{i=1}^n \frac{80r_i\pi }{1+r^2}\triangle r_i](https://tex.z-dn.net/?f=M%3D%5Csum_%7Bi%3D1%7D%5En%20%5Cfrac%7B80r_i%5Cpi%20%7D%7B1%2Br%5E2%7D%5Ctriangle%20r_i)
Therefore
![\sum_{i=1}^n \frac{80r_i\pi }{1+r^2}\triangle r_i=\int_0^5 \frac{80r\pi }{1+r^2}dr](https://tex.z-dn.net/?f=%5Csum_%7Bi%3D1%7D%5En%20%5Cfrac%7B80r_i%5Cpi%20%7D%7B1%2Br%5E2%7D%5Ctriangle%20r_i%3D%5Cint_0%5E5%20%5Cfrac%7B80r%5Cpi%20%7D%7B1%2Br%5E2%7Ddr)
![=40\pi[ln(1+r^2)]_0^5](https://tex.z-dn.net/?f=%3D40%5Cpi%5Bln%281%2Br%5E2%29%5D_0%5E5)
![=40\pi [ln(1+5^2)-ln(1+0^2)]](https://tex.z-dn.net/?f=%3D40%5Cpi%20%5Bln%281%2B5%5E2%29-ln%281%2B0%5E2%29%5D)
![=40\pi ln(26)](https://tex.z-dn.net/?f=%3D40%5Cpi%20ln%2826%29)
= 409.59 kg (approx)
Therefore the mass of the of the oil is 409.59 kg.