Answer:
Step-by-step explanation:
The equation of the line through the point 
& 
can be represented by:
Making m the subject;
∴
we need to carry out the equation of the line through (0,1) and (1,2)
i.e
y - 1 = m(x - 0)
y - 1 = mx
where;
m = 1
Thus;
y - 1 = (1)x
y - 1 = x ---- (1)
The equation of the line through (1,2) & (4,1) is:
y -2 = m (x - 1)
where;
∴
-3(y-2) = x - 1
-3y + 6 = x - 1
x = -3y + 7
Thus: for equation of two lines
x = y - 1
x = -3y + 7
i.e.
y - 1 = -3y + 7
y + 3y = 1 + 7
4y = 8
y = 2
Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7
∴
![\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy](https://tex.z-dn.net/?f=%5Ciint_D%208y%5E2%20%5C%20dA%20%3D8%20%5Cint%5E2_1%20%20%5Cbigg%20%28%20%5Bxy%5E2%5D%5E%7B-3y%2B7%7D_%7By-1%7D%20%5Cbigg%20%29%20%5C%20dy)
![\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy](https://tex.z-dn.net/?f=%5Ciint_D%208y%5E2%20%5C%20dA%20%3D8%20%5Cint%5E2_1%20%20%5Cbigg%20%28%20%5By%5E2%28-3y%2B7-y%2B1%29%5D%5Cbigg%20%29%20%5C%20dy)
![\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ([y^2(-4y+8)] \bigg ) \ dy](https://tex.z-dn.net/?f=%5Ciint_D%208y%5E2%20%5C%20dA%20%3D8%20%5Cint%5E2_1%20%20%5Cbigg%20%28%5By%5E2%28-4y%2B8%29%5D%20%5Cbigg%20%29%20%5C%20dy)
![\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1](https://tex.z-dn.net/?f=%5Ciint_D%208y%5E2%20%5C%20dA%20%3D8%20%5Cbigg%20%5B%5Cdfrac%7B%20-4y%5E4%7D%7B4%7D%2B%5Cdfrac%7B8y%5E3%7D%7B3%7D%20%5Cbigg%20%5D%5E2_1)
![\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1](https://tex.z-dn.net/?f=%5Ciint_D%208y%5E2%20%5C%20dA%20%3D8%20%5Cbigg%20%5B%20-y%5E4%2B%5Cdfrac%7B8y%5E3%7D%7B3%7D%20%5Cbigg%20%5D%5E2_1)
![\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]](https://tex.z-dn.net/?f=%5Ciint_D%208y%5E2%20%5C%20dA%20%3D8%20%5Cbigg%20%5B%20-2%5E4%2B%5Cdfrac%7B8%282%29%5E3%7D%7B3%7D%20%2B%201%5E4-%20%5Cdfrac%7B8%5Ctimes%20%281%29%5E3%7D%7B3%7D%5Cbigg%5D)
![\iint_D 8y^2 \ dA =8 \bigg [ -16+\dfrac{64}{3} + 1- \dfrac{8}{3}\bigg]](https://tex.z-dn.net/?f=%5Ciint_D%208y%5E2%20%5C%20dA%20%3D8%20%5Cbigg%20%5B%20-16%2B%5Cdfrac%7B64%7D%7B3%7D%20%2B%201-%20%5Cdfrac%7B8%7D%7B3%7D%5Cbigg%5D)
![\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]](https://tex.z-dn.net/?f=%5Ciint_D%208y%5E2%20%5C%20dA%20%3D8%20%5Cbigg%20%5B%20-15%2B%20%5Cdfrac%7B64-8%7D%7B3%7D%5Cbigg%5D)
![\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]](https://tex.z-dn.net/?f=%5Ciint_D%208y%5E2%20%5C%20dA%20%3D8%20%5Cbigg%20%5B%20-15%2B%20%5Cdfrac%7B56%7D%7B3%7D%5Cbigg%5D)
![\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{-45+56}{3}\bigg]](https://tex.z-dn.net/?f=%5Ciint_D%208y%5E2%20%5C%20dA%20%3D8%20%5Cbigg%20%5B%20%20%5Cdfrac%7B-45%2B56%7D%7B3%7D%5Cbigg%5D)
![\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{11}{3}\bigg]](https://tex.z-dn.net/?f=%5Ciint_D%208y%5E2%20%5C%20dA%20%3D8%20%5Cbigg%20%5B%20%20%5Cdfrac%7B11%7D%7B3%7D%5Cbigg%5D)
The greatest common factor of 15 and 20 is 5
Well, since the graph starts at (0,50), making our y-intercept 50, we can knock off D as an option.
The next step is to find the slope. To do this, find two points and subtract the y values over the x values. (Just make sure that whichever y value you start with - start with the same x value).
So we have >>> (50, 200) and (0,50) from earlier.
200 - 50 / 50 - 0 = 150/50 or 3/1, or 3.
This makes the correct answer A since the equation is represented as y = mx + b, where m is the slope and m has a value of 3.
Answer:
No solutions
Step-by-step explanation:
Answer:
(0,0)
Step-by-step explanation:
