2(2y^4 + 7y^3 - 4y^2 -20)
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Y^2
4y^2 - 40/y^2 + 14y - 8
2(-2y^4 - 7y^3 + 4y^2 + 20)
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Y^2
Answer:
The inverse is y = x + 5
Step-by-step explanation:
The general equation of a straight line is;
y = mx + c
In this case , the y-intercept is -5
So the partial equation is;
y = mx - 5
To get m, we use the x-intercept
The x-intercept coordinate is (5,0)
Insert this in the equation , we have;
0 = 5m-5
5m = 5
m = 5/5
m= 1
The equation of the line is thus;
y = x-5
So we want to find the inverse of this;
Replace x with d
y = d-5
Make d the subject of the formula
d = y + 5
replace d with x
x = y + 5
now replace x with y
So we have
y = x + 5
64 x 10 = 640
Easier way to simplify :
64 x 1 = 64 and then put the 0 at the end which = 640
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