If f(x) = integral of 1/ (sqrt of t^3 +2)dt, then f'1 = ?
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Greetings!
Answer:
2y = x - 6
Step-by-step explanation:
First, we must find the slope of the current equation.
This is the number in front of the x.
Seeing as this is -2x, the slope of this line is -2
When finding the slope of a line perpendicular, you need to find the
So, in this case it is:
The negatives cancel out which leave
So the gradient is
Now, to find the equation of a line, you need to use:
y - y1 = m(x - x1)
Where ya and x1 are the values in the coordinates (2 , -2)
So y1 = -2, x1 = 2, and m is a half. Plug these values in:
y - - 2 =
We need to get rid of the half so we multiply the whole equation by 2:
2y - - 4 = (x - 2)
The minus and the negative turn into a positive:
2y + 4 = x - 2
And now simply move the +4 over to the other side, making it a negative:
2y = -2 - 4 + x
Simplify:
2y = x - 6
<h3>So the equation of the line is 2y = x - 6</h3><h2>Hope this helps!</h2>
