(d) Find the value of k so that kx + 5y = 10 makes an angle of 30° with y-axis.

k = -8.66

Step-by-step explanation:

We have the equation:

k*x + 5*y = 10

First, let's rewrite this as a line in the slope-intercept form:

5*y = 10 - k*x

y = 10/5 - (k/5)*x

y = -(k/5)*x + 2

Now, we want to make an angle of 30° with respect to the y-axis.

Then we can think on a triangle rectangle, where we know one angle of 30°, the adjacent cathetus will be equal to the difference between the y-component and the y-intercept, and the opposite cathetus will be equal to the x-component.

To get this, we first need to evaluate the function in one given point, let's use x = 5. (we could use any value of x here, i use 5 because it is easier)

y = -(k/5)*5 + 2

y = -k + 2

then the cathetus of our triangle will be:

adjacent cathetus: (y - 2) = (-k + 2 - 2) = -k

opposite cathetus: x = 5.

Now we know the relation:

Tan(a) = (opposite cathetus)/(adjacent cathetus)

where a = 30°

Then:

Tan(30°) = 5/(-k)

-k = 5/tan(0.577) = 8.66

k = -8.66

This is the value of k that we wanted to find.

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50