Y=3x-5 complete missing value in the solution to the equation (1,_) i cant find the second point

1 answer:

6 0

$\bf y = 3x - 5\qquad \qquad \boxed{(\stackrel{x}{1}~~,~~\stackrel{y}{?})}\qquad \qquad y = 3(\stackrel{x}{1})-5\implies y = -2 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill (1,-2)~\hfill$

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How do you do this question?

Andrei [34K]

Answer:

Solution : Option B, or 9π

Step-by-step explanation:

We are given that y = x, x = 3, and y = 0.

Now assume we have a circle that models the given information. The radius will be x, so to determine the area of that circle we have πx². And knowing that x = 3 and y = 0, we have the following integral:

$\int _0^3$

So our set up for solving this problem, would be such:

$\int _0^3x^2\pi \:$

By solving this integral we receive our solution:

$\int _0^3x^2\pi dx,\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=> \pi \cdot \int _0^3x^2dx\\\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1}\\=> \pi \left[\frac{x^{2+1}}{2+1}\right]^3_0\\=> \pi \left[\frac{x^3}{3}\right]^3_0\\\mathrm{Compute\:the\:boundaries}: \left[\frac{x^3}{3}\right]^3_0=9\\\mathrm{Substitute:9\pi }$