I can’t do these right off hand but look them up on google and google will tell you exactly how to do and if you type in the problem it will explain how to the the problem as well , I hope you understand and this helped you.

4 0

3 years ago

Find the shaded area

REY [17]

The parabola y = x ² and the line x + y = 12 intersect for

x ² = 12 - x

x ² + x - 12 = 0

(x - 3) (x + 4) = 0

===> x = 3

so you can compute the area by using two integrals,

$\displaystyle \int_0^3 x^2\,\mathrm dx + \int_3^{12}(12-x)\,\mathrm dx$

Then the area you want is

$\displaystyle \frac{x^3}3\bigg|_0^3 + \left(12x-\frac{x^2}2\right)\bigg|_3^{12} = \left(\frac{3^3}3-\frac{0^3}3\right) + \left(12^2-\frac{12^2}2 - 12\times3 + \frac{3^2}2\right) \\\\ = \boxed{\frac{99}2}$

Alternatively, you can subtract the area bounded by y = x ², x + y = 12, and the y-axis in the first quadrant from the area of a triangle with height 12 (the y-intercept of the line) and length 12 (the x-intercept).

Such a triangle has area

1/2 × 12 × 12 = 72

and the area you want to cut away from this is given by a single integral,

$\displaystyle \int_0^3 ((12-x)-x^2)\,\mathrm dx = \int_0^3(12-x-x^2)\,\mathrm dx$

The integral has a value of

$\displaystyle \left(12x-\frac{x^2}2-\frac{x^3}3\right)\bigg|_0^3 = 12\times3 - \frac{3^2}2 - \frac{3^3}3 \\\\ = \frac{45}2$

and so the area of the shaded region is again 72 - 45/2 = 99/2.

7 0

3 years ago

A students pays of 18 an hour is 4 more than twice the amount of the student earns per hour at an internship enter an solve an e

svp [43]

Answer:

the hourly pay fro the internship is $7/hr

Step-by-step explanation:

The statement can be written mathematically as below

$18 = 4 + 2x$

the hourly rate of the internship is $x$

solving the equation, we have

$18 = 4 + 2x$

subtract 4 from both sides of the equation

$18 - 4 = 4 +2x -4$

$14 = 2x$

$x = 14/2$

$x = 7$

the hourly pay fro the internship is $7/hr

3 0

3 years ago

Read 2 more answers

Use the key features of the polynomial f(x) = -2x3 + 5x2 - 2x - 3 to describe its end behavior.

Andre45 [30]

Answer:

Step-by-step explanation:

First consider the parent function y = x^3 and its graph. The left side starts in Quadrant III and continues up through Quadrant I. A negative sign in front of the x^3 reflects this original graph in the x-axis; the graph now starts in Quadrant II and descends into Quadrant IV.

The end behavior of the given function is the same as that of the graph of y = -x^3;

The graph begins in Quadrant II and descends into Quadrant IV, down.

8 0

3 years ago