11 Alex earns $84 for 12 hours of work,

Mathematics

1 answer:

Vikentia [17]3 years ago

3 0

A) 84/12 = $7
B)12 * 7

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Find the value of x and y (i need this asap please)

disa [49]

For a right triangle with 60° angle, and hypotenuse h=10√3,
x=h*sin(60)=10√ 3 * √3/2 = 30/2=15
y=h*cos(60)=10√3 * (1/2) = 5√3

3 0

3 years ago

Convert the integral below to polar coordinates and evaluate the integral.

Lelechka [254]

Answer:

$\int\limits_{0}^{4/\sqrt{2}}\int\limits_{y}^{\sqrt{16-y^2}} xy \, dxdy = \int\limits_{0}^{\pi/4} \, \int\limits_{0}^{4} r^3 cos(\theta)\sin(\theta) \,\, drd\theta = 16$

Step-by-step explanation:

We are trying to evaluate this integral.

$\int\limits_{0}^{4/\sqrt{2}}\,\,\int\limits_{y}^{\sqrt{16-y^2}} xy \,\,dxdy$

The first thing that we have to do is understand this region in the plane.

$\{ (x,y) \in \mathbb{R} : 0\leq y \leq \frac{4}{\sqrt{2}} \,\, , y \leq x \leq \, \sqrt{16-y^2} \}$

If you graph it looks something like the photo I join.

Now we need to describe that same region in polar coordinates.

That same region in polar coordinates would be

$\{ (r,\theta) : \,\, 0 \leq \theta \leq \frac{\pi}{4} \,\,\, 0\leq r \leq 4 \}$

Now remember that when we do the polar transformation we use the following formula

$\int\limits_{a}^{b} \, \int\limits_{c}^{d} f(x,y) \,dxdy = \int\limits_{\theta_1}^{\theta_2} \, \int\limits_{r_1}^{r_2} r* f(rcos(\theta),rsin(\theta)) \,drd\theta$

Then our integral would be

$\int\limits_{0}^{4/\sqrt{2}}\int\limits_{y}^{\sqrt{16-y^2}} xy \, dxdy = \int\limits_{0}^{\pi/4} \, \int\limits_{0}^{4} r^3 cos(\theta)\sin(\theta) \,\, drd\theta = 16$