Please help me ASAP

Mathematics

2 answers:

Sedaia [141]3 years ago

8 0

Answer:

18°

Step-by-step explanation:

<u>The exterior angle of the polygon A is the difference of interior angles of pentagon and the square:</u>

180(5 - 2)/5 - 180(4 - 2)/4 =
540/5 - 360/4 =
108 - 90 =
18

Talja [164]3 years ago

6 0

Answer:

135 degrees

Step-by-step explanation:

360 - 90 = 270

270 / 2 = 135 degrees

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Suppose someone tells you that she has a triangle with sides having lengths 2.6, 8.1, and 8.6. Is this a right triangle? Why or

Elodia [21]

First you check for Pythagoras' Theorem.

Is 8.6² = 8.1² + 2.6² ?

8.1² + 2.6² = 65.61 + 6.76 = 72.37
8.6² = 73.96

8.1² + 2.6² ≠ 8.6²

Since Pythagoras' Theorem is not satisfied it is not a right angled triangle.

The longest side is 8.6, let's find the angle facing the longest side 8.6.

By cosine formula CosA = (b² + c² - a²) / (2bc)

CosA = (2.6² + 8.1² - 8.6²) / (2*2.6*8.1) = -1.59/42.12 = -0.0377

A = Cos inverse (-0.0377) = 92.1 degrees

So there is an angle is greater than 90 degrees.

8 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

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