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3 years ago

20 points

Mathematics

2 answers:

ioda3 years ago

5 0

So hmm notice the picture below

simply get the area of the segment, and subtract it from the area of the circle, what's left, is the difference

notice your circle, has a radius "r" of 18

mario62 [17]3 years ago

3 0

Answer:

Hence, Total area of shaded region= (243π+162) ft²

Step-by-step explanation:

Area of shaded region can be calculated as

Area of right triangle + area of the circle below the right triangle

Area of circle=πr² (360° is covered)

360°-90°=270°

i.e. (270÷360)°= 3/4 part of the circle is covered

So, area of circle below the right triangle= 3×π×18×18/4

= 243π

area of right triangle= base×height/2

= 18×18/2

= 162

Hence, total area of shaded region= (243π+162) ft²

Hence, option A is correct.

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valentinak56 [21]

Answer:

AAS Theorem

Step-by-step explanation:

we can see that

angle ZVY ≅ angle WVY

angle VZY ≅ angle VWY

side VY shared by both triangles.

by Angle- Angle- SIde (AAS), the triangles are congruent.

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3 years ago

What is two mixed numbers that have the sum of 3

Juli2301 [7.4K]

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3 years ago

Which of the following is the product of the rational expressions shown below? 7x/x-4•x/x+7

Gre4nikov [31]

<h2>The product of the rational expressions $\dfrac{7x}{x-4}.\dfrac{x}{x+7} = \dfrac{7x^2}{x^2+3x-28}$ .</h2>

Step-by-step explanation:

We have,

$\dfrac{7x}{x-4}.\dfrac{x}{x+7}$

To find, the product of the rational expressions $\dfrac{7x}{x-4}.\dfrac{x}{x+7}$ = ?

∴ $\dfrac{7x}{x-4}.\dfrac{x}{x+7}$

= $\dfrac{7x.x}{(x-4)(x+7)}$

= $\dfrac{7x^2}{(x(x+7)-4(x+7)}$

= $\dfrac{7x^2}{x^2+7x-4x-28}$

= $\dfrac{7x^2}{x^2+3x-28}$

Thus, the product of the rational expressions $\dfrac{7x}{x-4}.\dfrac{x}{x+7} = \dfrac{7x^2}{x^2+3x-28}$ .

7 0

3 years ago

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Ulleksa [173]

Answer:

Step-by-step explanation:

7 0

3 years ago

Solve sin 0 + 1 = cos20 on the interval 0 ≤ 0 < 2pi. Show work please!

yan [13]

Answer:

$\theta=\frac{\pi}{2},\frac{3\pi}{2}\frac{2\pi}{3}\frac{4\pi}{3}$

Step-by-step explanation:

You need 2 things in order to solve this equation: a trig identity sheet and a unit circle.

You will find when you look on your trig identity sheet that

$cos(2\theta)=1-2sin^2(\theta)$

so we will make that replacement, getting everything in terms of sin:

$sin(\theta)+1=1-2sin^2(\theta)$

Now we will get everything on one side of the equals sign, set it equal to 0, and solve it:

$2sin^2(\theta)+sin(\theta)=0$

We can factor out the sin(theta), since it's common in both terms:

$sin(\theta)(2sin(\theta)+1)=0$

Because of the Zero Product Property, either

$sin(\theta)=0$ or

$2sin(\theta)+1=0$

Look at the unit circle and find which values of theta have a sin ratio of 0 in the interval from 0 to 2pi. They are:

$\theta=\frac{\pi}{2},\frac{3\pi}{2}$

The next equation needs to first be solved for sin(theta):

$2sin(\theta)+1=0$ so

$2sin(\theta)=-1$ and

$sin(\theta)=-\frac{1}{2}$

Go back to your unit circle and find the values of theta where the sin is -1/2 in the interval. They are:

$\theta=\frac{2\pi}{3},\frac{4\pi}{3}$

7 0

3 years ago

*20 points*

20 points