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Damm [24]
3 years ago
15

How do I find the area of a circle?

Mathematics
1 answer:
pogonyaev3 years ago
8 0
How to find the area of a circle:

The area of a circle can be found by multiplying pi ( π = 3.14) by the square of the radius
If a circle has a radius of 4, its area is 3.14*4*4=50.24
If you know the diameter, the radius is 1/2 as large.
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What is the factorization of the expression below?
olchik [2.2K]

Answer: D. (x-6)(x-6)

Step-by-step explanation:

We have the following trinomial, which is also known as perfect square trinomial, because the first and the third term are squares:

x^{2}-12x+36  (1)

First term:x^{2}  

Third term:36=6^{2}  

If we factorize this expression we have:

(x-6)(x-6)=x^{2}-6x-6x+36=x^{2}-12x+36

As we can see, the trinomial (1) has two factors:

(x-6) and (x-6)

Therefore, the correct option is D.

8 0
3 years ago
What is the solution to the trigonometric inequality 2sin(x)+3&gt;sin ^2(x) over the interval
navik [9.2K]

The intervals that satisfy the given trigonometric Inequality are; 0 ≤ x < 3π/2 and 3π/2 < x ≤ 2π

<h3>How to solve trigonometric inequality?</h3>

We are given the trigonometric Inequality;

2 sin(x) + 3 > sin²(x)

Rearranging gives us;

sin²(x) - 2 sin(x) - 3 < 0

Factorizing this gives us;

(sin(x) - 3)(sin(x) + 1) < 0

Thus;

sin(x) - 3 = 0 or sin(x) + 1 = 0

sin(x) = 3 or sin(x) = -1

sin(x) = 3 is not possible because sin(x) ≤ 1.

Thus, we will work with;

sin(x) = -1 for the interval 0 ≤ x ≤ 2π radians.

Then, x = sin⁻¹(-1)

x = 3π/2.

Now, if we split up the solution domain into two intervals, we have;

from 0 ≤ x < 3π/2, at x = 0. Then;

sin²(0) - 2 sin(0) - 3

= 0² - 0 - 3

= -3 < 0

Thus, the interval 0 ≤ x < 3π/2 is true.

From 3π/2 < x ≤ 2π, take x = 2π. Then;

sin²(2π) - 2 sin(2π) - 3

= 0² - 0 - 3

= -3 < 0

Thus, the interval 3π/2 < x ≤ 2π is also true.

Read more about trigonometric inequality at; brainly.com/question/27862380

#SPJ1

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Someone plz help me :(
butalik [34]
You have it correct the answer is 20
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