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3 years ago

What are the center and radius of the circle given by the equation ?

Mathematics

1 answer:

valina [46]3 years ago

3 0

General formula of the circle

(x-a)² + (y - b)² = r²,

where (a,b) - coordinates of the center, r - radius of the circle.

(x-3)²+(y+4)²=4

(x-3)²+(y+4)²=2²

So, the center is (3, -4) and radius r = 2.

A. (3, –4); 2

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2 years ago

Help me pls becks or anyone I need to pass

Valentin [98]

Answer:

area of 105° segment = 74.2 in²

Step-by-step explanation:

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2 years ago

If sin(a) = cos(2a), then what is the value of a?

Mariulka [41]

Answer:

-90 , 30

Step-by-step explanation:

sina = cos(2a)

= 1 - 2(sina)^2

2(sina)^2 + sina -- 1 = 0

( sina + 1 )( 2sina - 1 ) = 0

sina = -1 or sina = 1/2

=> a = -90 degrees or => a = 30 degrees

3 0

3 years ago

Can you help me to solve this

stealth61 [152]

Answer:

$\text{Triangle 1:}\\\\x=30\sqrt{2},\\\\\text{Triangle 2:}\\x\approx 36.18,\\?\approx12.37$

Step-by-step explanation:

*Notes (clarified by the person who asked this question):

-The triangle on the right has a right angle (angle that appears to be a right angle is a right angle)

-The bottom side of the right triangle is marked with a question mark (?)

<u>Triangle 1 (triangle on left):</u>

Special triangles:

In all 45-45-90 triangles, the ratio of the sides is $x:x:x\sqrt{2}$ , where $x\sqrt{2}$ is the hypotenuse of the triangle. Since one of the legs is marked as $30$ , the hypotenuse must be $\boxed{30\sqrt{2}}$

It's also possible to use a variety of trigonometry to solve this problem. Basic trig for right triangles is applicable and may be the simplest:

$\cos 45^{\circ}=\frac{30}{x},\\x=\frac{30}{\cos 45^{\circ}}=\frac{30}{\frac{\sqrt{2}}{2}}=30\cdot \frac{2}{\sqrt{2}}=\frac{60}{\sqrt{2}}=\frac{60\cdot\sqrt{2}}{\sqrt{2}\cdot \sqrt{2}}=\frac{60\sqrt{2}}{2}=\boxed{30\sqrt{2}}\\$

<u>Triangle 2 (triangle on right):</u>

We can use basic trig for right triangles to set up the following equations:

$\cos 20^{\circ}=\frac{34}{x},\\x=\frac{34}{\cos 20^{\circ}}\approx \boxed{36.18}$ ,

$\tan 20^{\circ}=\frac{?}{34},\\?=34\tan 20^{\circ}\approx \boxed{12.37}$

We can verify these answers using the Pythagorean theorem. The Pythagorean theorem states that in all right triangles, the following must be true:

$a^2+b^2=c^2$ , where $c$ is the hypotenuse of the triangle and $a$ and $b$ are two legs of the triangle.

Verify $34^2+(34\tan20^{\circ})^2=\left(\frac{34}{\cos20^{\circ}}\right)^2\:\checkmark$

3 0

2 years ago

Is this in standard form ? if it isn't then what would it be in standard form ?

Assoli18 [71]

No...standard form cannot have fractions

2/3x - 1/3y = 2....multiply by common denominator of 3 to get rid of fractions
2x - y = 6 <== standard form

7 0

2 years ago

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