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valentina_108 [34]
3 years ago
8

Need help with this problem!!

Mathematics
1 answer:
Nezavi [6.7K]3 years ago
8 0

Answer:

Area:

4 x 4 = 16

Finding area of semi circle:

4 is your diameter so half of it is your radius which is 2 since half of 4 is 2!

2^2<---your radius being squared = 4

4(radius squared) x 3.14(pi) = 12.56

12.56 divided by 2 since its a semi circle is = 6.28

6.28 + 16 = 22.28 is your area

Perimeter is:

4 + 4 + 4 (all sides of a square are equal therefore one or two given lengths will be all the sides) = 12

Circumference:

Radius is 2,

2(you just always have to multiply this number when finding circumference) x 3.14(pi) x 2(radius), 2 x 3.14 x 2 = 12.56

12.56 divided by 2 = 6.28

6.28 + 12 = 18.28 is your perimeter.

Just a refresh:

Circumference Formula:

2(always use this number when finding circumference) x pi(3.14 or 22/7 depending on what they tell you to use for pi) x radius

Area of a Circle Formula:

Radius squared x pi(3.14 or 22/7 whatever they tell you to use for pi)

Another thing you should remember:

Whenever it gives you 1/4 of a circle or 1/3 or a semi circle or any fraction, REMEMBER TO DIVIDE BY THAT DENOMINATOR TO WHAT YOU GET FROM EITHER CIRCUMFERENCE OR AREA OF A CIRCLE!

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Step-by-step explanation:

The slope-intercept form of an equation of a line:

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Put the given y-intercept b = 7 and the coordinates of the point (-2, 1) to the equation:

1=-2m+7          <em>subtract 7 from both sides</em>

-6=-2m       <em>divide both sides by (-2)</em>

3=m\to m=3

We have the equation:

y=3x+7

Convert it to the general form Ax+By+C=0:

y=3x+7              <em>subtract 3x and 7 from both sides</em>

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3x-y+7=0

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If cos() = − 2 3 and is in Quadrant III, find tan() cot() + csc(). Incorrect: Your answer is incorrect.
nydimaria [60]

Answer:

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}

Step-by-step explanation:

Given

\cos(\theta) = -\frac{2}{3}

\theta \to Quadrant III

Required

Determine \tan(\theta) \cdot \cot(\theta) + \csc(\theta)

We have:

\cos(\theta) = -\frac{2}{3}

We know that:

\sin^2(\theta) + \cos^2(\theta) = 1

This gives:

\sin^2(\theta) + (-\frac{2}{3})^2 = 1

\sin^2(\theta) + (\frac{4}{9}) = 1

Collect like terms

\sin^2(\theta)  = 1 - \frac{4}{9}

Take LCM and solve

\sin^2(\theta)  = \frac{9 -4}{9}

\sin^2(\theta)  = \frac{5}{9}

Take the square roots of both sides

\sin(\theta)  = \±\frac{\sqrt 5}{3}

Sin is negative in quadrant III. So:

\sin(\theta)  = -\frac{\sqrt 5}{3}

Calculate \csc(\theta)

\csc(\theta) = \frac{1}{\sin(\theta)}

We have: \sin(\theta)  = -\frac{\sqrt 5}{3}

So:

\csc(\theta) = \frac{1}{-\frac{\sqrt 5}{3}}

\csc(\theta) = \frac{-3}{\sqrt 5}

Rationalize

\csc(\theta) = \frac{-3}{\sqrt 5}*\frac{\sqrt 5}{\sqrt 5}

\csc(\theta) = \frac{-3\sqrt 5}{5}

So, we have:

\tan(\theta) \cdot \cot(\theta) + \csc(\theta)

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \tan(\theta) \cdot \frac{1}{\tan(\theta)} + \csc(\theta)

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 + \csc(\theta)

Substitute: \csc(\theta) = \frac{-3\sqrt 5}{5}

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 -\frac{3\sqrt 5}{5}

Take LCM

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}

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