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3 years ago

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DO NOT BE DISCOURAGED IF YOU SEE AN ANSWER! PLEASE HELP ASAP WILL GIVE BRAINLIEST AND THANKS AND 5 STAR RATING TO THE CORRECT AN

SWERER, IN ORDER TO GET ALL 98 POINTS YOU MUST ANSWER ALL 5 QUESTIONS AND CORRECTLY THANK YOU!

1 answer:

nignag [31]3 years ago

8 0

See the attached picture:

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What number is in between 375 and 400?

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376,377,378,379,380,381,382,383,384,385,386,387,388,389,390,391,392,393,394,395,396,397,398,399,400
just kidding its 396.5
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3 years ago

Which greek geometer founded a philosophical society that devoted itself to study of mathematics

BlackZzzverrR [31]

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Pythagoras was the Greek geometer .

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1 year ago

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7,900 dollars is placed in an account and gains an annual interest rate of 5.5%. Which of the exponential equations below best m

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I don't the answer but the next person will answer this right.

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2 years ago

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

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

6 0

3 years ago

point T"(-2,5) is a vertex of triangle T"O"P". The original image was rotated 90° counterclockwise around the origin and rhen tr

emmainna [20.7K]

ANSWER

T(-2,1)

EXPLANATION

Let T(a,b) be the coordinates.

When this point is rotated 90° counterclockwise about the origin,

Then,

$T(a,b)\to \: T'( - b,a)$

If this point is then translated using the rule;

$(x,y)\to (x-1,y+7)$

then,

$T(a,b)\to \: T'( - b,a) \to T"( - b - 1,a + 7)$

It was given that, T"(-2,5)

This implies that,

-b-1=-2

-b=-2+1

-b=-1

b=1

a+7=5

a=5-7

a=-2

Therefore the coordinates of T are:

(-2,1)

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3 years ago

Read 2 more answers