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Pepsi [2]
3 years ago
12

Find the value in the ratio table...

Mathematics
1 answer:
azamat3 years ago
6 0
Try this way:
Note,
1. that 16/2=8 it means 10/2=5.
2. that 10*3=30, it means 16*3=48
Finally for whole table:
================
forks:     16   8   48
================
spoons: 10   5   30
================
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If you can mow 90 lawns in 45 hours, how long will it take to mow 12?
Alex777 [14]
<h3>Answer:  6 hours</h3>

Work Shown:

(90 lawns)/(45 hours) = (12 lawns)/(x hours)

90/45 = 12/x

2 = 12/x

2x = 12

x = 12/2

x = 6

It takes 6 hours to mow 12 lawns.

5 0
1 year ago
Please help me with this question.
Gelneren [198K]

Answer 1/4

Step-by-step explanation:

3/12 divided by 3 is 1/4

7 0
2 years ago
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The equation is equivalent to u = rst. <br><br> True/ False
Jlenok [28]

The equation r = u/st is equivalent to u = rst .

True

7 0
3 years ago
Does anyone know what the answer?
11111nata11111 [884]

Answer:

x = 7

Step-by-step explanation:

Hi,

7x - 7 = 4x + 14

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I hope this helps :)

5 0
3 years ago
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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}

6 0
3 years ago
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