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Elina [12.6K]
2 years ago
12

A book store and a movie theater are 6 kilometers apart along the same street. A florist is located between the bookstore and th

e theater on the same street. The florist is 2.5 kilometers from the theater. How far is the florist from the bookstore

Mathematics
1 answer:
devlian [24]2 years ago
4 0

The distance from florist to movie theater is about 3.5 km on the same street.

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

An independent variable is a variable that does not depend on other variables while a dependent variable is a variable that depends on other variables.

A book store and a movie theater are 6 kilometers apart. The florist is on the same street, 2.5 kilometers from the theater. Hence:

Distance from book store to movie theater = Distance from book store to florist + Distance from florist to movie theater

Hence:

6 km = 2.5 km + Distance from florist to movie theater

Distance from florist to movie theater = 3.5 km

The distance from florist to movie theater is about 3.5 km on the same street.

Find out more on equation at: brainly.com/question/2972832

#SPJ1

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Answer:

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Step-by-step explanation:

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3(2x+y)=(9)3                +<u>6x+3y=27</u>

                               1/16(16x)=(30)1/16  = x=1.875

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                                    1/-8(-8y)=(7)1/-8 = y=-0.875

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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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<em>Note:</em>

<em>Your questions sound a little unclear.</em>

<em>But, I am assuming your questions as:</em>

<em>''70% of x is 14 min. Find the value of x''.</em>

<em>So, I will solve assuming this statement which anyways would clear your concept. </em>

Answer:

The value of x = 20.

Step-by-step explanation:

Given

<em>''70% of x is 14 min. Find the value of x''.</em>

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divide both sides by 0.7

0.7x/0.7 = 14/0.7

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