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Alisiya [41]
3 years ago
15

How many candy bars must be sold to earn a profit of $600

Mathematics
2 answers:
Juliette [100K]3 years ago
8 0
Well, it depends how much is each candy bar?
attashe74 [19]3 years ago
3 0

Answer:

218

Step-by-step explanation:

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Identify two values that have a value less than 3 what is the answer to this question?
Digiron [165]

Answer:

the answer is 2.9 x 10^0 and 3.2 x 10^-2

Step-by-step explanation:

10^0 is equal to 1 so 2.9 x 10^0 is really 2.9 times 1 which is less than 3

if i could get brainliest answer that would be great!

10^-2 is a exponential that makes a number go down so 3.2 going down is less than 3

8 0
3 years ago
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What is the y-coordinate of the midpoint of the line segment of (-4,13) and (18,-3.5)
Nady [450]

The midpoint is (7,4.75) so the y-coordinate should be y=4.75. I hope that helps

4 0
3 years ago
Given the equation 2X +4/3 Y equals one and Y -9/13 x=9 by what Vector would you multiply the first equation so that combining t
Mila [183]

Answer: Option B

B. \frac{9}{26}

Step-by-step explanation:

We have the following equations:

2x + \frac{4}{3}y = 1       (1)

y -\frac{9}{13}x=9          (2)

Let us call "a" the coefficient of the variable x in the first equation and call "b" the coefficient of the variable x in the second equation.

Then we must multiply the number "a" by a value z such that when adding az + b the result is zero.a = 2

b = -\frac{9}{13}

So

2z-\frac{9}{13} = 0

We solve the equation for z

2z=\frac{9}{13}

z=\frac{9}{26}

The first equation must be multiplied by a value of \frac{9}{26}

4 0
3 years ago
Suppose that 65% of the defendants are truly guilty. Suppose also that juries vote a guilty person innocent with probability 0.2
Mashcka [7]

Answer:

<em><u>note:</u></em>

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5 0
3 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
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