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3 years ago

15

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

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

Read 2 more answers

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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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