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enot [183]
3 years ago
7

Last month, Brenna and Richard sold candy to raise money for their debate team. Richard sold 1/2 as much candy as Brenna did. If

Brenna sold 1/5 of a box of candy, how many boxes of candy did Richard sell?
Mathematics
1 answer:
AURORKA [14]3 years ago
6 0

Answer:

Richard sold \frac{1}{10} box of the candy.

Step-by-step explanation:

Given:

Number of boxes sold by Brenna = \frac15

We need to find the number of boxes sold by Richard.

Solution:

Now given:

Richard sold 1/2 as much candy as Brenna did.

So we can say that;

Number of boxes sold by Richard is equal to \frac{1}{2} multiplied by Number of boxes sold by Brenna.

framing in equation form we get;

Number of boxes sold by Richard = \frac12\times\frac15=\frac{1}{10}

Hence Richard sold \frac{1}{10} box of the candy.

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melamori03 [73]
When you are looking at a graph, a minimum point would be where the curve is decreasing, then begins to increase. Right at the point where it switches, the slope is a horizontal line, or 0. We can take the derivative is f(x), then look for all the x values where the slope (which is equal to the first derivative) is equal to zero.

f'(x) = 2 * -4sin(2x - pi)

The 2 comes from the derivative of the inside, 2x-pi.

So now set the derivative equal to 0.

-8sin(2x-pi) = 0

We can drop the -8 by dividing both sides by -8.

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This can be rewritten as arcsin(0) = 2x-pi

So when theta equals 0, what is the value of sin(theta)? At an angle of 0, there is just a horizontal line pointing to the right on the unit circle with length of 1. Sine is y/h, but there is no y value so it is just 0. If arcsin(0) = 0, we can now set 2x-pi = 0

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f(0) = -4

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f(pi) = -4

So the minimum points are at x=0 and x=pi




5 0
4 years ago
[(23 × 2.5) , 1 2] + 120<br> Please solve
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