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Svetlanka [38]
3 years ago
7

A cone-shaped container with a height of 6 inches and radius of 2 inches is filled with a substance that is worth $5 per cubic f

oot. find the total value of the substance in the container.
Mathematics
1 answer:
dimulka [17.4K]3 years ago
4 0
We know that

[volume of a cone]=pi*r²*h/3
r=2 in
h=6 in
[volume of a cone]=pi*2²*6/3------> 25.12 in³

1 ft³---------> 1728 in³
then
X------------> 25.12 in³
X=25.12/1728-----> x=0.014537 ft³

if 1 ft³--------------------> cost $5
0.014537 ft³-----------> X
X=0.014537*5-----> x=0.073

the answer is
<span>the total value of the substance in the container is $0.07</span>
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Every triangle’s inner angles’ addition equals to 180°
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Points A and B are midpoints of the sides of triangle QRS. Triangle Q R S is cut by line segment A B. Point A is the midpoint of
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Would appreciate the help ! ​
aleksandr82 [10.1K]

This is one pathway to prove the identity.

Part 1

\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{1}{\tan(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\cot(\theta) = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)(1-\cos(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\

Part 2

\frac{\sin^2(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)-\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-(\cos(\theta)-\cos^2(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-\cos(\theta)+\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\

Part 3

\frac{\sin^2(\theta)+\cos^2(\theta)-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1}{\sin(\theta)} = \frac{1}{\sin(\theta)} \ \ {\checkmark}\\\\

As the steps above show, the goal is to get both sides be the same identical expression. You should only work with one side to transform it into the other. In this case, the left side transforms while the right side stays fixed the entire time. The general rule is that you should convert the more complicated expression into a simpler form.

We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity \sin^2(\theta)+\cos^2(\theta) = 1 in the second to last step. I broke the steps into three parts to hopefully make it more manageable.

3 0
2 years ago
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