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Verdich [7]
4 years ago
11

Mckenzie and Cara both tried to find the missing side of the right triangle. Is either of them correct? Explain your reasoning.

Mathematics
1 answer:
hram777 [196]4 years ago
6 0
Since it is a right triangle, we can use the pythagorean theorem:

a^2+b^2=c^2

5^2 + 13^2 = c^2

25 + 169 = c^2

194 = c^2

c = sqrt(194)
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Problem 24

Part 1

\displaystyle \int \csc(x)\left(\sin(x)+\cot(x)\right)dx\\\\\displaystyle \int \frac{1}{\sin(x)}\left(\sin(x)+\frac{\cos(x)}{\sin(x)}\right)dx\\\\\displaystyle \int \frac{1}{\sin(x)}*\sin(x)+\frac{1}{\sin(x)}*\frac{\cos(x)}{\sin(x)}dx\\\\\displaystyle \int 1+\frac{\cos(x)}{\sin^2(x)}dx\\\\\displaystyle \int 1 dx+\int \frac{\cos(x)}{\sin^2(x)}dx\\\\\displaystyle x+C_1+\int \frac{1}{u^2}du \ \text{ ... where } u = \sin(x)\\\\\displaystyle x+C_1+\int u^{-2}du\\\\

Part 2

\displaystyle x+C_1+\frac{1}{1+(-2)}u^{-2+1}+C_2\\\\\displaystyle x+C_1+\frac{1}{-1}u^{-1}+C_2\\\\\displaystyle x+C_1-u^{-1}+C_2\\\\\displaystyle x+C_1-\frac{1}{u}+C_2\\\\\displaystyle x+C_1-\frac{1}{\sin(x)}+C_2\\\\\displaystyle x-\frac{1}{\sin(x)}+C_1+C_2\\\\\displaystyle x-\frac{1}{\sin(x)}+C\\\\\displaystyle x-\csc(x)+C\\\\

<h3>Answer:  x - csc(x) + C</h3>

Don't forget about the plus C constant

==========================================================

Problem 26

Fortunately, there aren't as many steps for this problem.

\displaystyle \int \frac{dy}{\csc(y)}\\\\\displaystyle \int \frac{1}{\csc(y)}dy\\\\\displaystyle \int \sin(y)dy\\\\\displaystyle -\cos(y)+C\\\\

<h3>Answer:  -cos(y)  + C</h3>
6 0
3 years ago
Two vehicles, a car and a truck, leave an intersection at the same time. The car heads east at an average speed of 30 miles per
svet-max [94.6K]
I don’t fully understand but you might be able to do something like

70t -30t = d, but I think that would be how much farther they are from the start point, but assuming they are driving in a straight line or would be a right triangle, so maybe The square root of 70t^2 + 30t^2
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umka21 [38]
The equation is 6+6=12
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3 years ago
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