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olasank [31]
3 years ago
12

Can someone help me?

Mathematics
1 answer:
nordsb [41]3 years ago
7 0
The base of the cylinder has area πr² = 36π ≈ 113.10 (assuming the radius is 6, and not the diameter, that's not clear to me).

If you take the paper wrapped around the cilinder, it would be a rectangular piece of height 12 and length equal to the circumference of the cilinder, being 2πr = 12π, so the area would be 12*12π=144π ≈ 452.39.

So in total this is 452.39+113.10 ≈  565.49, that's answer D.
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s2008m [1.1K]

Answer:

2.9

Step-by-step explanation:

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Two Trains left a station at the same time. One traveled north at a certain speed and the other traveled south at twice that spe
sashaice [31]
Let s = northbound train
Then
2s = southbound train
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Distance = time * speed
4s + 4(2s) = 600
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4s + 8s = 600
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s = 600/12
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s = 50 mph is the northbound train
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4(50) + 4(100) = 600
3 0
3 years ago
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Find the remaining trigonometric ratios of θ if csc(θ) = -6 and cos(θ) is positive
VikaD [51]
Now, the cosecant of θ is -6, or namely -6/1.

however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}

recall that 

\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad\qquad 
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\\\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\\\\\\
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\qquad \qquad 
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}

therefore, let's just plug that on the remaining ones,

\bf sin(\theta)=\cfrac{-1}{6}
\qquad\qquad 
cos(\theta)=\cfrac{\sqrt{35}}{6}
\\\\\\
% tangent
tan(\theta)=\cfrac{-1}{\sqrt{35}}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{\sqrt{35}}{1}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}

now, let's rationalize the denominator on tangent and secant,

\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}
3 0
3 years ago
The sum of three numbers is 36. The first number is two times the second number, and the second number is
MakcuM [25]

Answer:

o (16, 8, 12)

Step-by-step explanation:

1st number: 8*2 =16

2nd number:12-4=8

3rd number: given

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3 years ago
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attashe74 [19]

option A.55°

Answer:

Solution given:

<x+<y=<z

exterior angle of a triangle is equal to the sum of two opposite interior angle

4n-18+n+8=133-6n

5n+6n=133+10

11n=143

n=143/11

n=13

<u>exterior</u><u> </u><u>angle</u><u> </u><u>=</u><u>1</u><u>3</u><u>3</u><u>-</u><u>6</u><u>*</u><u>1</u><u>3</u><u>=</u><u>5</u><u>5</u><u>°</u>

<u>option</u><u> </u><u>A</u><u>.</u><u>5</u><u>5</u><u>°</u>

5 0
3 years ago
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