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NNADVOKAT [17]
3 years ago
15

If a 6 mile race is divided into 1 3 of a mile sections, how many sections are there in the race?

Mathematics
1 answer:
arlik [135]3 years ago
4 0
Hey there!

Here is your answer:

<u>The proper answer to this question is "18 sections".</u>

Reason:

<u>Divide 6 and 1/3</u>

<u>6×3=18</u>
<u>1×1=1</u>

<u>18/1=18</u>

Therefore it would be divided into 18 different sections.

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit
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Answer:

64.2

Step-by-step explanation:

Since a right triangle is equal to 90 degrees, you would subtract 25.8 from 90 to get your answer.

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Please help me for the love of God if i fail I have to repeat the class
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x is in quadrant II, so \sin x>0.

Recall that for any angle \alpha,

\sin^2\alpha+\cos^2\alpha=1

Then with the conditions determined above, we get

\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35

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Now recall the compound angle formulas:

\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta

\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta

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as well as the definition of tangent:

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Then

1. \sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}

2. \cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}

3. \tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}

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5. \cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}

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\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2

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Because x is in quadrant II, we know that \dfrac x2 is in quadrant I. Specifically, we know \dfrac\pi2, so \dfrac\pi4. In this quadrant, we have \tan\dfrac x2>0, so

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