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Kamila [148]
3 years ago
14

oel biked 9 miles east and then 12 miles north. If he biked back to his starting point using the most direct route, how many mil

es would he ride all together?
Mathematics
1 answer:
kvasek [131]3 years ago
6 0

Answer:

36 miles

Step-by-step explanation:

(Imagine a huge triangle on a perfectly open space-- not with streets or houses or cornfields.)

The dimensions of the two sides given are multiples of a classic Pythagorean Triple: 3, 4, 5.    3^{2} + 4^{2} = 5^{2}  

9 is 3x3, 12 is 3x4 so the remaining side would be 3x5  15

Add those 3 sides to get the total distance, 36 miles.

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During basketball practice, Mai attempted 40 free throws and was successful on 25% of them. How many successful free throws did
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10

Step-by-step explanation:

25% is the same as 0.25 so multiply 40 by 0.25.

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Can you help with both questions 4 and 5?
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Well since the triangle is split into 2 triangles and they both are the same size so BD on the bottom is 15 which means naturally BC must be 30. Five I'm not sure about
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Find the exact value of sin(cos^-1(4/5))
boyakko [2]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2762144

_______________


Let  \mathsf{\theta=cos^{-1}\!\left(\dfrac{4}{5}\right).}


\mathsf{0\le \theta\le\pi,}  because that is the range of the inverse cosine funcition.


Also,

\mathsf{cos\,\theta=cos\!\left[cos^{-1}\!\left(\dfrac{4}{5}\right)\right]}\\\\\\
\mathsf{cos\,\theta=\dfrac{4}{5}}\\\\\\ \mathsf{5\,cos\,\theta=4}


Square both sides and apply the fundamental trigonometric identity:

\mathsf{(5\,cos\,\theta)^2=4^2}\\\\
\mathsf{5^2\,cos^2\,\theta=4^2}\\\\
\mathsf{25\,cos^2\,\theta=16\qquad\qquad(but,~cos^2\,\theta=1-sin^2\,\theta)}\\\\
\mathsf{25\cdot (1-sin^2\,\theta)=16}

\mathsf{25-25\,sin^2\,\theta=16}\\\\
\mathsf{25-16=25\,sin^2\,\theta}\\\\
\mathsf{9=25\,sin^2\,\theta}\\\\
\mathsf{sin^2\,\theta=\dfrac{9}{25}}


\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{9}{25}}}\\\\\\
\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{3^2}{5^2}}}\\\\\\
\mathsf{sin\,\theta=\pm\,\dfrac{3}{5}}


But \mathsf{0\le \theta\le\pi,} which means \theta lies either in the 1st or the 2nd quadrant. So \mathsf{sin\,\theta} is a positive number:

\mathsf{sin\,\theta=\dfrac{3}{5}}\\\\\\
\therefore~~\mathsf{sin\!\left[cos^{-1}\!\left(\dfrac{4}{5}\right)\right]=\dfrac{3}{5}\qquad\quad\checkmark}


I hope this helps. =)


Tags:  <em>inverse trigonometric function cosine sine cos sin trig trigonometry</em>

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3 years ago
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Answer:

20.9 %, it is more than the national population 13.6%.

Step-by-step explanation:

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died number of persons = 46

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Which of the following statements about a linear function is never true?
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Answer:

D

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