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3 years ago

Answer the question

Mathematics

2 answers:

elena-14-01-66 [18.8K]3 years ago

8 0

D. Is the answer

Hope this helps

Nostrana [21]3 years ago

4 0

Answer:

82%

Step-by-step explanation:

Total amount of laps =

18×50=900

738/900×100=82

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A man pays Rs.1440 for income tax at the rate of Rs.2 per Rs.100. Find his income

Mashcka [7]

Answer:

so 2 rupees tax for every 100rupees

we need to find how many 100's in 1440 and multiply with 2 so first divide 1440by 100 you will get 14.4 and multiply with 2 so the income tax is 28.8 you need to pay total 172.8.

Hope you understand if you understand give me a like

8 0

3 years ago

Please help I’ll mark you as brainliest if correct

Nat2105 [25]

9514 1404 393

Answer:

the correct answer is marked

Step-by-step explanation:

You want k in the equation ...

scholars = k × teachers

Dividing by teachers, we get ...

k = scholars/teachers = 75/3 = 25

The constant of proportionality is 25.

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=cos%20%5B%20arctan%28%5Cfrac%7B12%7D%7B5%7D%29%20-%20arcsin%20%28%5Cfrac%7B-3%7D%7B5%7D%29%5D"

qwelly [4]

Answer: -16/65

Step-by-step explanation:

Drawing the right triangle (as attached) gives us that $\arctan \left(\frac{12}{5} \right)=\arcsin \left(\frac{12}{13} \right)$

Also, $-\arcsin \left(-\frac{3}{5} \right)=\arcsin \left(\frac{3}{5} \right)$

This means our original expression is equal to:

$\cos \left[\arcsin \left(\frac{12}{13} \right)+\arcsin \left(\frac{3}{5} \right) \right]$

Using the cosine addition formula, which states $\cos(a+b)=\cos a \cos b-\sin a \sin b$ , we get this itself is equal to:

$\cos \left(\arcsin \left(\frac{12}{13} \right) \right)\cos \left(\arcsin \left(\frac{3}{5} \right)\right)-\sin \left(\arcsin \left(\frac{12}{13} \right) \right)\sin \left(\arcsin \left(\frac{3}{5} \right)\right)$

Since $\sin^{2} \theta+\cos^{2} \theta=1$ , we know that:

$\sin^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)+\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=1\\\\\frac{144}{169} +\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=1\\\\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=\frac{25}{169}\\\\cos \left(\arcsin \left(\frac{12}{13} \right)\right)=\frac{5}{13}$

Similarly, cos(arcsin(3/5))=4/5.

This means the given expression is equal to:

$\left(\frac{5}{13} \right) \left(\frac{4}{5} \right)-\left(\frac{12}{13} \right) \left(\frac{3}{5} \right)\\\\\frac{20}{65}-\frac{36}{65}=\boxed{-\frac{16}{65}}$