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Dmitry_Shevchenko [17]

4 years ago

10

Lauren is the touring students at the library on Saturday if she is a library for a total of 6 hours and she helps each student

one at a time for 3/4 of an hour how many students does she tutor?

1 answer:

wariber [46]4 years ago

8 0

Lauren ‘tutored’(i think that’s what the question said idk) 4 students. take the 3/4 and multiply it by the value of time(6 hours). the calculated answer is 4.5, but you can’t half-tutor a student, and you can’t round up, so the answer is 4

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Please answer this<br>Hope you get along<br>I report What to answer

Llana [10]

What the hel you expect us to do all that for 5 kids points? Go ask you mom to do that for you. If you're asking US to do that, then it has to be 40 or 50 points.

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3 years ago

Is my answer correct??

Gwar [14]

No it isn’t correct, it’s the second one! Good luck!

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3 years ago

NO LINKS!!!!! Write 5 summary for the data below. Show your work.

ankoles [38]

Answer:

$lower \: quartile = \frac{1}{4} \times 8 \\ = {2}^{nd} \\ = 3$

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5 0

3 years ago

What are the Factors of 32 and 72?

Alexus [3.1K]

Answer:

The factors of 32 are 32, 16, 8, 4, 2, 1. The factors of 72 are 72, 36, 24, 18, 12, 9, 8, 6, 4, 3, 2, 1. The common factors of 32 and 72 are 8, 4, 2, 1, intersecting the two sets above.

plz mark brainliest :DD

4 0

3 years ago

Read 2 more answers

In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc

Troyanec [42]

Answer:

Step-by-step explanation:

(a)

Consider the following:

$A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°$

Use sine rule,

$\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}$

Again consider,

$\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Thus, the angle B is function of A is, $B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Now find $\frac{dB}{dA}$

Differentiate implicitly the function $\sin{B}=\sqrt{\frac{3}{2}}\sin{A}$ with respect to A to get,

$\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}$

b)

When $A=\frac{\pi}{4},B=\frac{\pi}{3}$ , the value of $\frac{dB}{dA}$ is,

$\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}$

c)

In general, the linear approximation at x= a is,

$f(x)=f'(x).(x-a)+f(a)$

Here the function $f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

At $A=\frac{\pi}{4}$

$f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}$

And,

$f'(A)=\frac{dB}{dA}=\sqrt{3}$ from part b

Therefore, the linear approximation at $A=\frac{\pi}{4}$ is,

$f(x)=f'(A).(x-A)+f(A)\\\\=f'(\frac{\pi}{4}).(x-\frac{\pi}{4})+f(\frac{\pi}{4})\\\\=\sqrt{3}.[x-\frac{\pi}{4}]+\frac{\pi}{3}$

d)

Use part (c), when $A=46°$ , B is approximately,

$B=f(46°)=\sqrt{3}[46°-\frac{\pi}{4}]+\frac{\pi}{3}\\\\=\sqrt{3}(1°)+\frac{\pi}{3}\\\\=61.732°$

8 0

3 years ago