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3 years ago

Weather

Mathematics

1 answer:

prisoha [69]3 years ago

5 0

The total snowfall for listed 10 days is 349 cm.

Further explanation:

The sum of snow fell each day will give us the total snowfall for 10 days.

Given:

Snow fell on 10 days:

14 cm, 13 cm, 7 cm, 3 cm, 11 cm, 14 cm,

28 cm, 108 cm, 84 cm, 67 cm

So,

$Total\ snowfall\ for\ 10\ days\ listed=14+13+7+3+11+14+28+108+84+67\\=349\ cm$

The total snowfall for listed 10 days is 349 cm.

Keywords: Addition, Word problems

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Solve this equation<br><br><br> -3(6 + 6b) + 4 = -104

yulyashka [42]

Answer:

b = 5.

Step-by-step explanation:

-3(6 + 6b) + 4 = -104

-18 - 18b + 4 = -104

-18b = -104 + 18 - 4

-18b = -90

b = -90 / -18

b = 5.

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3 years ago

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

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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}
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Again consider,

$\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
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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,

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

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

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

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,

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

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

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

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3 years ago

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solmaris [256]

Let's assume two variables x and y which represent the local and international calls respectively.

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from equation 1:-
x=852 -y (sub in equation 2)
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51.12 -0.06 y +0.15 y =69.84 (subtracting both sides by 51.12)
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3 years ago

Select the multiplication equation that could represent the following question: how many 3/8s are in 5/4

muminat

Answer:

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Step-by-step explanation:

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