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3 years ago

Aslam is planning to do social service for 3 hours a day. How much time does he spend on social service in 2 months?

Mathematics

1 answer:

ladessa [460]3 years ago

7 0

Answer:

if the month has 30 days: 180 hours

if the month has 31 days: 186 hours

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Can you solve this equation?

Leya [2.2K]

Answer:

$10 + 10 + 10 = 30 \\ 10 + 5 + 5 = 20 \\ 5 + 4 + 4 = 13 \\ 10 + 5 + 2 = 17$

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3 years ago

Read 2 more answers

Three interior angles of a polygon are,100degrees,120degrees,and 108degrees.Find the remaining two sides,if one of them is three

k0ka [10]

Answer:

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Step-by-step explanation:

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3 years ago

What is (64-6squared)÷(5+2)you are welcome

lina2011 [118]

I believe the correct answer is 1.08
64-6= 58
58 squared= 7.6(or 7.61)
5+2= 7
7.6÷7= 1.08

4 0

4 years ago

A sphere has a diameter of 14.5 inches.

dlinn [17]

Answer:

≈ 660.5 in²

Step-by-step explanation:

The surface area (A) of a sphere is calculated using the formula

A = 4πr² ← r is the radius

Given diameter = 14. 5 then radius = 7.25, thus

A = 4π × 7.25² = 4π × 52.5625 ≈ 660.5 in²

4 0

3 years ago

How do I find the integral<br> ∫10(x−1)(x2+9)dxint10/((x-1)(x^2+9))dx ?

defon

$\int\frac{10}{(x-1)(x^2+9)}\ dx=(*)\\\\\frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}=\frac{A(x^2+9)+(Bx+C)(x-1)}{(x-1)(x^2+9)}\\\\=\frac{Ax^2+9A+Bx^2-Bx+Cx-C}{(x-1)(x^2+9)}=\frac{(A+B)x^2+(-B+C)x+(9A-C)}{(x-1)(x^2+9)}\\\Updownarrow\\10=(A+B)x^2+(-B+C)x+(9A-C)\\\Updownarrow\\A+B=0\ and\ -B+C=0\ and\ 9A-C=10\\A=-B\ and\ C=B\to9(-B)-B=10\to-10B=10\to B=-1\\A=-(-1)=1\ and\ C=-1$

$(*)=\int\left(\frac{1}{x-1}+\frac{-x-1}{x^2+9}\right)\ dx=\int\left(\frac{1}{x-1}-\frac{x+1}{x^2+9}\right)\ dx\\\\=\int\frac{1}{x-1}\ dx-\int\frac{x+1}{x^2+9}\ dx=\int\frac{1}{x-1}-\int\frac{x}{x^2+9}\ dx-\int\frac{1}{x^2+9}\ dx=(**)\\\\\#1\ \int\frac{1}{x-1}\ dx\Rightarrow\left|\begin{array}{ccc}x-1=t\\dx=dt\end{array}\right|\Rightarrow\int\frac{1}{t}\ dt=lnt+C_1=ln(x-1)+C_1$

$\#2\ \int\frac{x}{x^2+9}\ dx\Rightarrow \left|\begin{array}{ccc}x^2+9=u\\2x\ dx=du\\x\ dx=\frac{1}{2}\ du\end{array}\right|\Rightarrow\int\left(\frac{1}{2}\cdot\frac{1}{u}\right)\ du=\frac{1}{2}\int\frac{1}{u}\ du\\\\\\=\frac{1}{2}ln(u)+C_2=\frac{1}{2}ln(x^2+9)+C_2$

$\#3\ \int\frac{1}{x^2+9}\ dx=\int\frac{1}{x^2+3^2}\ dx=\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3\\\\therefore:\\\\\#1;\ \#2;\ \#3\Rightarrow(**)=ln(x-1)+C_1-\frac{1}{2}ln(x^2+9)+C_2-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3$

$\boxed{=ln(x-1)-\frac{1}{2}ln(x^2+9)-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C}$

4 0

4 years ago