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butalik [34]
3 years ago
12

For each fraction if you multiply both the numerator and the denominator by the same number is the resulting fraction equivalent

to the original fraction ?
Mathematics
1 answer:
lukranit [14]3 years ago
5 0
Yes, it is. 

example : 
1/2 * 2/2 = 2/4...so 1/2 and 2/4 are equivalent because 2/4 reduces to 1/2

1/3 * 3/3 = 3/9...so 3/9 is equivalent to 1/3 because 3/9 can reduce to 1/3
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y2-y1 over x2-x1 (you can check my work if unsure)
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How many three-person relay teams can be chosen from six students?
oksano4ka [1.4K]
2 because 6 divided by 3 is 2

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3 years ago
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What is the surface area of the sphere shown below with a radious of 6
Bumek [7]

Surface Area of a Sphere is...

4\Pi r^2

Now...

\begin{gathered} 4\Pi r^2 \\ 4\cdot\text{ }\Pi\cdot6^2 \\ =\text{ 4}\cdot\Pi\cdot36 \\ =\text{ 144}\Pi\text{ square unit} \end{gathered}

The answer is Option C

4 0
1 year ago
Do you know the inequality? Please explain and show answer :)
DerKrebs [107]
If u have a open circle, then the inequality has no equal sign. But if it is a closed circle, the inequality has an equal sign.
Shading to the left means " less then " .
Shading to the right means " greater then " .

(1) u have an open circle on -7 with shading to the left....
     x < -7

(2) u have a closed circle on 4.5 with shading to the left....
    x < = 4.5 (thats less then or equal)

(3) u have an open circle on -5 with shading to the right...
     x > -5

(4) u have an open circle on 1.5 with shading to the right...
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4 0
3 years ago
Find the indefinite integral. (Note: Solve by the simplest method—not all require integration by parts. Use C for the constant o
umka2103 [35]

Answer:

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-1/2  ln⁡(1+x^2 )+C

Step-by-step explanation:

∫▒〖1st .2nd dx=1st∫▒〖2nd dx〗-∫▒〖(derivative of 1st) dx∫▒〖2nd dx〗〗〗

Let 1st=arctan⁡(x)

And 2nd=1

∫▒〖arctan⁡(x).1 dx=arctan⁡(x) ∫▒〖1 dx〗-∫▒〖(derivative of arctan(x))dx∫▒〖1 dx〗〗〗

As we know that  

derivative of arctan(x)=1/(1+x^2 )

∫▒〖1 dx〗=x

So  

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-∫▒〖(1/(1+x^2 ))dx.x〗…………Eq1

Let’s solve ∫▒(1/(1+x^2 ))dx by substitution now  

Let 1+x^2=u

du=2xdx

Multiply and divide ∫▒〖(1/(1+x^2 ))dx.x〗 by 2 we get

1/2 ∫▒〖(2/(1+x^2 ))dx.x〗=1/2 ∫▒(2xdx/u)  

1/2 ∫▒(2xdx/u) =1/2 ∫▒(du/u)  

1/2 ∫▒(2xdx/u) =1/2  ln⁡(u)+C

1/2 ∫▒(2xdx/u) =1/2  ln⁡(1+x^2 )+C

Putting values in Eq1 we get

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-1/2  ln⁡(1+x^2 )+C  (required soultion)

3 0
3 years ago
Read 2 more answers
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