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Amiraneli [1.4K]
3 years ago
5

Plz help brainlest if correct

Mathematics
1 answer:
Elodia [21]3 years ago
5 0

Answer:

Help with what....????

Step-by-step explanation:

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The angle measures of any triangle = 180 adding 27 and 63. = 90 leaving 90 degrees for the remaining angle therefore the triangle is a right triangle by definition if a triangle has a 90 degree angle it's a right triangle
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In a right-angled triangle, the hypotenuse is 113 and a leg is 112.
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Answer: 15

Step-by-step explanation: Use the formula: a^{2}+b^{2}=c^{2}

Remember c^{2} should always be the hypotenuse or the longest side.

112^{2}+b^{2}=113^{2} ---> evaluate the exponents

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b^{2}=225 ---> take the square root of both sides

b=15

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Shannon is paid £1150 a month after deductions.what is her net annual salary?(a year=12 months )
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£13800

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1150*12=£13800

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3 years ago
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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
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