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Tresset [83]
3 years ago
12

2) Find the value of x. (pic below) A. 180 degrees B. 58 degrees C. 122 degrees

Mathematics
2 answers:
Romashka [77]3 years ago
8 0

Answer:

122 degrees, its not a straight line its obtuse.

N76 [4]3 years ago
5 0

Answer:

A.

Step-by-step explanation:

I think

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100 students were surveyed for their pets at home. There are 36 students with dogs, 33 students with cats, 30 students with fish
DedPeter [7]

Answer:

A: the answer would be 1 i think for part b: 6 part C:72 part D:36 and part E :72

Step-by-step explanation:

hope this helps i think im right im not 100% sure but that is kinda confusing

7 0
3 years ago
Read 2 more answers
Divide of a right angle into two parts such that the number of English minutes in one to the
stira [4]

Answer:

  • π/3 radians, 3600 English minutes
  • π/6 radians, 333,333 1/3 French seconds

Step-by-step explanation:

An English minute is (1/60)(π/180) radian = π/10800 radian.

A French second is (1/100)(1/100)(π/200) radian = π/2000000 radian.

So, the ratio of angle sizes in common units will be ...

  27(π/10800) : 2500(π/2000000) = (2/800) : (1/800) = 2 : 1

Since the total of these ratio units is 2+1 = 3, we need to divide the right angle into 3 parts. The English part will be 2 of those; the French part will be 1 of them. A 1/3 part of a right angle is π/6 radians, so the right angle division will be ...

  English : French = π/3 radians : π/6 radians

  = 3600 English minutes : 333,333 1/3 French seconds

  = 27 : 2500

5 0
3 years ago
PLEASE HELP!!!!! <br> What is x?
Gelneren [198K]

Answer: 84 degrees

Step-by-step explanation: I'm not quite sure how to explain this problem, sadly, but you basically add up the degrees together in the big triangle all together, and then figure out the rest.

5 0
3 years ago
Find F"(x) if f(x) = cot (x)
hammer [34]

f(x)=\cot x\implies f'(x)=-\csc^2x\implies\boxed{f''(x)=2\csc^2x\cot x}

If you don't know the first derivative of \cot, but you do for \sin and \cos, you can derive the former via the quotient rule:

\cot x=\dfrac{\cos x}{\sin x}

\implies(\cot x)'=\dfrac{\sin x(-\sin x)-\cos x(\cos x)}{\sin^2x}=-\dfrac1{\sin^2x}=-\csc^2x

or if you know the derivative of \tan:

\cot x=\dfrac1{\tan x}

\implies(\cot x)'=-(\tan x)^{-2}\sec^2x=-\dfrac{\sec^2x}{\tan^2x}=-\dfrac{\frac1{\cos^2x}}{\frac{\sin^2x}{\cos^2x}}=-\dfrac1{\sin^2x}=-\csc^2x

As for the second derivative, you can use the power/chain rules:

(-\csc^2x)'=-2\csc x(\csc x)'=-2\csc x(-\csc x\cot x)=2\csc^2x\cot x

or if you don't know the derivative of \csc,

\csc x=\dfrac1{\sin x}

\implies(-\csc^2x)'=\left(-(\sin x)^{-2}\right)'=2(\sin x)^{-3}(\sin x)'=\dfrac{2\cos x}{\sin^3x}

which is the same as the previous result since

\csc^2x\cot x=\dfrac1{\sin^2x}\dfrac{\cos x}{\sin x}=\dfrac{\cos x}{\sin^3x}

4 0
3 years ago
2. Sean burns 540 calories per hour when he runs. How many calories does he
olchik [2.2K]

Answer:

540 times the hours he runs

Step-by-step explanation:

Simply just 540 times the hours he runs

8 0
3 years ago
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