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sineoko [7]
3 years ago
14

HELPPPP Cos x cosy = 1/2(sin (x+y)+ sin (x-y)) true or false

Mathematics
1 answer:
julsineya [31]3 years ago
8 0

Answer:

Step-by-step explanation:

REcall that sin(a+b) = sin(a)cos(b)+cos(a)sin(b). Then

sin(a+b)+sin(a-b) = sin(a)cos(b)+cos(a)sin(b) + sin(a)cos(b) -cos(a)sin(b) = 2sin(a)cos(b). Then

sin(a)cos(b) = (1/2)(sin(a+b)+sin(a-b))

So it is false

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Answer:

93+(68+7)= 168

Step-by-step explanation:

You have to answer the problem in the parenthesis and then add 93.

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A sprinkler is set to water the section of lawn represented by the shaded region in the circle below
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1/2 cup of milk per 2 cups of flour. Find the unit rate.
mestny [16]

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Step-by-step explanation:

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3 years ago
Integrate the following
enot [183]

I suppose you mean to have the entire numerator under the square root?

\displaystyle\int_2^4\frac{\sqrt{x^2-4}}{x^2}\,\mathrm dx

We can use a trigonometric substitution to start:

x=2\sec t\implies\mathrm dx=2\sec t\tan t\,\mathrm dt

Then for x=2, t=\sec^{-1}1=0; for x=4, t=\sec^{-1}2=\frac\pi3. So the integral is equivalent to

\displaystyle\int_0^{\pi/3}\frac{\sqrt{(2\sec t)^2-4}}{(2\sec t)^2}(2\sec t\tan t)\,\mathrm dt=\int_0^{\pi/3}\frac{\tan^2t}{\sec t}\,\mathrm dt

We can write

\dfrac{\tan^2t}{\sec t}=\dfrac{\frac{\sin^2t}{\cos^2t}}{\frac1{\cos t}}=\dfrac{\sin^2t}{\cos t}=\dfrac{1-\cos^2t}{\cos t}=\sec t-\cos t

so the integral becomes

\displaystyle\int_0^{\pi/3}(\sec t-\cos t)\,\mathrm dt=\boxed{\ln(2+\sqrt3)-\frac{\sqrt3}2}

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3 years ago
What is the value of 0.5 + 0.3? Write your answer as a fraction in lowest terms.
Nikitich [7]

Answer:

4/5

Step-by-step explanation:

=.8

.8 into a fraction is 4/5

7 0
3 years ago
Read 2 more answers
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