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forsale [732]
3 years ago
13

I'LL MARK AS BRAINLIEST ( ◜‿◝ )♡

Mathematics
1 answer:
Ymorist [56]3 years ago
5 0

Answer:

You can follow the following method, there are other ways to do this too.

Step-by-step explanation:

Taking RHS,

(cosx-sinx+cosx+sinx)/(cos^2x-sin^2) [Taking LCM of den.]

2cosx/cos2x

Taking LHS,

sin2x/cos2x*1/sinx

2sinxcosx/cos2x*1/sinx

2cosx/cos2x

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Function Operations Picture attached
sergeinik [125]

Answer: Second option

Step-by-step explanation:

You need to make the multiplication of the function c(x)=\frac{5}{x-2} and the function d(x)=x+3. Then:

(cd)(x)=(\frac{5}{x-2})(x+3)

You need to apply the Distributive property:

(cd)(x)=\frac{5(x+3)}{x-2}\\\\(cd)(x)=\frac{5x+15}{x-2}

Therefore, the domain will be all the number that make the denominator equal to zero.

Then, make the denominator equal to zero and solve for x:

x-2=0\\x=2

Therefore, the domain is: All real values of x except x=2

6 0
3 years ago
If a function f(x)=x^3 has the domain {-3,0,2,4} what is it's range
torisob [31]
27, 0, 8, 64
All u need to do is cube all the numbers in the domain
4 0
3 years ago
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solve & show ur steps PLEASE Laura jogs 1/5 kilometer in 1/6 hour. What is Lara's rate in kilometers per hour? write your an
velikii [3]
The picture is how I solved it and it will be correct

4 0
3 years ago
Anne plans to increase the prices of all the items in her store by 5%. To the nearest cent, how much will the artist save if the
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4 years ago
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Find the area of the shaded regions below. Give your answer as a completely simplified exact value in terms of π (no approximati
pochemuha

!

Answer:   8 (Pi - sqrt(3))

Discussion:

The area of the shaded region is that of the semicircle minus the area of the triangle..

Area of semicircle = 1/2 * Pi * R^2        

    Where R^2 is the square of the radius of the circle. In our case, R ( = OC)

     = 4 so the semicircle area is

    (1/2) * Pi * (4^2) = (1/2) * Pi * 16 = 8 Pi

Area of triangle.

   First of all, angle ACB is a right angle ( i.e. 90 degrees).

     * This is the Theorem of Thales from elementary Plane Geometry. *

  so by Pythagoras

    AC^2 + BC^2 = AB^2

 But CB = 4 (given) and AB = 4*2 = 8 ( the diameter is twice the radius).

 Substituting these in Pythagoras gives

    AC^2 + 4^2 = 8^2 or

    AC^2 = 8^2 - 4^2- = 64 - 16 = 48

    Hence AC = sqrt(48) = sqrt (16*3) = 4 * sqrt(3)

We are almost done! The area of the triangle is given by

   (1/2) b * h = (1/2)  BC * AC = (1/2) 4 * (4 * sqrt(3)) =  8 sqrt(3)

We conclude the area area of the shaded part is

  8 PI - 8 sqrt(3)   = 8 (Pi - sqrt(3))

Note that sqrt(3) is approx  1.7 so (PI - sqrt(3)) is a positive number, as it better well be!

8 0
3 years ago
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