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dalvyx [7]
3 years ago
8

Write the equation for the inverse of the function y=cos2x

Mathematics
2 answers:
MrMuchimi3 years ago
6 0
The answer is basically A
viva [34]3 years ago
3 0

we are given

y=cos(2x)

Since, we have to find inverse function

so, we will use following steps

step-1:

Set y=f(x)

It is already there

step-2:

Switch x and y

x=cos(2y)

step-3:

Solve for y

x=cos(2y)

we can take cos^-1 both sides

cos^{-1}(x)=cos^{-1}(cos(2y))

now, we can simplify it

cos^{-1}(x)=2y

Divide both sides by 2

y=\frac{cos^{-1}(x)}{2}

so, option-D........Answer

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Why does PEMDAS mean?
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Step-by-step explanation:

This is the order that you should solve your expressions that are given to you (of course, you still have to follow the left -> right rule afterwards).

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Factor x^2+ 5x + 6<br> pleaseee
alekssr [168]

Answer:

( x + 2 ) ( x  +3 )

6 0
3 years ago
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I am confused about how to do this problem and I don't think it should be hard, but I just don't know how to approach it, so if
solniwko [45]

Answer:

a) (0.555, 0) and (6, 0)

b) r = -3 and r = 1.8

c) (0.875, 0.676)

d) (0, 1.235)

Step-by-step explanation:

Set each term in the numerator and denominator equal to 0 and find r.

In the numerator:

r = 7/8, 5/9, or 6

In the denominator:

r = 9/5, 7/8, or -3

Zeros in the numerator that aren't in the denominator are r-intercepts.

Zeros in the denominator that aren't in the numerator are vertical asymptotes.

Zeros in both the numerator and the denominator are holes.

a) (0.555, 0) and (6, 0)

b) r = -3 and r = 1.8

c) Evaluate m(r) at r = 7/8.  To do that, first divide out the common term (-8r + 7) from the numerator and denominator.

m(r) = (-9r+5)² (r−6)² / ( (-5r+9)² (r+3)² )

m(⅞) = (-9×⅞+5)² (⅞−6)² / ( (-5×⅞+9)² (⅞+3)² )

m(⅞) = (-23/8)² (-41/8)² / ( (37/8)² (31/8)² )

m(⅞) = (-23)² (-41)² / ( (37)² (31)² )

m(⅞) = 0.676

The hole is at (0.875, 0.676).

d) Evaluate m(r) at r = 0.

m(0) = (-9×0+5)² (0−6)² / ( (-5×0+9)² (0+3)² )

m(0) = (5)² (-6)² / ( (9)² (3)² )

m(0) = 1.235

The m(r)-intercept is (0, 1.235).

5 0
3 years ago
How to evaluate higher powers of i
Arisa [49]

Answer:

i^0 = 1

i^1 = i

i^2 = -1

i^3 = -i

i^4 = 1

i^5 = i

i^6 = -1

i^7 = -i

i^8 = 1

i^9 = i

Step-by-step explanation:

Anything raised to 0 is 1

Anything raised to 1 is itself

I just always knew i^2 = -1

I put this rest into the calculator honestly

But if you look at the stuff from i^2, you'll notice there's a pattern (-1, -i, 1, i) and it repeats. I bet if you raise i to the 10th, 11th, 12th, and 13th power, it is going to repeat the same pattern. I hope this helps in any way.

4 0
3 years ago
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