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daser333 [38]
3 years ago
13

Which function, g or h, is the inverse function for function f

Mathematics
1 answer:
Tanzania [10]3 years ago
6 0

Answer:

2,0

Step-by-step explanation:

chhxhxxhhxh

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How to solve this trigonometric equation cos3x + sin5x = 0
mrs_skeptik [129]

Answer:

  x = {nπ -π/4, (4nπ -π)/16}

Step-by-step explanation:

It can be helpful to make use of the identities for angle sums and differences to rewrite the sum:

  cos(3x) +sin(5x) = cos(4x -x) +sin(4x +x)

  = cos(4x)cos(x) +sin(4x)sin(x) +sin(4x)cos(x) +cos(4x)sin(x)

  = sin(x)(sin(4x) +cos(4x)) +cos(x)(sin(4x) +cos(4x))

  = (sin(x) +cos(x))·(sin(4x) +cos(4x))

Each of the sums in this product is of the same form, so each can be simplified using the identity ...

  sin(x) +cos(x) = √2·sin(x +π/4)

Then the given equation can be rewritten as ...

  cos(3x) +sin(5x) = 0

  2·sin(x +π/4)·sin(4x +π/4) = 0

Of course sin(x) = 0 for x = n·π, so these factors are zero when ...

  sin(x +π/4) = 0   ⇒   x = nπ -π/4

  sin(4x +π/4) = 0   ⇒   x = (nπ -π/4)/4 = (4nπ -π)/16

The solutions are ...

  x ∈ {(n-1)π/4, (4n-1)π/16} . . . . . for any integer n

5 0
2 years ago
What is the area of a sector that has a radius of 25 mm and has an angle measure of 50°?
Nimfa-mama [501]

The area of the sector is 86.81π mm²

<u>Explanation:</u>

Given:

Radius of the sector, 25mm

Angle, α = 50°

Area of the sector, A = ?

We know:

Area of the sector = \frac{\alpha }{360} X \pi r^2

On substituting the value, we get:

A = \frac{50}{360} X \pi X (25)^2\\\\A = 86.81 \pi mm^2

Therefore, the area of the sector is 86.81π mm²

3 0
3 years ago
F(x) =x^6-3x+9<br> g(x)=3x^3+2x^2-4x-9<br> Find (f-g)(x)
enyata [817]

Step-by-step explanation:

(f-g)(x)=

x ^{2}  - 3x + 9 - (3 {x}^{3}  + 2 {x}^{2}   - 4x - 9) =  \\ x ^{2}  - 3x + 9 - 3 {x}^{3}  - 2 {x}^{2}  + 4x  + 9 = \\  - 3 {x}^{3}  - x ^{2}  + x + 18

7 0
3 years ago
Can someone help with this
kari74 [83]

Answer:

I think D option is write hope you help

5 0
2 years ago
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