What condition does a function need to meet in order to be invertible?

Mathematics

1 answer:

pickupchik [31]4 years ago

4 0

Answer:

A function must be "one-to-one" in order to be invertable meaning that each y-value is associated with only one x-value. You can easily check to see if a function is invertable by using the "horizontal line test" where you run your finger up and down the graph and if at any point your finger is touching two or more points the graph has failed the test.

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33)

Damm [24]

Answer:

33 is C

Step-by-step explanation:

I just did the test. not sure about the other ones tho

7 0

3 years ago

Find the orthogonal trajectories of the family of curves. (Use C for any needed constant.)

Dominik [7]

Answer:

y = Ax^2

Step-by-step explanation:

below is the detailed solution

To determine the orthogonal trajectories of the family of curves

X^2 + 2y^2 = 17k^2

we have to differentiateX^2 + 2y^2 = 17k^2 with respect to x

= 2x + 4y dy/dx = 0

Hence : dy/dx = - x/2y

we have to determine the negative reciprocal

dy/dx = 2y/x ----------- 1

integrate equation 1

∫dy/2y = ∫dx/x

= 1/2 log y = log x + log c

log y = 2logx + 2logc

log y = logx^2 + logC^2

therefore : y = Ax^2 ; where C^2 = A

8 0

3 years ago

Find the sum of the three smallest positive values of θ such that <img src="https://tex.z-dn.net/?f=4%5Ccos%5E2%282%5Ctheta-%5Cp

Harlamova29_29 [7]

Answer:

∅= 13π/12

Step-by-step explanation:

Solve until you get cos(___) on one side and a value on the other side.

I got cos(2∅-π) = ±√(3)/2

I know that the only cos() value that gives me ±√(3)/2 are -5π/6, -π/6, and π/6 from looking at the unit circle; those 3 are the smallest value that would give me ±√(3)/2.

So i solve for what value of 2∅-π would give me -5π/6, -π/6, and π/6

by solving for

2∅₁-π = -5π/6, 2∅₂-π = -π/6, and 2∅₃-π = π/6

I got ∅₁=π/12, ∅₂=5π/12, ∅₃=7π/12.

The question wanted the sum of the three smallest value so i added it up and got 13π/12.

And thanks for peer review. Good catch on my mistakes.