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Olegator [25]
3 years ago
11

F^-1(-10) + f(-6) Please help!!!!

Mathematics
1 answer:
Brums [2.3K]3 years ago
4 0

Answer:

-10/f-6f

Step-by-step explanation:

simplify it the equation

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To determine whether the inverse of a function is a function you can perform the horizontal line test.
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Answer:

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Step-by-step explanation:

If a horizontal line intersects the graph of a function in all places at exactly one point (the horizontal line test), the inverse of the function is also a function.

For example, the inverse of a hyperbola (like ƒ(x) =1/x) is a function, because every horizontal line intersects with the graph at exactly one point.

However, the inverse of a parabola (like ƒ(x) = x²) is not a function, because a horizontal line intersects with the graph at two points.

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Just take a look at the picture and say the answer no explaining
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16-25

Step-by-step explanation:

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2 years ago
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There are 9 red and 6 green marbles in a bag
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Answer:

there are 15 marbles in the bag

Step-by-step explanation:

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Help!! 50 points and brainliest!
Viktor [21]

Answer:

Second choice:

x=2t

y=4t^2+4t-3

Fifth choice:

x=t+1

y=t^2+4t

Step-by-step explanation:

Let's look at choice 1.

x=t+1

y=t^2+2t

I'm going to subtract 1 on both sides for the first equation giving me x-1=t. I will replace the t in the second equation with this substitution from equation 1.

y=(x-1)^2+2(x-1)

Expand using the distributive property and the identity (u+v)^2=u^2+2uv+v^2:

y=(x^2-2x+1)+(2x-2)

y=x^2+(-2x+2x)+(1-2)

y=x^2+0+-1

y=x^2

So this not the desired result.

Let's look at choice 2.

x=2t

y=4t^2+4t-3

Solve the first equation for t by dividing both sides by 2:

t=\frac{x}{2}.

Let's plug this into equation 2:

y=4(\frac{x}{2})^2+4(\frac{x}{2})-3

y=4(\frac{x^2}{4})+2x-3

y=x^2+2x-3

This is the desired result.

Choice 3:

x=t-3

y=t^2+2t

Solve the first equation for t by adding 3 on both sides:

x+3=t.

Plug into second equation:

y=(x+3)^2+2(x+3)

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

y=(x^2+6x+9)+(2x+6)

y=(x^2)+(6x+2x)+(9+6)

y=x^2+8x+15

Not the desired result.

Choice 4:

x=t^2

y=2t-3

I'm going to solve the bottom equation for t since I don't want to deal with square roots.

Add 3 on both sides:

y+3=2t

Divide both sides by 2:

\frac{y+3}{2}=t

Plug into equation 1:

x=(\frac{y+3}{2})^2

This is not the desired result because the y variable will be squared now instead of the x variable.

Choice 5:

x=t+1

y=t^2+4t

Solve the first equation for t by subtracting 1 on both sides:

x-1=t.

Plug into equation 2:

y=(x-1)^2+4(x-1)

Distribute and use the binomial square identity used earlier:

y=(x^2-2x+1)+(4x-4)

y=(x^2)+(-2x+4x)+(1-4)

y=x^2+2x+-3

y=x^2+2x-3.

This is the desired result.

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3 years ago
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What are x and 8 in this lineat equation? 8/x+5=50/25
MatroZZZ [7]

Answer:

x= -8/3

Step-by-step explanation:

4 0
3 years ago
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