Show how to make addend the next tens number

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

Read 2 more answers

Are exponential function one to one. How can you tell? What does this tell you about their inverses

aleksandrvk [35]

See in the explanation

<h2>Explanation:</h2>

<h3>1. Are exponential function one to one. How can you tell?</h3>

A function $f$ is one-to-one if each value of $x$ corresponds to exactly one value of $y$ .

To demonstrate this, we take the Horizontal Line Test that states:

<em>A function $f$ has an inverse function if and only if there is no any horizontal line that intersects the graph of $f$ at more than one point.</em>

As you can see in the first figure, the horizontal line $y=3$ (the green one) intersects the graph of the exponential function $y=e^x$ (the red one) in just one point. If you take every horizontal line $y=y_{0}$ with $y_{0}$ any real number, you will find that every line intersects the exponential function in just one point. Therefore, this function is one-to-one

<h3>2. What does this tell you about their inverses?</h3>

Another important thing is that:

A function has an inverse function if and only if is one-to-one.

As we have demonstrated that exponential functions are one-to-one by Horizontal Line Test, then we conclude exponential functions have inverse functions. The domain of the inverse function is the range of the original one and the range of the inverse function is the domain of the original one. The inverse of $y=e^x$ is $y=ln(x)$ whose graph is the second figure below.