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3 years ago

Does anyone know how to graph this please help

Mathematics

1 answer:

Anna71 [15]3 years ago

8 0

Just like the Vertex Formula you do this:

y = 3(x - 4)³

This is where -H gives you the OPPOSITE terms of what they really are, and K gives you the NORMAL term, so just apply that to here. Whenever you are doing a phase shift, you put it in parentheses using the Vertex Formula, but giving the opposite term of what it really is:

Y = (X - H)² + K

Y = (X - H)³ + K

Now here is the graph. You see how I shifted it 4 units away from the origin?

So, find your roots [zeros (x-intercept)], plug in 0 for y and solve for x. When you do this, you will get 4 = x. You could also look at the graph and can tell that the ONLY x-intercept is 4. So, either way, you will get the same result.

I am joyous to assist you anytime.

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Answer:

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Step-by-step explanation:

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4 years ago

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3 years ago

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3 years ago

Solve the given initial-value problem. x' = 1 2 0 1 − 1 2 x, x(0) = 2 7

Ilia_Sergeevich [38]

I'll go out on a limb and guess the system is

$\mathbf x'=\begin{bmatrix}\frac12&0\\1&-\frac12\end{bmatrix}\mathbf x$

with initial condition $\mathbf x(0)=\begin{bmatrix}2&7\end{bmatrix}^\top$ . The coefficient matrix has eigenvalues $\lambda$ such that

$\begin{vmatrix}\frac12-\lambda&0\\1&-\frac12-\lambda\end{vmatrix}=\lambda^2-\dfrac14=0\implies\lambda=\pm\dfrac12$

The corresponding eigenvectors $\eta$ are such that

$\lambda=\dfrac12\implies\begin{bmatrix}\frac12-\frac12&0\\1&-\frac12-\frac12\end{bmatrix}\eta=\begin{bmatrix}0&0\\1&-1\end{bmatrix}\eta=\begin{bmatrix}0\\0\end{bmatrix}$
$\implies\eta=\begin{bmatrix}1\\1\end{bmatrix}$

$\lambda=-\dfrac12\implies\begin{bmatrix}\frac12+\frac12&0\\1&-\frac12+\frac12\end{bmatrix}\eta=\begin{bmatrix}1&0\\1&0\end{bmatrix}\eta=\begin{bmatrix}0\\0\end{bmatrix}$
$\implies\eta=\begin{bmatrix}0\\1\end{bmatrix}$

So the characteristic solution to the ODE system is

$\mathbf x(t)=C_1\begin{bmatrix}1\\1\end{bmatrix}e^{t/2}+C_2\begin{bmatrix}0\\1\end{bmatrix}e^{-t/2}$

When $t=0$ , we have

$\begin{bmatrix}2\\7\end{bmatrix}=C_1\begin{bmatrix}1\\1\end{bmatrix}+C_2\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}C_1\\C_1+C_2\end{bmatrix}$

from which it follows that $C_1=2$ and $C_2=5$ , making the particular solution to the IVP

$\mathbf x(t)=2\begin{bmatrix}1\\1\end{bmatrix}e^{t/2}+5\begin{bmatrix}0\\1\end{bmatrix}e^{-t/2}$

$\mathbf x(t)=\begin{bmatrix}2e^{t/2}\\2e^{t/2}+5e^{-t/2}\end{bmatrix}$

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4 years ago

A large university accepts 70% of the students who apply. Of the students the university accepts, 30% actually enroll. If 30,000

Irina-Kira [14]

Answer:

630

Step-by-step explanation:

you need to do a 2 step equation, first finding the amount of students that get accepted, and then use the info you just got solving how many people actually enroll.

formula:

part part %

---------- = -----------

total total %

5 0

3 years ago

Does anyone know how to graph this please help​

Does anyone know how to graph this please help