Answer:
<em>Q'</em> = (-4, -2)
<em>R' </em>= (4, -2)
<em>S'</em> = (4, 2)
<em>T'</em> = (-4, 2)
Step-by-step explanation:
First, we can create a matrix to scale this rectangle by putting all the x coordinates in the top row and all the y coordinates in the bottom row and multiply it by 4.
Our initial matrix to multiply:
![4\left[\begin{array}{cccc}-1&1&1&-1\\-2&-2&-1&-1\end{array}\right]](https://tex.z-dn.net/?f=4%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D-1%261%261%26-1%5C%5C-2%26-2%26-1%26-1%5Cend%7Barray%7D%5Cright%5D)
Moved to the origin:
![4\left[\begin{array}{cccc}-1&1&1&-1\\-0.5&-0.5&0.5&0.5\end{array}\right]](https://tex.z-dn.net/?f=4%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D-1%261%261%26-1%5C%5C-0.5%26-0.5%260.5%260.5%5Cend%7Barray%7D%5Cright%5D)
Multiplied by four:
![\left[\begin{array}{cccc}-4&4&4&-4\\-2&-2&2&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D-4%264%264%26-4%5C%5C-2%26-2%262%262%5Cend%7Barray%7D%5Cright%5D)
This gives us the points of
<em>Q'</em> = (-4, -2)
<em>R' </em>= (4, -2)
<em>S'</em> = (4, 2)
<em>T'</em> = (-4, 2)
which are our answers.
<u>Given</u>:
Juan has been saving for a new bike for two months. So far, he has saved $150 for the bike.
The money that he has saved so far is 30% of the cost of the bike.
We need to determine the total cost of the bike.
<u>Total cost of the bike:</u>
Let x denote the total cost of the bike.
The total cost of the bike is given by

The value of 30% can be written as 
Thus, we get;

Multiplying both sides by
, we have;

Simplifying, we have;


Thus, the value of x is 500.
Hence, the total cost of the bike is $500.
132* as it’s on the same parallel line
I think it’s (4,2), If not I am so sorry.
Hello!
This is a problem about the general solution of a differential equation.
What we can first do here is separate the variables so that we have the same variable for each side (ex.
with the
term and
with the
term).


Then, we can integrate using the power rule to get rid of the differentiating terms, remember to add the constant of integration, C, to at least one side of the resulting equation.

Then here, we just solve for
and we have our general solution.
![y=\sqrt[3]{\frac{1}{2}x^2-x+C}](https://tex.z-dn.net/?f=y%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B1%7D%7B2%7Dx%5E2-x%2BC%7D)
We can see that answer choice D has an equivalent equation, so answer choice D is the correct answer.
Hope this helps!