The function g(x) is g(x)= (3x)^2
<h3>How to solve for g(x)?</h3>
The complete question is in the image
From the graph in the image, we have:
f(x) = x^2
The function f(x) is stretched by a factor of 3 to form g(x).
This means that:
g(x) = f(3x)
So, we have:
g(x)= (3x)^2
Hence, the function g(x) is g(x)= (3x)^2
Read more about function transformation at:
brainly.com/question/10222182
#SPJ1
Answer:
4/1
Step-by-step explanation:
To get from one dot to the next, you go up by four and to the right by one, making the slope 4/1
Write the system in augmented-matrix form:
![\left[\begin{array}{ccc|c}2&2&4&16\\5&-2&3&-1\\1&2&-3&-9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D2%262%264%2616%5C%5C5%26-2%263%26-1%5C%5C1%262%26-3%26-9%5Cend%7Barray%7D%5Cright%5D)
Multiply through row 1 by 1/2:
![\left[\begin{array}{ccc|c}1&1&2&8\\5&-2&3&-1\\1&2&-3&-9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%262%268%5C%5C5%26-2%263%26-1%5C%5C1%262%26-3%26-9%5Cend%7Barray%7D%5Cright%5D)
Add -1(row 1) to row 3, and add -5(row 1) to row 2:
![\left[\begin{array}{ccc|c}1&1&2&8\\0&-7&-7&-41\\0&1&-5&-17\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%262%268%5C%5C0%26-7%26-7%26-41%5C%5C0%261%26-5%26-17%5Cend%7Barray%7D%5Cright%5D)
Swap rows 2 and 3:
![\left[\begin{array}{ccc|c}1&1&2&8\\0&1&-5&-17\\0&-7&-7&-41\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%262%268%5C%5C0%261%26-5%26-17%5C%5C0%26-7%26-7%26-41%5Cend%7Barray%7D%5Cright%5D)
Add -7(row 2) to row 3:
![\left[\begin{array}{ccc|c}1&1&2&8\\0&1&-5&-17\\0&0&-42&-160\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%262%268%5C%5C0%261%26-5%26-17%5C%5C0%260%26-42%26-160%5Cend%7Barray%7D%5Cright%5D)
Multiply through row 3 by -1/42:
![\left[\begin{array}{ccc|c}1&1&2&8\\0&1&-5&-17\\0&0&1&\frac{80}{21}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%262%268%5C%5C0%261%26-5%26-17%5C%5C0%260%261%26%5Cfrac%7B80%7D%7B21%7D%5Cend%7Barray%7D%5Cright%5D)
Add 5(row 3) to row 2:
![\left[\begin{array}{ccc|c}1&1&2&8\\0&1&0&\frac{43}{21}\\0&0&1&\frac{80}{21}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%262%268%5C%5C0%261%260%26%5Cfrac%7B43%7D%7B21%7D%5C%5C0%260%261%26%5Cfrac%7B80%7D%7B21%7D%5Cend%7Barray%7D%5Cright%5D)
Add -1(row 2) and -2(row3) to row 1:
![\left[\begin{array}{ccc|c}1&0&0&-\frac53\\0&1&0&\frac{43}{21}\\0&0&1&\frac{80}{21}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%26-%5Cfrac53%5C%5C0%261%260%26%5Cfrac%7B43%7D%7B21%7D%5C%5C0%260%261%26%5Cfrac%7B80%7D%7B21%7D%5Cend%7Barray%7D%5Cright%5D)
So the solution to the system is
6u + 3u = 18
9u = 18
u = 18/9
u = 2
Answer:
So if you're having difficulty visualizing the differences, you can rewrite them so they all have the same denominator
7/12 = 35/60
1/2 = 30/60
2/3 = 40/60
4/10 = 24/60
1/6 = 10/60
So now it's easier to see how they compare to each other.
10/60, 24/60, 30/60, 35/60, 40/60
And then simplify them or just refer to what they were before
1/6, 4/10, 1/2, 7/12, 2/3
I hope this helps!
