Answer:
No, they are not equivalent.
Step-by-step explanation:
There are two ways to check this.
1. You can multiply the means and extremes. If the ratios are equivalent, they products of the means and extremes will be equal.
Multiply the means (2 and 7) and multiply the extremes (6 and 3)
14 ≠ 18
2. You could also check these by dividing them.
6/2 = 7/3
3 ≠ 2.3
Since neither one of the checks are correct, 6:2 and 7:3 are not equivalent.
I really hope this helps =)
Answer = 8
Because:
c(c + 3) - c(c-4) = 9c - 16
c^2 + 3c - c^2 + 4c = 9c - 16
3c + 4c = 9c - 16
7c = 9c - 16
7c + 16 = 9c
16 = 2c
8 = c
Answer:


z = 2
Step-by-step explanation:
Given equations are
x - 2y - z = 2
x + 3y - 2z = 4
-x + 2y + 3z = 2
from the given equations the augmented matrix can be written as
![\left[\begin{array}{ccc}1&-2&-1:2\\1&3&-2:4\\-1&2&3:2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-2%26-1%3A2%5C%5C1%263%26-2%3A4%5C%5C-1%262%263%3A2%5Cend%7Barray%7D%5Cright%5D)

![=\ \left[\begin{array}{ccc}1&-2&-1:2\\0&5&-1:2\\0&0&2:4\end{array}\right]](https://tex.z-dn.net/?f=%3D%5C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-2%26-1%3A2%5C%5C0%265%26-1%3A2%5C%5C0%260%262%3A4%5Cend%7Barray%7D%5Cright%5D)

![=\ \left[\begin{array}{ccc}1&-2&-1:2\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\0&0&2:4\end{array}\right]](https://tex.z-dn.net/?f=%3D%5C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-2%26-1%3A2%5C%5C0%261%26%5Cdfrac%7B-1%7D%7B5%7D%3A%5Cdfrac%7B2%7D%7B5%7D%5C%5C0%260%262%3A4%5Cend%7Barray%7D%5Cright%5D)

![=\ \left[\begin{array}{ccc}1&0&-1-\dfrac{2}{5}:2+\dfrac{4}{5}\\\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\\\0&0&2:4\end{array}\right]](https://tex.z-dn.net/?f=%3D%5C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%26-1-%5Cdfrac%7B2%7D%7B5%7D%3A2%2B%5Cdfrac%7B4%7D%7B5%7D%5C%5C%5C%5C0%261%26%5Cdfrac%7B-1%7D%7B5%7D%3A%5Cdfrac%7B2%7D%7B5%7D%5C%5C%5C%5C0%260%262%3A4%5Cend%7Barray%7D%5Cright%5D)
![=\ \left[\begin{array}{ccc}1&0&\dfrac{-7}{5}:\dfrac{14}{5}\\\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\\\0&0&2:4\end{array}\right]](https://tex.z-dn.net/?f=%3D%5C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%26%5Cdfrac%7B-7%7D%7B5%7D%3A%5Cdfrac%7B14%7D%7B5%7D%5C%5C%5C%5C0%261%26%5Cdfrac%7B-1%7D%7B5%7D%3A%5Cdfrac%7B2%7D%7B5%7D%5C%5C%5C%5C0%260%262%3A4%5Cend%7Barray%7D%5Cright%5D)

![=\ \left[\begin{array}{ccc}1&0&\dfrac{-7}{5}:\dfrac{14}{5}\\\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\\\0&0&1:2\end{array}\right]](https://tex.z-dn.net/?f=%3D%5C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%26%5Cdfrac%7B-7%7D%7B5%7D%3A%5Cdfrac%7B14%7D%7B5%7D%5C%5C%5C%5C0%261%26%5Cdfrac%7B-1%7D%7B5%7D%3A%5Cdfrac%7B2%7D%7B5%7D%5C%5C%5C%5C0%260%261%3A2%5Cend%7Barray%7D%5Cright%5D)

![=\ \left[\begin{array}{ccc}1&0&0:\dfrac{14}{5}+\dfrac{7}{5}\\\\0&1&0:\dfrac{2}{5}+\dfrac{1}{5}\\\\0&0&1:2\end{array}\right]](https://tex.z-dn.net/?f=%3D%5C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%3A%5Cdfrac%7B14%7D%7B5%7D%2B%5Cdfrac%7B7%7D%7B5%7D%5C%5C%5C%5C0%261%260%3A%5Cdfrac%7B2%7D%7B5%7D%2B%5Cdfrac%7B1%7D%7B5%7D%5C%5C%5C%5C0%260%261%3A2%5Cend%7Barray%7D%5Cright%5D)
So, from the above augmented matrix, we can write


z = 2
Answer:
f(-5)=19, so t= -5
Step-by-step explanation:
We begin with the function f(t). To find when it equals 19, we can set the function equal to 19 and then solve for t
