The dude on top of me is correct
I would have each block be 1/6 of a yard
You could technically have any value you want, but for me 1/6 is easiest because 1/2 and 1/3 will scale up to this like so
1/2 = (1/2)*(3/3) = 3/6
1/3 = (1/3)*(2/2) = 2/6
The diagram below might help if you're still stuck on why I picked 1/6.
Answer:
-5n + 8
Step-by-step explanation:
<u>1. What is the difference?</u>
The sequence goes down by 5. This means that the formula will have -5n in it.
<u>2. Work out the term before the first term.</u>
The first term is 3, and we know that the sequence goes up by 5. So, to get the term before 3 we would add 5.
3 + 5 = 8 Remember, this is positive 8. (+8)
<u>3. Put that term at the end of the equation. </u>
We have -5n already and we just worked out the term before the first one which is positive 8 - so put that at the end of the equation.
-5n + 8 This is our answer!
Just to prove it works:
<em>Substitute: n = term</em>
Lets see if we can get the 3rd term which is -7. (n = 3)
-5(3) + 8
-15 + 8 = -7
See, it works!
Answer:
C. ![\left[\begin{array}{ccc}-24&0&-3\\8&-48&56\\25&-6&10\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-24%260%26-3%5C%5C8%26-48%2656%5C%5C25%26-6%2610%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The given matrices are;
![F=\left[\begin{array}{cc}-2&0\\0&8\\2&1\end{array}\right]](https://tex.z-dn.net/?f=F%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%260%5C%5C0%268%5C%5C2%261%5Cend%7Barray%7D%5Cright%5D)
and
![C=\left[\begin{array}{ccc}12&0&\frac{3}{2}\\1&-6&7\end{array}\right]](https://tex.z-dn.net/?f=C%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D12%260%26%5Cfrac%7B3%7D%7B2%7D%5C%5C1%26-6%267%5Cend%7Barray%7D%5Cright%5D)
![FC=\left[\begin{array}{cc}-2&0\\0&8\\2&1\end{array}\right]\left[\begin{array}{ccc}12&0&\frac{3}{2}\\1&-6&7\end{array}\right]](https://tex.z-dn.net/?f=FC%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%260%5C%5C0%268%5C%5C2%261%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D12%260%26%5Cfrac%7B3%7D%7B2%7D%5C%5C1%26-6%267%5Cend%7Barray%7D%5Cright%5D)
![FC=\left[\begin{array}{ccc}12(-2)+0(1)&0(-2)+-6(0)&\frac{3}{2}(-2)+7(0)\\12(0)+8(1)&0(0)+8(-6)&0(\frac{3}{2})+8(7)\\2(12)+1(1)&2(0)+1(-6)&2(\frac{3}{2})+1(7)\end{array}\right]](https://tex.z-dn.net/?f=FC%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D12%28-2%29%2B0%281%29%260%28-2%29%2B-6%280%29%26%5Cfrac%7B3%7D%7B2%7D%28-2%29%2B7%280%29%5C%5C12%280%29%2B8%281%29%260%280%29%2B8%28-6%29%260%28%5Cfrac%7B3%7D%7B2%7D%29%2B8%287%29%5C%5C2%2812%29%2B1%281%29%262%280%29%2B1%28-6%29%262%28%5Cfrac%7B3%7D%7B2%7D%29%2B1%287%29%5Cend%7Barray%7D%5Cright%5D)
This simplifies to;
![FC=\left[\begin{array}{ccc}-24&0&-3\\8&-48&56\\25&-6&10\end{array}\right]](https://tex.z-dn.net/?f=FC%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-24%260%26-3%5C%5C8%26-48%2656%5C%5C25%26-6%2610%5Cend%7Barray%7D%5Cright%5D)
The correct answer is C
