Is it asking you to tell them what the numbers that the letters substitute are??
If so o might be able to help with that
The answer would be 3n^2 + 2.
This can be found/proven by replacing "n" with term number (1,2,3,4...), then solving to get the final number. For example 3 * 1^2 + 2. You would first do 1^2, which is 1. Next, you would multiply 1 by 3, to get 3. Finally, you'd and the 2 to get 5. 5 is the 1st term, and you can use this same equation to get the rest of the terms you need.
I hope this helps!
Answer:
![\left[\begin{array}{cccc}-12&-13&13&|15\\7&-10&-3&|11\\7&14&5&\:\:\:|-5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D-12%26-13%2613%26%7C15%5C%5C7%26-10%26-3%26%7C11%5C%5C7%2614%265%26%5C%3A%5C%3A%5C%3A%7C-5%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The system of equations is;
-12x-13y +13z =15
7x-10y-3z = 11
7x+14y +5z = -5
The coefficient matrix is ![\left[\begin{array}{ccc}-12&-13&13\\7&-10&-3\\7&14&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-12%26-13%2613%5C%5C7%26-10%26-3%5C%5C7%2614%265%5Cend%7Barray%7D%5Cright%5D)
The constant matrix is ![\left[\begin{array}{c}15\\11\\-5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D15%5C%5C11%5C%5C-5%5Cend%7Barray%7D%5Cright%5D)
The augmented matrix is
Make a substitution:
Then the system becomes
![\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bu%7D%7D%7Bu-v%7D%2B%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bu%7D%7D%7Bu%2Bv%7D%3D%5Cdfrac%7B81%7D%7B182%7D%5C%5C%5C%5C%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bv%7D%7D%7Bu-v%7D-%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bv%7D%7D%7Bu%2Bv%7D%3D%5Cdfrac1%7B182%7D%5Cend%7Bcases%7D)
Simplifying the equations gives
![\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bu%5E4%7D%7D%7Bu%5E2-v%5E2%7D%3D%5Cdfrac%7B81%7D%7B182%7D%5C%5C%5C%5C%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bv%5E4%7D%7D%7Bu%5E2-v%5E2%7D%3D%5Cdfrac1%7B182%7D%5Cend%7Bcases%7D)
which is to say,
![\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bu%5E4%7D%7D%7Bu%5E2-v%5E2%7D%3D%5Cdfrac%7B81%5Ctimes4%5Csqrt%5B3%5D%7Bv%5E4%7D%7D%7Bu%5E2-v%5E2%7D)
![\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81](https://tex.z-dn.net/?f=%5Cimplies%5Csqrt%5B3%5D%7B%5Cleft%28%5Cdfrac%20uv%5Cright%29%5E4%7D%3D81)
Substituting this into the new system gives
![\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1](https://tex.z-dn.net/?f=%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bv%5E4%7D%7D%7B%28%5Cpm27v%29%5E2-v%5E2%7D%3D%5Cdfrac1%7B182%7D%5Cimplies%5Cdfrac1%7Bv%5E2%7D%3D1%5Cimplies%20v%3D%5Cpm1)
Then
(meaning two solutions are (7, 13) and (-7, -13))
Answer:
Step-by-step explanation:
Solving for x means you have to factor. First factor out the GCF of 2 to get:
and now we'll factor using the regular old method of ac and then factoring by grouping. In our polynomial, a = 3, b = 1, c = -6. Therefore, a times c is 3 * -6 which is -18. We need some combinations of the factors of 18 that will add to give us 1, the b term in the middle. The factors of 18 are:
1, 18
2, 9
3, 6 and that's it. Hm...it seems that won't work, so let's throw this into the quadratic formula, going back to the original and a = 6, b = 2 and c = -12:
and
and
and
and
which finally simplifies to
No wonder that didn't factor using the traditional method of factoring! We could have found that out by finding first the value of the discriminant, but oh well!
