let's multiply both sides in each equation by the LCD of all fractions in it, thus doing away with the denominator.
![\begin{cases} \cfrac{1}{2}x+\cfrac{1}{3}y&=7\\\\ \cfrac{1}{4}x+\cfrac{2}{3}y&=6 \end{cases}\implies \begin{cases} \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{6}}{6\left( \cfrac{1}{2}x+\cfrac{1}{3}y \right)=6(7)}\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{12}}{12\left( \cfrac{1}{4}x+\cfrac{2}{3}y\right)=12(6)} \end{cases}\implies \begin{cases} 3x+2y=42\\ 3x+8y=72 \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%20%5Ccfrac%7B1%7D%7B2%7Dx%2B%5Ccfrac%7B1%7D%7B3%7Dy%26%3D7%5C%5C%5C%5C%20%5Ccfrac%7B1%7D%7B4%7Dx%2B%5Ccfrac%7B2%7D%7B3%7Dy%26%3D6%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Bcases%7D%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B6%7D%7D%7B6%5Cleft%28%20%5Ccfrac%7B1%7D%7B2%7Dx%2B%5Ccfrac%7B1%7D%7B3%7Dy%20%5Cright%29%3D6%287%29%7D%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B12%7D%7D%7B12%5Cleft%28%20%5Ccfrac%7B1%7D%7B4%7Dx%2B%5Ccfrac%7B2%7D%7B3%7Dy%5Cright%29%3D12%286%29%7D%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Bcases%7D%203x%2B2y%3D42%5C%5C%203x%2B8y%3D72%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \stackrel{\textit{using elimination}}{ \begin{array}{llll} 3x+2y=42&\times -1\implies &\begin{matrix} -3x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~-2y=&-42\\ 3x+8y-72 &&~~\begin{matrix} 3x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~+8y=&72\\ \cline{3-4}\\ &&~\hfill 6y=&30 \end{array}} \\\\\\ y=\cfrac{30}{6}\implies \blacktriangleright y=5 \blacktriangleleft \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Busing%20elimination%7D%7D%7B%20%5Cbegin%7Barray%7D%7Bllll%7D%203x%2B2y%3D42%26%5Ctimes%20-1%5Cimplies%20%26%5Cbegin%7Bmatrix%7D%20-3x%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~-2y%3D%26-42%5C%5C%203x%2B8y-72%20%26%26~~%5Cbegin%7Bmatrix%7D%203x%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%2B8y%3D%2672%5C%5C%20%5Ccline%7B3-4%7D%5C%5C%20%26%26~%5Chfill%206y%3D%2630%20%5Cend%7Barray%7D%7D%20%5C%5C%5C%5C%5C%5C%20y%3D%5Ccfrac%7B30%7D%7B6%7D%5Cimplies%20%5Cblacktriangleright%20y%3D5%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \stackrel{\textit{substituting \underline{y} on the 1st equation}~\hfill }{3x+2(5)=42\implies 3x+10=42}\implies 3x=32 \\\\\\ x=\cfrac{32}{3}\implies \blacktriangleright x=10\frac{2}{3} \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \left(10\frac{2}{3}~~,~~5 \right)~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bsubstituting%20%5Cunderline%7By%7D%20on%20the%201st%20equation%7D~%5Chfill%20%7D%7B3x%2B2%285%29%3D42%5Cimplies%203x%2B10%3D42%7D%5Cimplies%203x%3D32%20%5C%5C%5C%5C%5C%5C%20x%3D%5Ccfrac%7B32%7D%7B3%7D%5Cimplies%20%5Cblacktriangleright%20x%3D10%5Cfrac%7B2%7D%7B3%7D%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%5Cleft%2810%5Cfrac%7B2%7D%7B3%7D~~%2C~~5%20%5Cright%29~%5Chfill)
Answer:
5.3 hours
Step-by-step explanation:
Given:
3 hours = 8lbs. turkey
?? hours = 14lbs. turkey
Do proportions:
=
<-- Cross Multiply
42 = 8x <-- Solve for x
x = 5.25
Round to the nearest tenth:
x = 5.3 hours
Answer:
4
Step-by-step explanation:
The question is not clear. You have indicated the original function as 12sin(0) - 9sin²(0)
If so, the solution is trivial. At 0, sin(0) is 0 so the solution is 0
However, I will assume you meant the angle to be
rather than 0 which makes sense and proceed accordingly
We can find the maximum or minimum of any function by finding the first derivate and setting it equal to 0
The original function is

Taking the first derivative of this with respect to
and setting it equal to 0 lets us solve for the maximum (or minimum) value
The first derivative of
w.r.t
is

And setting this = 0 gives

Eliminating
on both sides and solving for
gives us
Plugging this value of
into the original equation gives us

This is the maximum value that the function can acquire. The attached graph shows this as correct
Answer:
Hope this helps! If you have any questions feel free to ask.
Step-by-step explanation: