Simultaneous equations can be solved using inverse matrix operation.
The complete steps of Jacob's solution are:
![\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]^{-1} \cdot \left[\begin{array}{cc}4&1\\-2&3\end{array}\right]\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14}\left[\begin{array}{cc}3&-1\\2&4\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%261%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D%5E%7B-1%7D%20%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%261%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cfrac%7B1%7D%7B14%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%26-1%5C%5C2%264%5Cend%7Barray%7D%5Cright%5D%20%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%26-22%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{cc}4&1\\-2&3\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%261%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D%20%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%26-22%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14} \left[\begin{array}{c}28&-84\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cfrac%7B1%7D%7B14%7D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D28%26-84%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}2&-6\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%26-6%5Cend%7Barray%7D%5Cright%5D)
We have:


Calculate the determinant of ![\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%261%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D)



So, the inverse matrix becomes
![A = \frac{1}{14}\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7B1%7D%7B14%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%261%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D)
Replace the first column with
to calculate the value of x
![x = \frac{1}{14}\left[\begin{array}{cc}2&1\\-22&3\end{array}\right]](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B1%7D%7B14%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%261%5C%5C-22%263%5Cend%7Barray%7D%5Cright%5D)
So, we have:




Replace the second column with
to calculate the value of y
![y = \frac{1}{14}\left[\begin{array}{cc}4&2\\-2&-22\end{array}\right]](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B1%7D%7B14%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%262%5C%5C-2%26-22%5Cend%7Barray%7D%5Cright%5D)
So, we have:




Hence, the complete process is:
![\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]^{-1} \cdot \left[\begin{array}{cc}4&1\\-2&3\end{array}\right]\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14}\left[\begin{array}{cc}3&-1\\2&4\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%261%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D%5E%7B-1%7D%20%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%261%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cfrac%7B1%7D%7B14%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%26-1%5C%5C2%264%5Cend%7Barray%7D%5Cright%5D%20%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%26-22%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{cc}4&1\\-2&3\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%261%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D%20%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%26-22%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14} \left[\begin{array}{c}28&-84\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cfrac%7B1%7D%7B14%7D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D28%26-84%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}2&-6\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%26-6%5Cend%7Barray%7D%5Cright%5D)
Read more about matrices at:
brainly.com/question/11367104
the answer is in your question lol.
its pi, which is
≈ 3.14 feet ^3
^^^ thats
ur answer
Answer:
Step-by-step explanation:
If you round it to the nearest whole number, then 17.526 becomes 18 because if the first digit after the decimal(in this case,5) is 5 or greater you round to the next whole number.
If you round it to the nearest tenth, it would result in 17.5, because if the second digit after the decimal (in this case, 2) is 5 or greater you round to the next tenth of decimal.
Hope this helps!
Answer:
3/10
Step-by-step explanation:
To find the fraction of guests that ordered chicken, we must evaluate:
1 - 2/5 - 3/10
= 1 - 4/10 - 3/10
= 1 - 7/10
= 3/10
Step-by-step answer:
assuming the true-false test have equal probabilities (each 0.5), we can use the binomial probability to calculate the sum of probabilities of getting 10, 11 or 12 questions correctly out of 12.
p=probability of success = 0.5
N=number of questions
x = number of correct answers
then
P(x) = C(N,x)(p^x)((1-p)^(N-x))
where C(N,x) = N!/(x!(N-x)!) = number of combinations of taking x objects out of N.
P(10) = C(12,10)(0.5^10)((1-0.5)^2) = 33/2048 = 0.01611
P(11) = C(12,11)(0.5^11)((1-0.5)^1) = 3/1024 = 0.00293
P(12) = C(12,12)(0.5^12)((1-0.5)^0) = 1/4096 = 0.00024
for a total probability of 79/4096 = 0.01929