7:14 or simplified 1:2 (use simplified)
Find total amount of fruit by adding
4+7+3=14
7 pears to 14 total fruits in ratio form is 7:14 which simplifies to 1:2
Answer:
![A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]](https://tex.z-dn.net/?f=A%5E%7B-1%7D%3D%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B1%7D%7B9%7D%20%26%20%5Cfrac%7B4%7D%7B27%7D%20%26%20-%20%5Cfrac%7B2%7D%7B27%7D%20%5C%5C%5C%5C%20%5Cfrac%7B8%7D%7B9%7D%20%26%20%5Cfrac%7B5%7D%7B27%7D%20%26%20%5Cfrac%7B11%7D%7B27%7D%20%5C%5C%5C%5C%20-%20%5Cfrac%7B4%7D%7B9%7D%20%26%20%5Cfrac%7B2%7D%7B27%7D%20%26%20-%20%5Cfrac%7B1%7D%7B27%7D%20%5Cend%7Barray%7D%20%5Cright%5D)
Step-by-step explanation:
We want to find the inverse of ![A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7D%201%20%26%200%20%26%20-2%20%5C%5C%5C%5C%204%20%26%201%20%26%203%20%5C%5C%5C%5C%20-4%20%26%202%20%26%203%20%5Cend%7Barray%7D%20%5Cright%5D)
To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.
So, augment the matrix with identity matrix:
![\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 4&1&3&0&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%26-2%261%260%260%20%5C%5C%5C%5C%204%261%263%260%261%260%20%5C%5C%5C%5C%20-4%262%263%260%260%261%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 1 multiplied by 4 from row 2
![\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%26-2%261%260%260%20%5C%5C%5C%5C%200%261%2611%26-4%261%260%20%5C%5C%5C%5C%20-4%262%263%260%260%261%5Cend%7Barray%7D%5Cright%5D)
- Add row 1 multiplied by 4 to row 3
![\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%26-2%261%260%260%20%5C%5C%5C%5C%200%261%2611%26-4%261%260%20%5C%5C%5C%5C%200%262%26-5%264%260%261%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 2 multiplied by 2 from row 3
![\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%26-2%261%260%260%20%5C%5C%5C%5C%200%261%2611%26-4%261%260%20%5C%5C%5C%5C%200%260%26-27%2612%26-2%261%5Cend%7Barray%7D%5Cright%5D)
![\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%26-2%261%260%260%20%5C%5C%5C%5C%200%261%2611%26-4%261%260%20%5C%5C%5C%5C%200%260%261%26-%20%5Cfrac%7B4%7D%7B9%7D%26%5Cfrac%7B2%7D%7B27%7D%26-%20%5Cfrac%7B1%7D%7B27%7D%5Cend%7Barray%7D%5Cright%5D)
- Add row 3 multiplied by 2 to row 1
![\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%260%26%5Cfrac%7B1%7D%7B9%7D%26%5Cfrac%7B4%7D%7B27%7D%26-%20%5Cfrac%7B2%7D%7B27%7D%20%5C%5C%5C%5C%200%261%2611%26-4%261%260%20%5C%5C%5C%5C%200%260%261%26-%20%5Cfrac%7B4%7D%7B9%7D%26%5Cfrac%7B2%7D%7B27%7D%26-%20%5Cfrac%7B1%7D%7B27%7D%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 3 multiplied by 11 from row 2
![\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%260%26%5Cfrac%7B1%7D%7B9%7D%26%5Cfrac%7B4%7D%7B27%7D%26-%20%5Cfrac%7B2%7D%7B27%7D%20%5C%5C%5C%5C%200%261%260%26%5Cfrac%7B8%7D%7B9%7D%26%5Cfrac%7B5%7D%7B27%7D%26%5Cfrac%7B11%7D%7B27%7D%20%5C%5C%5C%5C%200%260%261%26-%20%5Cfrac%7B4%7D%7B9%7D%26%5Cfrac%7B2%7D%7B27%7D%26-%20%5Cfrac%7B1%7D%7B27%7D%5Cend%7Barray%7D%5Cright%5D)
As can be seen, we have obtained the identity matrix to the left. So, we are done.
Answer:
The answer is C.
Step-by-step explanation:
Hit 'em with the Law of Sines.
sin(A)/a = sin(B)/b.
Let's say x is equal to "A", thus 5 is "a".
sin(x)/5 = sin(B)/b.
We could go for the obvious choice for "B", which would be the 90 degrees shown. To solve for the hypotenuse which will be "b", let's use the Pythagorean Theorem:
a^2 + b^2 = c^2
5^2 + 20^2 = c^2
25 + 400 = 425
sqrt(425) = about 20.6, which we can now substitute "b" with.
sin(x)/5 = sin(90)/20.6
sin(x)/5 = 1/20.6
sin(x)/5 = 0.04854...
sin(x) = 0.2427...
You can plug in sin^-1(0.2427) into your calculator, and you should end up with something like 14.047... which equates to answer choice C.
Answer:
<h3>
second term minus first term </h3>
Step-by-step explanation:
Difference is a result of subtracting (not dividing!)
To calculate difference between two terms we have to subtract the previous from the next:
The Prove that two non-zero vectors are collinear if and only if one vector is a scalar multiple of the other is given below.
<h3>What are the proves?</h3>
1. To know collinear vectors:
∧ ⁻a ║ ⁻a
If ⁻b = ∧ ⁻a
then |⁻b| = |∧ ⁻a|
So one can say that line ⁻b and ⁻a are collinear.
2. If ⁻a and ⁻b are collinear
Assuming |b| length is 'μ' times of |⁻a |
Then | 'μ' ⁻a| = | 'μ' ⁻a|
So ⁻b = 'μ' ⁻a
Learn more about vectors from
brainly.com/question/25705666
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