Idk seriously i suck at math but i use photomath it is a really good app u should try it
The problem is translated into
14=_÷56
56÷14=4
100÷4=2
So the answer is 25%
Eliminate
.
![u + v = (3x - 4y) + (x + 4y) = 4x \implies x = \dfrac{u+v}4](https://tex.z-dn.net/?f=u%20%2B%20v%20%3D%20%283x%20-%204y%29%20%2B%20%28x%20%2B%204y%29%20%3D%204x%20%5Cimplies%20x%20%3D%20%5Cdfrac%7Bu%2Bv%7D4)
Eliminate
.
![u - 3v = (3x - 4y) - 3 (x + 4y) = -16y \implies y = \dfrac{3v-u}{16}](https://tex.z-dn.net/?f=u%20-%203v%20%3D%20%283x%20-%204y%29%20-%203%20%28x%20%2B%204y%29%20%3D%20-16y%20%5Cimplies%20y%20%3D%20%5Cdfrac%7B3v-u%7D%7B16%7D)
The Jacobian for this change of coordinates is
![J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} \dfrac14 & \dfrac14 \\\\ -\dfrac1{16} & \dfrac3{16} \end{bmatrix}](https://tex.z-dn.net/?f=J%20%3D%20%5Cbegin%7Bbmatrix%7D%20x_u%20%26%20x_v%20%5C%5C%20y_u%20%26%20y_v%20%5Cend%7Bbmatrix%7D%20%3D%20%5Cbegin%7Bbmatrix%7D%20%5Cdfrac14%20%26%20%5Cdfrac14%20%5C%5C%5C%5C%20-%5Cdfrac1%7B16%7D%20%26%20%5Cdfrac3%7B16%7D%20%5Cend%7Bbmatrix%7D)
with determinant
![\det(J) = \dfrac14\cdot\dfrac3{16} - \dfrac14\cdot\left(-\dfrac1{16}\right) = \dfrac1{16}](https://tex.z-dn.net/?f=%5Cdet%28J%29%20%3D%20%5Cdfrac14%5Ccdot%5Cdfrac3%7B16%7D%20-%20%5Cdfrac14%5Ccdot%5Cleft%28-%5Cdfrac1%7B16%7D%5Cright%29%20%3D%20%5Cdfrac1%7B16%7D)
The absolute value of a number is the distance the number is from zero on the number line. Only one number is 0 units from 0. It is zero. For all other distances, there are always two numbers. Example: The two numbers a distance of 1 unit from zero are -1 and 1. Example: The two numbers a distance of 4.5 units from zero are -4.5 and 4.5.
In your case, the two numbers a distance of 1.75 units from zero are -1.75 and 1.75. Find -1.75 between -1.5 and -2 and place a dot there. Find 1.75 between 1.5 and 2 and place a dot there.