Answer:
I would use a proportion for each ingredient:
Cup sugar
4:10=½:x (10×½)÷4= 1,25 cup sugar
Cup brown sugar:
4:10=¼:x (10×¼)÷4= 0.625 cup brown sugar
Peanut butter
4:10=⅔:x (10×⅔)÷4= 1,65 cup peanut butter
Cup oats
2×¼ = ½
4:10=½:x (10×½)÷4= 1,25 cup oats
Cup milk
4:10=¾:x (10×¾)÷4= 1,875 cup milk
Tsp salt
4:10=1:x (10×1)÷4= 2,5 tsp salt
2 tbsp cocoa
4:10=2:x (10×2)÷4= 5 tbsp cocoa
1 tsp vanilla
4:10=1:x (10×1)÷4= 2,5 tsp vanilla
2 eggs
4:10=2:x (10×2)÷4= 5 eggs
3 cups flour
4:10=3:x (10×3)÷4= 7,5 cups flour
12 oz nuts
4:10=12:x (10×12)÷4= 30 oz nuts
Answer:
x=4-√79, x=4+√79
Step-by-step explanation:
The exponents in the prime factorization are, 1, 2, and 1.
If the coefficient matrix has a pivot in each column, it means that it is shaped like this:
![A=\left[\begin{array}{cccc}a_{1,1}&a_{1,2}&a_{1,3}&a_{1,4}\\0&a_{2,2}&a_{2,3}&a_{2,4}\\0&0&a_{3,3}&a_{3,4}\\0&0&0&a_{4,4}\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7Da_%7B1%2C1%7D%26a_%7B1%2C2%7D%26a_%7B1%2C3%7D%26a_%7B1%2C4%7D%5C%5C0%26a_%7B2%2C2%7D%26a_%7B2%2C3%7D%26a_%7B2%2C4%7D%5C%5C0%260%26a_%7B3%2C3%7D%26a_%7B3%2C4%7D%5C%5C0%260%260%26a_%7B4%2C4%7D%5Cend%7Barray%7D%5Cright%5D)
So, the correspondant system
will look like this:
![\left[\begin{array}{cccc}a_{1,1}&a_{1,2}&a_{1,3}&a_{1,4}\\0&a_{2,2}&a_{2,3}&a_{2,4}\\0&0&a_{3,3}&a_{3,4}\\0&0&0&a_{4,4}\end{array}\right]\cdot \left[\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right] = \left[\begin{array}{c}b_1\\b_2\\b_3\\b_4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7Da_%7B1%2C1%7D%26a_%7B1%2C2%7D%26a_%7B1%2C3%7D%26a_%7B1%2C4%7D%5C%5C0%26a_%7B2%2C2%7D%26a_%7B2%2C3%7D%26a_%7B2%2C4%7D%5C%5C0%260%26a_%7B3%2C3%7D%26a_%7B3%2C4%7D%5C%5C0%260%260%26a_%7B4%2C4%7D%5Cend%7Barray%7D%5Cright%5D%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx_1%5C%5Cx_2%5C%5Cx_3%5C%5Cx_4%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Db_1%5C%5Cb_2%5C%5Cb_3%5C%5Cb_4%5Cend%7Barray%7D%5Cright%5D)
This turn into the following system of equations:
The last equation is solvable for 
: we easily have
Once the value for is known, we can solve the third equation for 
:
(recall that is now known)
The pattern should be clear: you can use the last equation to solve for . Once it is known, the third equation involves the only variable . Once
Answer:
Hi, I believe this is reflection then rotation. I haven't studied this in some time but from what I remember I think that would be the right answer. I hope I helped, have a nice day :)
