Plz help when possible

Mathematics

1 answer:

BabaBlast [244]3 years ago

8 0

Answer: G

Step-by-step explanation:

The total bottle sold on monday is so i'm gonna assume that 40 people were in the cafeteria that day. The cafeteria had 80 customers, which is double the amount of customers from yesterday. If we want to make predictions from monday's data, we should double all the values in the table (so basically apple juice is now 22, cranberry is 14, etc.) so it's more fitting to the number of customers on tuesday.

apple - 22

cranberry - 14

orange - 36

pineapple - 8

G is not supported because it says that 6 times more apple juice will be sold than cranberry. So if we look at the [modified} table, cranberry is WAY over half the amount of apple juice. This means that G's statement is not supported by the table

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Jalia has $500. She spends $220 on a pair of Yeezys for Christmas. Later she realizes that

telo118 [61]

Answer:

145

Step-by-step explanation:

500-220=280

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3 0

3 years ago

Question 5 (5 points)

Tpy6a [65]

Answer:

The solution of the system of equations is, (1,-1,2)

Step-by-step explanation:

Given system equation;

x + 5y - 3z = -10

-5x + 6y – 5z = -21

-x + 8y - 8z = -25

Matrix form is written as;

$\left[\begin{array}{ccc}1&5&-3\\-5&6&-5\\-1&8&-8\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}-10\\-21\\-25\end{array}\right] \\\\\\det. = 1\left[\begin{array}{cc}\\6&-5\\8&-8\end{array}\right] -5\left[\begin{array}{cc}\\-5&-5\\-1&-8\end{array}\right] -3\left[\begin{array}{cc}\\-5&6\\-1&8\end{array}\right] \\\\\\det. = 1(-8) -5(35)-3(-34)= -8 - 175+ 102 = -81$

Cofactor;

$First \ row \left[\begin{array}{cc}+\\ 6&-5\\\ 8&-8\end{array}\right \left\begin{array}{cc}-\\ -5&-5\\-1&-8\end{array}\right \left\begin{array}{cc}+\\-5&6\\-1&8\end{array}\right] = [-8 \ \ -35 \ \ -34]\\\\\\\ Second \ row \left[\begin{array}{cc}-\\ 5&-3\\\ 8&-8\end{array}\right \left\begin{array}{cc}+\\ 1&-3\\-1&-8\end{array}\right \left\begin{array}{cc}-\\1&5\\-1&8\end{array}\right] = [16\ \ -11 \ \ -13]\\\\\\$

$Third \ row \left[\begin{array}{cc}+\\ 5&-3\\\ 6&-5\end{array}\right \left\begin{array}{cc}-\\ 1&-3\\-5&-5\end{array}\right \left\begin{array}{cc}+\\1&5\\-5&6\end{array}\right]= [-7 \ \ 20\ \ 31]$

$Cofactor =$ $\left[\begin{array}{ccc}-8&-35&-34\\16&-11&-13\\-7&20&31\end{array}\right]$

$inverse \ matrix =-\frac{1}{81} \left[\begin{array}{ccc}-8&16&-7\\-35&-11&20\\-34&-13&31\end{array}\right] \\\\\\$

Solution of the matrix:

$\left[\begin{array}{c}x\\y\\z\end{array}\right] = -\frac{1}{81} \left[\begin{array}{ccc}-8&16&-7\\-35&-11&20\\-34&-13&31\end{array}\right] X \left[\begin{array}{c}-10\\-21\\-25\end{array}\right] = \left[\begin{array}{c}\frac{-8*-10 }{-81 } +\frac{16*-21 }{-81 } + \frac{-7*-25 }{-81 }\\\\\frac{-35*-10 }{-81 } +\frac{-11*-21 }{-81 }+ \frac{20*-25 }{-81 }\\\\\frac{-34*-10 }{-81 }+ \frac{-13*-21 }{-81 }+ \frac{31*-25 }{-81 }\end{array}\right] \\\\\$

$\left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}\frac{-81}{-81} \\\\\frac{81}{-81} \\\\\frac{-162}{-81} \end{array}\right] = \left[\begin{array}{c}1\\-1\\2\end{array}\right]$