Answer:
Part A:
Gwen has already burned 100 calories when Tristan and Keith started, and he continues to burn 10 calories per minute. The equation for Gwen is
y = 100 + 10x
Part B:
Tristan is burning 125 calories every 10 minutes. His burning calories rate is 12.5 calories per minute. The equation for Tristan is
y = 12.5x
Part C:
Keith is burning 300 calories in 30 minutes. His rate per minute is 10 calories per minute. The equation for Keith is
y = 10x
Part D:
Just follow the instructions and do the thing for the app. You can then answer the questions it is very easy.
Part E:
Do the same for E after you do Part D.
Step-by-step explanation:
Hope it helps!
<h2>
The required solution is x = 6 and y = 11 </h2>
Step-by-step explanation:
Given system of equations are
x+5y = 11 and x-y =5
![X=\left[\begin{array}{c}x\\y\end{array}\right]](https://tex.z-dn.net/?f=X%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D)
and ![B= \left[\begin{array}{c}11\\5\end{array}\right]](https://tex.z-dn.net/?f=B%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D11%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
∴AX=B
![adj A = \left[\begin{array}{cc}{-1}&{-5}\\{-1}&1\end{array}\right]](https://tex.z-dn.net/?f=adj%20A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%7B-1%7D%26%7B-5%7D%5C%5C%7B-1%7D%261%5Cend%7Barray%7D%5Cright%5D)

∴
So,![A^{-1} =\frac{ \left[\begin{array}{cc}{-1}&{-5}\\{-1}&1\end{array}\right]}{-6}](https://tex.z-dn.net/?f=A%5E%7B-1%7D%20%3D%5Cfrac%7B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%7B-1%7D%26%7B-5%7D%5C%5C%7B-1%7D%261%5Cend%7Barray%7D%5Cright%5D%7D%7B-6%7D)
![A^{-1} ={ \left[\begin{array}{c \c} {{\frac{1}{6} }}&{\frac{5}{6}}\ \\ {{\frac{1}{6} }}&{\frac{-1}{6}} \end{array}\right]}](https://tex.z-dn.net/?f=A%5E%7B-1%7D%20%3D%7B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%20%5Cc%7D%20%20%7B%7B%5Cfrac%7B1%7D%7B6%7D%20%7D%7D%26%7B%5Cfrac%7B5%7D%7B6%7D%7D%5C%20%5C%5C%20%20%7B%7B%5Cfrac%7B1%7D%7B6%7D%20%7D%7D%26%7B%5Cfrac%7B-1%7D%7B6%7D%7D%20%5Cend%7Barray%7D%5Cright%5D%7D)

⇒![\left[\begin{array}{c}x\\y\end{array}\right] ={ \left[\begin{array}{c \c} {{\frac{1}{6} }}&{\frac{5}{6}}\ \\ {{\frac{1}{6} }}&{\frac{-1}{6}} \end{array}\right]} \times \left[\begin{array}{c}11\\5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%7B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%20%5Cc%7D%20%20%7B%7B%5Cfrac%7B1%7D%7B6%7D%20%7D%7D%26%7B%5Cfrac%7B5%7D%7B6%7D%7D%5C%20%5C%5C%20%20%7B%7B%5Cfrac%7B1%7D%7B6%7D%20%7D%7D%26%7B%5Cfrac%7B-1%7D%7B6%7D%7D%20%5Cend%7Barray%7D%5Cright%5D%7D%20%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D11%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
⇒![\left[\begin{array}{c}x\\y\end{array}\right] ={ \left[\begin{array}{c} {6}\\ {11} \end{array}\right]}](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%7B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D%20%20%7B6%7D%5C%5C%20%20%7B11%7D%20%5Cend%7Barray%7D%5Cright%5D%7D)
∴ x= 6 and y = 11
The required solution is x = 6 and y = 11
1/6^4= 1/1296
I think this is the answer
Answer:
A,B,C
Step-by-step explanation:
in these quadrilaterals, the diagonals bisect each other paralellogram, rectangle, rhombus, square
in these quadrilaterals, the diagonals are congruent rectangle, square, isosceles trapezoid
in these quadrilaterals, each of the diagonals bisects a pair of opposite angles rhombus, square
The answer would be 2.5.
hope this helps (;