Answer:
Let the base be p
Hypotenuse = 2p +6
Perpendicular = 2p + 4
By Pythagoras theoram
(2p+6)^2 = (2p+4)^2 +p^2
=> 4p^2 +36 + 24p = 4p^2 + 16 +16p +p^2
=> 36+ 24p = p^2 + 16p + 16
=> p^2 - 8p - 20 = 0
=> p^2 - 10p +2p - 20 = 0
=> p(p-10) +2(p-10) = 0
=> (p-10)(p+2) = 0
p = 10 and - 2
Length can't be negative
So,
p = 10
Base = 10
Perpendicular = 24
Hypotenuse = 26
Answer:
![\left[\begin{array}{ccc}5&7\\5&-8\\3&-9\end{array}\right] \left[\begin{array}{ccc}c\\d\end{array}\right] =\left[\begin{array}{ccc}-16\\3\\-15\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%267%5C%5C5%26-8%5C%5C3%26-9%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dc%5C%5Cd%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-16%5C%5C3%5C%5C-15%5Cend%7Barray%7D%5Cright%5D)
This tells us that:
![A=\left[\begin{array}{ccc}5&7\\5&-8\\3&-9\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%267%5C%5C5%26-8%5C%5C3%26-9%5Cend%7Barray%7D%5Cright%5D)
![b=\left[\begin{array}{ccc}-16\\3\\-15\end{array}\right]](https://tex.z-dn.net/?f=b%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-16%5C%5C3%5C%5C-15%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
So we are saying we have scalars, c and d, such that:
![c\left[\begin{array}{ccc}5\\5\\ 3\end{array}\right]+d\left[\begin{array}{ccc}7\\-8\\-9\end{array}\right]=\left[\begin{array}{ccc}-16\\3\\-15\end{array}\right]](https://tex.z-dn.net/?f=c%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C5%5C%5C%203%5Cend%7Barray%7D%5Cright%5D%2Bd%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%5C%5C-8%5C%5C-9%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-16%5C%5C3%5C%5C-15%5Cend%7Barray%7D%5Cright%5D)
.
So we want to find a way to express this as:
Ax=b where x is the scalar vector, ![\left[\begin{array}{ccc}c\\d\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dc%5C%5Cd%5Cend%7Barray%7D%5Cright%5D)
.
So we can write this as:
Answer:
y = -5x + 20
Step-by-step explanation:
Lets start with the first given values for x and y
(2,10)
Lets try out the first equation
y = 5x + 4
y = 5(2) + 4
y = 14
This equation wouldn't work because the y value is not 10 when the x value is 2
y = 5x + 20
y = 5(2) + 20
y = 30
This equation doesn't work either
y = -5x + 20
y = -5(2) + 20
y = -10 + 20
y = 10
This would be the correct equation since your y value is 10 when the x value is 2
X=6
y=7
6+7
= 13
1/2(6) + 7
= 3 +7
=10
Answer:
13. (1) x + 8x = 32
14. $337.5
Step-by-step explanation:
14. $270 x 0.25 = 67.5
270 + 67.5 = 337.5
I am sorry. I only got these-
