The answer you are looking for is 50h+25
Hope that this helps!
Answer:
Step-by-step explanation:
given is a system of linear equations in 3 variables as
This can be represented in matrix form as
AX=B Or
![\left[\begin{array}{ccc}-1&-4&2\\1&2&-1\\1&1&-1\end{array}\right] *\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}-10\\11\\14\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26-4%262%5C%5C1%262%26-1%5C%5C1%261%26-1%5Cend%7Barray%7D%5Cright%5D%20%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-10%5C%5C11%5C%5C14%5Cend%7Barray%7D%5Cright%5D)
So solution set
X would be 
|A|=-1(-1)+4(0)+2(-1)=--1
Cofactors of A are
-1 0 -1
-2 -1 -3
0 1 2
So inverse of A is
1 2 0
0 1 -1
1 3 -2
Solution set would be
x=12
y=-3
z=-5
Answer:
6
Step-by-step explanation:
the mean is 6 so is the median, can I have brainliest plz and thanks if yes
This revolves around exact trig values - no easy way to say this, you just need to memorise them. They are there for sin cos and tan, but I will give you the main tan ones below - note this is RADIANS (always work in them when you can, everything is better):
tan0: 0
tanpi/6: 1/sqrt(3)
tanpi/4: 1
tanpi/3: sqrt(3)
tanpi/2: undefined
Now we just need to equate -2pi/3 to something we understand. 2pi/3 is 1/3 of the way round a circle, so -2pi/3 is 1/3 of the way round the circle going backwards (anticlockwise), so on a diagram we already know it's in the third quadrant of the circle (somewhere between pi and 3pi/2 rads).
We also know it is pi/3 away from pi, so we are looking at sqrt(3) or -sqrt(3) because of those exact values.
Now we just need to work out if it's positive or negative. You can look up a graph of tan and it'll show that the graph intercepts y at (0,0) and has a period of pi rads. Therefore between pi and 3pi/2 rads, the values of tan are positive. Therefore, this gives us our answer of sqrt(3).
