V = LWH
<span>V = 50 * 1 * 1</span>
Answer: See below.
Step-by-step explanation:
The formatting is likely incorrect for these two options:
y=(x−1)2, and
y = ( x − − 1 ) 2
I'll assume they were meant to be:
y=(x−1)^2, and
y = ( x + 1 )^2
In this case they would be non-linear, since y depends on the value of 
Use a "^" sign to indicate raised to a power: x^2 means
.
y=(x−1)2 means y = 2x - 2
y=(x−1)^2 means y = 
B: isosceles triangle
Because there are two equal sides
A rhombus is a parallelogram with equal sides

firat we need to get two equation with two varibles let us work on x,y
so adding the first and last one will yield

now we since we used the first and third we need to use the second to get a correct system.let us multiply the third by 2 then add the second and third

now we have two equation with the variables x and y

you can solve it algebraically but you can see that the only solution possible is y=0 and x=-1 we have the values for x and y let us choose one of the three main equation and substitute to get z let us pick the first equation 5x-2y+z=-1-->5(-1)-2(0)+z=-1---->-5+z=-1-------->z=4
to make sure the system works let us check by substituting into the three equations
the first one will be 5x-2y+z=-1--->5(-1)-2(0)+4=-1---->-5+4=-1--->-1=-1 first equation holds
the second equation 3x+y+2z=6---->2(-1)+0+2(4)=6--->-2+8=-6--->-6=-6 second equation holds
the third equation x-3y-z=-5----->-1-3(0)-4=-5---->-1-4=-5--->-5=-5
our third equation also holds which makes our solution correct
x=-1,y=0,z=4