Answer:
First choice.
Step-by-step explanation:
You could plug in the choices to see which would make all the 3 equations true.
Let's start with (x=2,y=-6,z=1):
2x+y-z=-3
2(2)+-6-1=-3
4-6-1=-3
-2-1=-3
-3=-3 is true so the first choice satisfies the first equation.
5x-2y+2z=24
5(2)-2(-6)+2(1)=24
10+12+2=24
24=24 is true so the first choice satisfies the second equation.
3x-z=5
3(2)-1=5
6-1=5
5=5 is true so the first choice satisfies the third equation.
We don't have to go any further since we found the solution.
---------Another way.
Multiply the first equation by 2 and add equation 1 and equation 2 together.
2(2x+y-z=-3)
4x+2y-2z=-6 is the first equation multiplied by 2.
5x-2y+2z=24
----------------------Add the equations together:
9x+0+0=18
9x=18
Divide both sides by 9:
x=18/9
x=2
Using the third equation along with x=2 we can find z.
3x-z=5 with x=2:
3(2)-z=5
6-z=5
Add z on both sides:
6=5+z
Subtract 5 on both sides:
1=z
Now using the first equation along with 2x+y-z=-3 with x=2 and z=1:
2(2)+y-1=-3
4+y-1=-3
3+y=-3
Subtract 3 on both sides:
y=-6
So the solution is (x=2,y=-6,z=1).
Step-by-step explanation:
The domain is the possible x values of a function.
Since our function is a reciprocal function, our denominator can't equal 0. Note x=0, is a vertical asymptote.
So our domain is
All Reals except 0,
or (-oo,0) U (0, oo).
Answer:
2 (real) solutions.
Step-by-step explanation:
A quadratic always has two solutions, whether they are real or complex.
Sometimes the solution is complex, involving complex numbers (2 complex), sometimes they are real and distinct (2 real), and sometimes they are real and coincident (still two real, but they are the same).
In the case of
x^2+3x = 3, or
x² + 3x -3 = 0
we apply the quadratic formula to get
x = (-3 +/- sqrt(3^2+4(1)(3))/2
to give the two solutions
{(sqrt(21)-3)/2, -(sqrt(21)+3)/2,}
both of which are real.
Neither p or z ISNT A NUMBER SO DO ONE WITH A NUMBER THAT QUESTIPON IS WRONG
EEEEEEE
Answer:
10,20,40
Step-by-step explanation:
They all have a factor of 2,10, and 1
