
<h3>Answer: d. w = -6</h3>
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).
Answer:
20°
Step-by-step explanation:
∡EFJ = 70°
∡JFK = 90°
So, ∡KFG = 180 - ∡JFK - ∡EFJ = 180 - 90 - 70 = 20°
Answer:
The values of
so that
have vertical asymptotes are
,
,
,
,
.
Step-by-step explanation:
The function cosecant is the reciprocal of the function sine and vertical asymptotes are located at values of
so that function cosecant becomes undefined, that is, when function sine is zero, whose periodicity is
. Then, the vertical asymptotes associated with function cosecant are located in the values of
of the form:
, 
In other words, the values of
so that
have vertical asymptotes are
,
,
,
,
.
Answer:
FALSE, (2, 9) is not a solution to the set of inequalities given.
Step-by-step explanation:
Simply replace x by 2 and y by 9 in the inequalities and see if the inequality is true or not:
irst inequality:

so thi inequality is verified as true since 9 is larger or equal than 8
Now the second inequality:

This is FALSE since 9 is larger than 4 (not smaller)
Therefore the answer to the question is FALSE, (2, 9) is not a solution to the set of inequalities given.