Help with all of these problems

Sergio [31]

$4w-2(1-w)=-38\qquad|\text{use distributive property}\\\\4w+(-2)(1)+(-2)(-w)=-38\\\\4w-2+2w=-38\qquad|\text{add 2 to both sides}\\\\4w+2w=-36\\\\6w=-36\qquad|\text{divide both sides by 6}\\\\\boxed{w=-6}$

<h3>Answer: d. w = -6</h3>

4 0

3 years ago

What is the solution to this system of equations?

Ede4ka [16]

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).

8 0

3 years ago

I NEED HELP PLEASE 15 POINTS

ra1l [238]

Answer:

20°

Step-by-step explanation:

∡EFJ = 70°

∡JFK = 90°

So, ∡KFG = 180 - ∡JFK - ∡EFJ = 180 - 90 - 70 = 20°

8 0

2 years ago

Pre cacl pleaseeeee help. the options are in the attachments as well

ivanzaharov [21]

Answer:

The values of $x$ so that $y = \csc x$ have vertical asymptotes are $-2\pi$ , $-\pi$ , $0$ , $\pi$ , $2\pi$ .

Step-by-step explanation:

The function cosecant is the reciprocal of the function sine and vertical asymptotes are located at values of $x$ so that function cosecant becomes undefined, that is, when function sine is zero, whose periodicity is $\pi$ . Then, the vertical asymptotes associated with function cosecant are located in the values of $x$ of the form:

$x = 0\pm \pi\cdot i$ , $\forall \,i\in \mathbb{N}_{O}$