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Vilka [71]
3 years ago
5

What are the solutions to the equation 2x^2 - 2x - 12 = 0?

Mathematics
1 answer:
pantera1 [17]3 years ago
5 0
2(x)^2 - 2(x) - 12 = ?
  
2(x)^2 - 2(x) - 12

(x)^2 - 6 - x

Multiply ↪1 (-6) = -6

1 + -6 = -5 , right?

-3 + 2 = -1, right?

After we have pulled out the like terms we have to add/subtract!   x(x) - 3

Search for the common denominator
2(x) - 3

You can add up 4 for your terms
x + 2 (x - 3)

Make sure you solve this↪ 2 = 0

x + 2 = ?
x + 2 = 0
We have found the first; x = -2
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Since, this one has a single variable, I'll make the step easier and quicker

Solve this part ↪    x-3 = 0 
If we had the number 3 to your sides we would have found the outcome of the next x
                    
So, the second x is 3 

So, therefore your answer would have to be x =-2 ; x = 3 (most likely option C.)



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Solve for x, y, and z
zloy xaker [14]

Answer:

The solution is x=1,y=2,z=3

Step-by-step explanation:

The given system of equations is ;\

2x−3y+4z=8...(1)

3x+4y−5z=−4...(2)

4x−5y+6z=12...(3)

Make x the subject in equation (1)

x=\frac{8+3y-4z}{2}...(4)

Put equation (4) into equation (2) and (3)

3(\frac{8+3y-4z}{2})+4y-5z=-4

Multiply through by;

3(8+3y-4z)+8y-10z=-8

Expand;

24+9y-12z+8y-10z=-8

Simplify;

17y-22z=-32...(5)

Equation (4) in (3)

4(\frac{8+3y-4z}{2})-5y+6z=12

2(8+3y-4z)-5y+6z=12

16+6y-8z-5y+6z=12

y-2z=-4

y=2z-4...(6)

Put equation (6) into equation (5)

17(2z-4)-22z=-32

34z-68-22z=-32

34z-22z=-32+68

12z=36

z=3

Put z=3 into equation (6)

y=2(3)-4=2

Put y=2 and z=3 into equation 4

x=\frac{8+3(2)-4(3)}{2}=1

The solution is x=1,y=2,z=3

3 0
2 years ago
Can anyone solve this
Minchanka [31]
E because 8*8=64 so 8/49*8/49 = 64/49
8 0
3 years ago
A rectangle and a square have the same area. The width of the rectangle is 2 in less than the side of the square and the length
lina2011 [118]
The area of a rectangle is A=LW, the area of a square is A=S^2.

W=S-2 and L=2S-3

And we are told that the areas of each figure are the same.

S^2=LW, using L and W found above we have:

S^2=(2S-3)(S-2)  perform indicated multiplication on right side

S^2=2S^2-4S-3S+6  combine like terms on right side

S^2=2S^2-7S+6  subtract S^2 from both sides

S^2-7S+6=0  factor:

S^2-S-6S+6=0

S(S-1)-6(S-1)=0

(S-6)(S-1)=0, since W=S-2, and W>0, S>2 so:

S=6 is the only valid value for S.  Now we can find the dimensions of the rectangle...

W=S-2 and L=2S-3  given that S=6 in

W=4 in and L=9 in

So the width of the rectangle is 4 inches and the length of the rectangle is 9 inches.


5 0
3 years ago
Find the distance between the points A and B given below.
GalinKa [24]

Answer:

use formula

Step-by-step explanation:

distance between the points A and B=

√((x2-x1)²+(y2-y1)²)

5 0
3 years ago
Solve for y hdgkgsydkydutskgxtisit
Feliz [49]

You said  X = (5y+1) / (y-2)

Multiply each side by (y-2):

X(y-2) = (5y+1)

Eliminate parentheses:

Xy - 2X = 5y + 1

Add  2X  to each side:

Xy = 5y + 1 + 2X

Subtract  5y  from each side:

(X-5)y = 1 + 2X

Divide each side by  (X-5):

<em>y = (1+2X) / (X-5)  </em>

7 0
2 years ago
Read 2 more answers
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