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Sauron [17]
3 years ago
11

Factor the expression 12x^2 + 14x - 6. You must show your work.

Mathematics
1 answer:
KatRina [158]3 years ago
6 0

2(3x - 1)(2x + 3) Correct answer

2(6x^2 + 7x - 3)

2(6x^2 + 9x - 2x - 3)

2(3x •(2x + 3)- 2x - 3)

2(2x + 3)•(3x - 1)

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Triangle JKL is transformed to create triangle J'K'L'. The angles in both triangles are shown.
Viefleur [7K]

The correct statement is AngleL = 25° AngleL' = 25°.

We have given that,

Triangle JKL is transformed to create triangle J'K'L'.

We have to choose the correct option.

<h3>What is the information about the transformed triangle?</h3>

The following information should be considered:

In a rigid transformation, the image & pre-image are congruent.

Reflection, translation, and rotation are rigid transformations.

In a non-rigid transformation, the image and pre-image are similar.

Dilation is a non rigid transformation.

In a rigid or nonrigid transformation, the corresponding angles are the same.

If the corresponding sides are the same, then it is a rigid transformation.

If the corresponding sides are proportional, then it is a nonrigid transformation.

It can be a rigid or a nonrigid transformation based on whether the corresponding side lengths have the same measures.

Therefore option 3 is correct.

To learn more transformation of the triangle visit:

brainly.com/question/17429689

#SPJ1

5 0
1 year ago
Find a parametrization of the line in which the planes x + y + z = -6 and y + z = -8 intersect.
rewona [7]

Answer:

L(x,y) = (2,-8,0) + (0,-1,1)*t

Step-by-step explanation:

for the planes

x + y + z = -6  and y + z = -8

the intersection can be found subtracting the equation of the planes

x + y + z - ( y + z ) = -6 - (-8)

x= 2

therefore

x=2

z=z

y= -8 - z

using z as parameter t and the point (2,-8,0) as reference point , then

x= 2

y= -8 - t

z= 0 + t

another way of writing it is

L(x,y) = (2,-8,0) + (0,-1,1)*t

6 0
3 years ago
Is there any tutors orrr
Papessa [141]
There’s tutors for math i just asked a tutor for help lol
4 0
2 years ago
Read 2 more answers
Sebastian solved the radical equation y + 1 = but did not check his solution. (y + 1)2 = y2 + 2y + 1 = –2y – 3 y2 + 4y + 4 = 0 (
Softa [21]

Answer:

There are no true solutions to the equation.

Step-by-step explanation:

<u><em>The correct equation is</em></u>

y+1=\sqrt{-2y-3}

Solve for y

squared both sides

(y+1)^2=(-2y-3)

(y^2+2y+1)=(-2y-3)

y^2+2y+1+2y+3=0

y^2+4y+4=0

(y+2)(y+2)=0

y=-2

<em>Verify</em>

substitute the value of y in the original expression

-2+1=\sqrt{-2(-2)-3}

-1=1 ----> is not true

therefore

There are no true solutions to the equation.

4 0
3 years ago
Simplify the expression 18x4y3
Fed [463]
18*4y^3\\\&#10;Just \ Multiply\ The \ Coefficients.&#10;\\\ 72y^3
3 0
3 years ago
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