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Alex777 [14]
3 years ago
9

JKL has vertices J(3, -5) K(-8, -1) and L(5,1). JKL will be reflected across the line y= -1 to form J'K'L'. Which quadrant will

NOT contain a vertex of J'K'L'
Mathematics
1 answer:
katen-ka-za [31]3 years ago
4 0

Answer:

The correct answer is second quadrant.

Step-by-step explanation:

Vertices of J ≡ (3 , -5) which shows J is in fourth quadrant. When J is reflected across the line y = -1, the point J moves to first quadrant as J' with coordinates (3 , 3).

Vertices of K ≡ (-8 , -1) which shows K is in third quadrant. When K is reflected across the line y = -1, the point K remains where it is (in the third quadrant) as it is on the line y = -1 only.

Vertices of L ≡ (5 , 1) which shows L is in first quadrant. When L is reflected across the line y = -1, the point L moves to fourth quadrant as L' with coordinates (5 , -3).

Thus each of the first, fourth and third quadrant contains a vertex except the second quadrant.

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Solve for x in the equation.
oksano4ka [1.4K]
x^2 + 20x + 100 = 36

(x + 10)^2 = 36

x + 10 = \pm \sqrt{36}

x + 10 = 6   or   x + 10 = -6

x = -4   or   = -16
5 0
2 years ago
Negative 5X +3 equals 18
andrey2020 [161]
-5x +3 =18
Subtract 3 from each side
-5x= 15
Divide by -5 
x= -3 
6 0
3 years ago
Pls pls pls answer! I will mark brainliest!!!! tysm<br> xy = 72<br> (x + 2)(y - 4) = 128
zheka24 [161]

Answer:

x=2 and y=36

Step-by-step explanation:

you have two equations

1) xy=72 and 2) (x+2)(y-4)=128

solve for one of the variables in equation 2), lets do y

(x+2)(y-4)=128, divide by (x+2) to both sides and add 4 to isolate variable y

therefore, y=(128/(x+2))+4

then plug in y into equation 1)

so x(our new y)=72

x( 128/(x+2) +4 )=72

distribute the x so you will get

(128/(x+2))x +4x=72

we are now solving for x, so isolate the x by subtracting 4x and multiplying by (x+2) to get

128x=(x+2)(72-4x)

multiply by distribution and combine like terms

-4x^2 -64x +144 =0

solve for x by quadratic formula, you'll get x=2 or x=-18

substitute back into equation 1), (2)(y)=72, solve for y to get 36

then you can check by substituting back into equation 2)

(2+2)(36-4)=128

(4)(32)=128, 128=128

6 0
3 years ago
The perimeter of a rectangular house is 122fr the width is 5ft less than the length find the length and width
Step2247 [10]
Hello,

Let's w the width , l the length, p the perimeter.

w=l-5
p=2(l+w)=2(l+l-5)=4l-10

4l-10=122
==>4l=122+10
==>4l=132
==>l=132/4
==>l=33
So w=33-5=28

Proof:
2*(33+28)=2*61=122

3 0
2 years ago
Read 2 more answers
Question 2:
yarga [219]

Answer:

Step-by-step explanation:

f(x)= x^2- 6x + 8 = x² -4x - 2x + 4*2 = x(x-4) - 2(x-4) = (x-4)(x-2)

Real Zeros of this function are 2 and 4

D) g(x) = 2x^2– 12x + 16 = 2 (x^2- 6x + 8)

6 0
2 years ago
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