Which statements are true about reflections? Check all that apply.

Please help I will mark brainlist if correct

love history [14]

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Equation
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$y - 9 = - \dfrac{1}{2} (x + 2)$

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Rearrange the equation
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$y - 9 = - \dfrac{1}{2} (x + 2) \\ \\
 y - 9 = - \dfrac{1}{2}x - 1 \\ \\
 y - 9 + 9 = - \dfrac{1}{2}x - 1 + 9 \ \Longleftarrow \text{add 9 to both side} \\ \\
 y = -\dfrac{1}{2}x + 8$

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slope of the perpendicular line
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slope = 2

(*the slope of the perpendicular line is the negative reciprocal of the slope of line)

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Substitute slope into equation y = mx + c
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y = 2x + c

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Find y-intercept
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y = 2x + c
At point (0, 4)
4 = 2(0) + c
c = 4

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Substitute y-intercept into the equation y = 2x + c
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y = 2x + 4

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Answer: y = 2x + 4
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3 0

3 years ago

Which is the solution of x=2+√x-2? x=2 x=3 x=2 or x=3 no solution

erik [133]

No Solution
x=2+ Square root of x-2
-2
x-2= Square root of x-2
2s cancel out.
x = Square root of x
Or you can just graph it and see no solutions

6 0

3 years ago

I JUST NEED HELP PLEASE!! AHH!!~∆~

mafiozo [28]

Answer:

Step-by-step explanation:

take theta as reference angle

using sin rule

sin theta=opposite/hypotenuse

sin theta=5/13

sin theta=0.38

theta=sin 23

therefore the value of theta is 23 degree

8 0

2 years ago

What is the probability of flipping 4 coins and having exactly three of them land on tails?

Elodia [21]

3/4 coins are heads:
THHH
HTHH
HHTH
HHHT

4 chances

2 options of heads/tails with 4 coins

2*2*2*2= 16

4/16 :)

4 0

3 years ago

Now imagine that instead of walking along the path 1→2→3→4→1, ann walks 80 meters on a straight line 33∘ north of east starting

stealth61 [152]

Answer:

Anna's walk as a vector representation is $80\cos 33^{\circ}\hat{i}+80 \sin33^{\circ}\hat{j}$ and refer attachment.

Step-by-step explanation:

Let the origin be the point 1 from where Ann start walking.

Ann walks 80 meters on a straight line 33° north of the east starting at point 1 as shown in figure below,

Resolving into the vectors, the vertical component will be 80Sin33° and Horizontal component will be 80Cos33° as shown in figure (2)

Ann walk as a vector representation is $80\cos 33^{\circ}\hat{i}+80 \sin33^{\circ}\hat{j}$

Thus, Anna's walk as a vector representation is $80\cos 33^{\circ}\hat{i}+80 \sin33^{\circ}\hat{j}$

7 0

3 years ago

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