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cestrela7 [59]
3 years ago
13

Solve 2x - 11 = k for x.

Mathematics
1 answer:
denis23 [38]3 years ago
3 0
2x + -11 = k

Reorder the terms:
-11 + 2x = k

Solving
-11 + 2x = k

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '11' to each side of the equation.
-11 + 11 + 2x = 11 + k

Combine like terms: -11 + 11 = 0
0 + 2x = 11 + k
2x = 11 + k

Divide each side by '2'.
x = 5.5 + 0.5k

Simplifying
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2) Line segment MK has endpoints at (2, 3) and (5, ?4). Segment M'K' is the reflection of MK over the y-axis. Which statement de
Marianna [84]

Answer:

Option C -M'K' is the same length as MK

Step-by-step explanation:

Given : Line segment MK has endpoints at (2, 3) and (5,4)

               M'K' is the reflection of MK over the y-axis

By definition of reflection: reflection of point (x,y) across the the y-axis is the point (-x,y)

which implies M'K' has end points (-2,3) and (-5,4)

Now, we find the length of MK

let (x_1,y_1)=(2,3)\\\\(x_2,y_2)=(5,4)

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

⇒ d=\sqrt{(2-5)^2+(4-3)^2}

⇒d=\sqrt{9+1}

⇒d=\sqrt{10}   ....(1)

Now, we find the length of M'K'

let (x_1^{'},y_1^{'})=(-2,3)\\\\(x_2^{'},y_2^{'})=(-5,4)

d^{'}=\sqrt{(x_2^{'}-x_1^{'})^2+(y_2^{'}-y_1^{'})^2}

⇒ d^{'}=\sqrt{(-2+5)^2+(3-4)^2}

⇒d^{'}=\sqrt{9+1}

⇒d^{'}=\sqrt{10} .....(2)

from (1) and (2) we simply show that the length of MK and M'K' is equal

we can also refer the figure attached for reflection of MK and M'K'

therefore, Option C is correct


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3 years ago
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Answer:

y = -5x + 4/3

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What is the slope of a line that is parallel to the line y = x + 2?
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