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Nastasia [14]
3 years ago
12

Evaluate 12x - 1 when x = 4

Mathematics
2 answers:
Slav-nsk [51]3 years ago
6 0
12(4)-1

48-1

47 is the answer
liberstina [14]3 years ago
3 0

12x - 1 when x = 4 is

12(4) - 1 = 47

Hope this helps

-AaronWiseIsBae

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Billy is building a frame for his father's Garden in the shape of an isosceles trapezoid. The length of the parallel sides are 1
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C

Step-by-step explanation:

Hey There!

So if you didn't know in an isosceles trapezoid the non parallel sides are congruent

So given the base lengths and the perimeter we subtract the given length from the perimeter and divide that by 2

40-10=40

30-14=16

16/2=8

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Please help! I don't wanna fail!
ahrayia [7]

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32.55

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3 0
3 years ago
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


3 0
4 years ago
Read 2 more answers
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