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2 years ago

Find M<1 A. 12.71 degrees B. 15 degrees C. 74 degrees D. 106 degrees

Mathematics

1 answer:

Daniel [21]2 years ago

5 0

It is 74 degrees because those two equations equal each other and you must cancel them out which equals x=15. plug 15 in to one of the equations and you have the answer.

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Pattern x starts with the number 0 and follows the rule "add 2". pattern y starts with the number 30 and follows the rule "subtr

saul85 [17]

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2 years ago

Find the approximate perimeter of ABC plotted below.

maksim [4K]

Answer:

B. 21.2

Step-by-step explanation:

Perimeter of ∆ABC = AB + BC + AC

A(-4, 1)

B(-2, 3)

C(3, -4)

✔️Distance between A(-4, 1) and B(-2, 3):

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

$AB = \sqrt{(-2 - (-4))^2 + (3 - 1)^2} = \sqrt{(2)^2 + (2)^2)}$

$AB = \sqrt{4 + 4}$

$AB = \sqrt{16}$

AB = 4 units

✔️Distance between B(-2, 3) and C(3, -4):

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

$BC = \sqrt{(3 - (-2))^2 + (-4 - 3)^2} = \sqrt{(5)^2 + (-7)^2)}$

$BC = \sqrt{25 + 49}$

$BC = \sqrt{74}$

BC = 8.6 units (nearest tenth)

✔️Distance between A(-4, 1) and C(3, -4):

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

$AC = \sqrt{(3 - (-4))^2 + (-4 - 1)^2} = \sqrt{(7)^2 + (-5)^2)}$

$AC = \sqrt{47 + 25}$

$AC = \sqrt{74}$

AC = 8.6 units (nearest tenth)

Perimeter of ∆ABC = 4 + 8.6 + 8.6 = 21.2 units

8 0

3 years ago

FASTTT HURRYYYY PLEASEEEEEE The pair of figures is similar. Find the missing side.

Vladimir [108]

The answer is x=12 :)

6 0

2 years ago

please let us to do with it and it was the only one that I can get the best way for you and I am a bit more than one person who

Lelechka [254]

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8 0

3 years ago

given the two points (-24,7) and (30,25) a. What is an equation passing through the points? b. Is (51, 33) also on the same line

kicyunya [14]

Part 1) Given the two points (-24,7) and (30,25) a. What is an equation passing through the points?

step 1
find the slope m
m=(y2-y1)/(x2-x1)---->m=(25-7)/(30+24)----> m=18/54----> m=1/3

step 2
wit m=1/3 and the point (30,25)
find the equation of the line
y-y1=m*(x-x1)-----> y-25=(1/3)*(x-30)--->y=(1/3)*x-10+25
y=(1/3)*x+15

the answer Part 1) is
y=(1/3)*x+15

Part 2) Is (51, 33) also on the same line?
if the point (51.33) is on the line y=(1/3)*x+15
then
for x=51 the value of y must be 33
for x=51
y=(1/3)*51+15----> y=17+15----> y=32
32 is not 33
so
the point does not belong to the given line

the answer Part 2) is
the point does not belong to the given line

see the attached figure

7 0

2 years ago