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Nookie1986 [14]
3 years ago
11

If line HC and line GD are parallel, which statement is true?

Mathematics
1 answer:
olasank [31]3 years ago
3 0

Answer:

the answer is d

Step-by-step explanation:

hopefully this helps :)

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Alonzo will participate in a charity walkathon.Her Mother will donate $10 for Alonzo participation plus and additional $2 for ea
kiruha [24]

Answer:

$40 total, $30 with just the mile amount.

Step-by-step explanation:

$10 + $2(15) = $40 total

6 0
3 years ago
WILL MARK BRAINLIEST WILL REPORT IF U GIVE ME A LINK TO UKNOWN THING
meriva

Answer:

I think the answer will be x⩽4

5 0
3 years ago
Help please (And dont say that its 60,60, and 120. Thats wrong I already tried it)
Ainat [17]

The value of angle M<1 is 59.

The value of angle M<2 is 61.

The value of angle M<3 is 119.

<h3>What are the values of the angles?</h3>

The value of M3 can be determined by subtracting 61 from 180. This is because angles along a straight line is equal to 180.

180 - 61 = 119.

The value of M2 is equal to 61. This is because alternate angles are equal.

The value of M3 can be determined by subtracting 61, 60 from 180. This is because sum of angles in a triangle is equal to 180.

180 - 60 - 61 = 59

To learn more about triangles, please check: brainly.com/question/20694080

#SPJ1

7 0
2 years ago
What is the area of triangle ABC?
Elina [12.6K]

Answer:

\circ\ \ 36\sqrt{3}

Step-by-step explanation:

CD=12\times \sin(60)  =12\times \frac{\sqrt{3} }{2}

<u><em>Important Note</em></u>:………………………………………………………………………………………

ABC is an equilateral triangle because it has 2 angles of measure 60

===========================================================

According to the note AB = 12

<u><em>Final answer</em></u> :

\text{Area of triangle ABC} =\frac{AB\times CD}{2}

                                  =\frac{12\times 6\sqrt{3} }{2}

                                  =\frac{72\sqrt{3} }{2}

                                  =36\sqrt{3}

7 0
2 years ago
The given line passes through the points (0, -3) and (2, 3).
puteri [66]

The equation, in point-slope form of the line that is  parallel to the given line and passes through the point  (-1, -1) is y + 1 = 3(x + 1).

<u>Solution:</u>

Given that, a line passes through (0, -3) and (2, 3).

We have to find the line equation which is parallel to above line and  passes through (-1, -1).

Now, let us find the slope of the given line.

\text { slope } m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}, \text { where }\left(x_{1}, y_{1}\right) \text { and }\left(x_{2}, y_{2}\right) \text { are points on that line }

\text { Then, } m=\frac{3-(-3)}{2-0}=\frac{3+3}{2}=\frac{6}{2}=3

So, slope of given line is 3,  

Then, slope of required line is also 3, as slopes of parallel lines are equal.

Then, required line equation in point – slope form is given as:

\begin{array}{l}{y-y_{1}=m\left(x-x_{1}\right)} \\\\ {y-(-1)=3(x-(-1))} \\\\ {\rightarrow y+1=3(x+1)}\end{array}

Hence, the line equation is y + 1 = 3(x + 1).

6 0
3 years ago
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