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nevsk [136]
3 years ago
9

I NEED THIS ASAP!!! What is GC?

Mathematics
1 answer:
umka21 [38]3 years ago
6 0

9514 1404 393

Answer:

  D.  14

Step-by-step explanation:

Point G divides each median into parts that have the ratio ...

  short : long = 1 : 2

Then ...

  GD : GC = 1 : 2 = 7 : 14 . . . . . . . . multiply the ratio by 7

GC = 14

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Can anybody please help me out with this question I would really appreciate it so much . Please and thank you .
mr Goodwill [35]

Answer:

x=3 MT=36 MH=18

Step-by-step explanation:

The two triangles are the same, so 7x+8=10x-1

3x=9

x=3

MT=12x3=36

MH=36/2=18

3 0
3 years ago
I need answers ASAP!!
lesya [120]
What is the question?

7 0
3 years ago
Read 2 more answers
A stairway rises 6 feet 4 inches over a horizontal distance of 8 feet 6 inches. What is the diagonal length of the stairway
natali 33 [55]

Answer:

10.6 ft

Step-by-step explanation:

to find the diagonal length of the stairway we can use the pythagorean theorem or a^2 + b^2 = c^2

6'4" is also 6 1/3' or 19/3'

8'6" is also 8 1/2' or 17/2'

now square them both

361/9 + 289/4

make them have the same denominator

1444/36 + 2601/36

add them

4045/36

take the square root

10.6000524 is your answer

10.6 ft

7 0
2 years ago
Help please don’t understand.
sergij07 [2.7K]
Okay so first, area is length times width, so 1/4x*x would be first = 1/4(x^2)=64. Then area or x = 16 or -16

perimeter would be width times two plus length times two. I recommend using math papa to calculate these numbers.

4 0
3 years ago
Write the equation of a line Parallel to the given line and passes through points (-4, -3); (2, 3). Hints find the slope by usin
julia-pushkina [17]

First, we obtain the gradient (slope) of the first parallel line

\text{gradient, m}_1\text{ = }\frac{y_2-y_1}{x_2-x_1}\text{ = }\frac{3-(-3)}{2-(-4)}=\frac{6}{6}\text{ = 1}

Recall that since both lines are parallel, we have that,

m_1=m_2

Thus

m_2\text{ = 1}

Hence, we can find the equation of the parallel line given that it passes through the points (-4, -3)

Using

\begin{gathered} y\text{ = mx + c} \\ \text{where m = m}_2\text{ = 1} \\ \text{and x = -4 ,  y = -3, we have} \\ -3\text{ = 1(-4) + c} \\ -3\text{ = -4 + c} \\ 4\text{ - 3 = c} \\ c\text{ = 1} \\ \text{Thus, the equation of the line is y = (1)x + 1} \\ y\text{ = x + 1} \end{gathered}

5 0
1 year ago
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