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3 years ago

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M is the centroid of △DEF. There is a triangle DEF in which J, K, and L are the points on the side DE, side EF, and side DF resp

ectively. Segment DK, segment EL, and segment FJ intersect each other at point M. The length of DM is 8, the length of MJ is 2y, the length of EM is 6, and the length of FM is 2x. What is an expression for FJ? Select all that apply. A. 2x + 2y B. 3x C. 6y D. 3y

2 answers:

yanalaym [24]3 years ago

8 0

Answer:

B

Step-by-step explanation:

Mumz [18]3 years ago

3 0

Answer:

B

Step-by-step explanation:

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If a triangle has sides 4, 6 and 10 and a 2nd similar triangle has a perimeter of 60, find the 2nd largest side of the 2nd trian

algol13

The first triangle has a perimeter of 4+6+10 = 20.

The 2nd longest side is 6/20 = 3/10 of the perimeter. In the larger triangle, the 2nd longest side will have length (3/10)*60 = 18.

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I made 10 cups of granola that we eat in ⅔ cup servings. How many servings will we have out of the 10 cups of granola?

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3 years ago

Explain how to get that answer!!

ra1l [238]

We need to simplify $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$

First lets factor $\sqrt{14x^3}$

$\sqrt{14x^3}$ = $\sqrt{14} \sqrt{x^3}$
$\sqrt{14} = \sqrt{2} \sqrt{7}$ by applying the radical rule $\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}$
$\sqrt{x^3} = x^{3/2}$ By applying the radical rule $\sqrt[n]{x^m} = x^{m/n}$

So
$\sqrt{14x^3}$ = $\sqrt{14} \sqrt{x^3}$ = $\sqrt{2} \sqrt{7}x^{3/2}$

Now let's factor $\sqrt{18x}$
By applying the radical rule $\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}$ ,
$\sqrt{18x} = \sqrt{18} \sqrt{x}$
$\sqrt{18} = \sqrt{2} * 3$

So $\sqrt{18x}$ = $\sqrt{2}*3 \sqrt{x}$

So $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$ = $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x} }$

We know that $\sqrt[n]{x} = x^{1/n}$ so $\sqrt{x} = x^{1/2}$

We now have $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x}} = \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}}$

We know that $\frac{x^a}{x^b} = x^{a-b}$
So $\frac{x^{3/2}}{x^{1/2}} = x^{3/2 - 1/2} = x$

We now got $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}} = \frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3}
$

We can notice that the numerator and the denominator both got √2 in a multiplication, so we can simplify them, and we get:
$\frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3} = \frac{ \sqrt{7}x }{3}$

All in All, we get $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$ = $\frac{ \sqrt{7}x }{3}$

Hope this helps! :D

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3 years ago

will ran 4 miles on his first day of training the next day he ran 1/3 that distance how far did he run the second day

tekilochka [14]

He ran 4/3 of a mile the second day

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3 years ago

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Plz help!! Math isn’t my thing. I’ll give brainliest.

vladimir2022 [97]

Answer:

Here is your answer

Hope it helps

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3 years ago

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