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Aliun [14]
3 years ago
14

Triangle T R S has centroid Z. Lines are drawn from each point to the midpoint of the opposite side to form line segments T W, R

V, and S U.
In triangle TRS, VZ = 6 inches. What is RZ?

3 inches
6 inches
12 inches
18 inches
Mathematics
2 answers:
solong [7]3 years ago
6 0

Answer:

C?

Step-by-step explanation:

From eyeballing it, I can see that RZ is definetely not the first two. and RZ looks about exactly double the length of VZ. Hope this helps. :)

mina [271]3 years ago
6 0

Answer:

C. 12 inches

Step-by-step explanation:

May I have brainliest please? :)

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Boris made $306 for 18 hours of work. At the same rate, how many hours would he have to work to make $119?
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On an architect's plan for a house, the scale reads ½ in. = 5 ft. how long is the actual length of a room that's 2 in. long on t
saw5 [17]
1/2 in. = 5 ft
so., 1 in. = 5 * 2 = 10 ft

Now, 2 in = 10 * 2 = 20 ft

In short, Your Answer would be Option D

Hope this helps!
6 0
3 years ago
How do I go about solving (27x^3/8y^9)^5/3? And what is the role of the numerator and denominator?
MrRissso [65]
\left( \frac{27x^3}{8y^9}\right)^ \frac{5}{3}  \\\\\\ =\left( \frac{(3x)^3}{(2y^3)^3}\right)^ \frac{5}{3} \\\\\\ =  \frac{(3x)^{3 \times  \frac{5}{3} }}{(2y^3)^{3 \times  \frac{5}{3} }} \\\\\\ =\frac{(3x)^5}{(2y^3)^{5 }} \\\\\\ =\frac{243x^5}{32y^{15}}

Now, If the exponent was negative like you asked....

\left( \frac{27x^3}{8y^9}\right)^ {-\frac{5}{3}} \\\\\\ =\left( \frac{8y^9}{27x^3}\right)^ {\frac{5}{3}}\\\\\\ =\left( \frac{(2y^3)^3}{(3x)^3}\right)^ \frac{5}{3} \\\\\\ = \frac{(2y^3)^{3 \times \frac{5}{3} }}{(3x)^{3 \times \frac{5}{3} }} \\\\\\ =\frac{(2y^3)^{5 }}{(3x)^5} \\\\\\ =\frac{32y^{15}}{243x^5}

5 0
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The sum of the ages of the father and his son is 42 years. The product of their ages is 185. Find the age of the father and the
Iteru [2.4K]

Let f and s be the ages of the father and the son. We have

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From the first equation we derive

f=42-s

Substitute this expression for f in the second equation and we have

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Since the sum of the ages must be 42, the solutions would imply

s=5 \implies f=37,\quad s=37\implies f=5

We can only accept the first solution, since the second would imply a son older than his father!

4 0
4 years ago
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