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Annette [7]
3 years ago
5

Consider triangle GHJ.

Mathematics
2 answers:
gayaneshka [121]3 years ago
8 0
5 with square root 3 
bulgar [2K]3 years ago
5 0

we know that

<u>Applying the Pythagorean Theorem</u>

JG^{2} =HJ^{2}+HG^{2}

solve for HJ

HJ^{2}=JG^{2}-HG^{2}

in this problem we have

JG=10\ units\\ HG= 5\ units

Substitute the values in the formula above

HJ^{2}=10^{2}-5^{2}

HJ^{2}=100-25

HJ^{2}=75

HJ=\sqrt{75}=5\sqrt{3} \ units

therefore

<u>the answer is the option</u>

B. 5 sq. root 3 units

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Y=4x+8x, Solve for x.<br><br> It's x=y/12, I just need to know how to work it out.
Scrat [10]

Step-by-step explanation:

<h2>Answer:-</h2>

\tt{y = 4x+8x}

First, we need to add up the x.

We know that 4+8 = 12, so

\tt{4x+8x=12x}

Now, equating it to y,

\tt{y = 12x}

Now, if I multiply \tt{\dfrac{1}{12}} on both sides,

I get,

\tt{\dfrac{y}{12}= \dfrac{\cancel{12}x}{\cancel{12}}}

Now, as 12 got cancelled , I got the final answer as

\boxed{\tt{x =\dfrac{y}{12}}}

Hope it helps :)

4 0
2 years ago
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stira [4]
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5 0
3 years ago
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A group of friends went fishing. They each caught 20 fish but not all of them were keepers. Which person was able to keep 3/5 of
Romashka-Z-Leto [24]

Answer:

Those who were able to keep 12 fishes

Step-by-step explanation:

Given

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8 0
2 years ago
Find the cube roots of 125(cos 288° + i sin 288°)
kow [346]
Let r(cos O + i sin O)  be a cube root of 125(cos 288 + i sin 288)
then
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so r^3 = 125  and  cos 3O + i sin 3O  =  cos 288 + i sin 288

so r  = 5  and 3O = 288 + 360p and O = 96 +  120p

so one cube root is   5 (cos 96 + i sin 96)

Im a little rusty at this stuff Its been a long time.

Im not sure of the other 2 roots

sorry cant help you any more


3 0
2 years ago
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almond37 [142]

Answer:

I can't understand the questions

3 0
3 years ago
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