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Karolina [17]
3 years ago
8

Please help and explain

Mathematics
1 answer:
Nitella [24]3 years ago
5 0
Comment
The two triangles (ABC and ADE) can be shown to be similar (which I'll do below). Their corresponding parts are also similar. 
Step One
Prove Similarity.
A is common to both triangles              Construction
<ABC = <ADE                                       Given as right angles.

By the AA Theorem both triangles are similar.

Step Two
Find x

AD is to AB as DE is to x
AD = 3 + 4 = 7
AB = 3
DE = 6
7 / 3 = 6/x Cross multiply
7x = 3 * 6
7x = 18 Divide by 7
x = 18/7
x = 2.571 <<<<< answer

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How many times greater is the value of the 3 in 4,367 than the value of the 3 in 39?​
Lesechka [4]

Answer: Roughly 111.97

Step-by-step explanation:

1. 39 divided by 3 is 13.  

2. 4367 divided by 3 is not evenly distributed. Rounding, it would work out to be roughly 1455.66 

3. 1455.66 divided by 13 is roughly 111.97 (Again, this does not distribute evenly)

4. To check your work, multiply 111.97 times 13 and you should get somewhere around 1455.66 which is (4367 divided by 3).

Hopefully this helps! Feel free to mark brainliest! :)

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3 years ago
Which expression is equivalent to 10√5<br>A. √500<br>B.√105<br>C.√50<br>D.√15
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3 years ago
Cubed root x cubed root x2​
Yuki888 [10]

Answer:

Final answer is \sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x.

Step-by-step explanation:

Given problem is \sqrt[3]{x}\cdot\sqrt[3]{x^2}.

Now we need to simplify this problem.

\sqrt[3]{x}\cdot\sqrt[3]{x^2}

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}

Apply formula

\sqrt[n]{x^p}\cdot\sqrt[n]{x^q}=\sqrt[n]{x^{p+q}}

so we get:

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{1+2}}

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{3}}

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x

Hence final answer is \sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x.

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