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MA_775_DIABLO [31]
2 years ago
7

Can you find both the value of x and y? It is a 30-60-90 triangle

Mathematics
1 answer:
Juli2301 [7.4K]2 years ago
3 0

Answer:

x = 8*root3

y = 16

Step-by-step explanation:

You can use tangent and cosine of 60° to find x and y, BUT THERE'S A SHORTCUT. You pointed out that its a 30°-60°-90° triangle. That is a special right triangle. Here's the shortcut:

the longest side (hypotenuse) is double the shortest side. We have an 8 on the shortest side, so the hypotenuse (longest side) is 16. This ALWAYS works for 30°-60°-90° triangles!

Another shortcut:

The longer leg is the shorter leg × root3.

This ALWAYS works on 30°-60°-90° triangles.

So on your triangle, the short side is 8, that means the longer leg is 8×root3.

ta-daaaa!

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3 years ago
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\boxed{5 \cdot \sqrt{2}  \cdot \sqrt[6]{5} }

Step-by-step explanation:

\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }

\sqrt{\sqrt[3]{10} } \implies (10^\frac{1}{3} )^\frac{1}{2} =10^\frac{1}{6} =\sqrt[6]{10}

\therefore \sqrt{\sqrt[3]{10} }=\sqrt[6]{10}

\text{Solving }\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }

250=2 \cdot 5^3

\sqrt[3]{250}=\sqrt[3]{2\cdot 5^3}=5  \sqrt[3]{2}

Once

\sqrt[6]{2}  \cdot \sqrt[6]{5} = \sqrt[6]{10}

We have

5  \sqrt[3]{2} \cdot \sqrt[6]{2}  \cdot \sqrt[6]{5}

We can proceed considering the common base of exponentials

\sqrt[3]{2}  \cdot \sqrt[6]{2}  =  2^{\frac{1}{3}} \cdot  2^{\frac{1}{6} }  = 2^{\frac{3}{6} } = 2^{\frac{1}{2} }=\sqrt{2}

Therefore,

5  \sqrt[3]{2} \cdot \sqrt[6]{2}  \cdot \sqrt[6]{5} = 5 \cdot \sqrt{2}  \cdot \sqrt[6]{5}

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3 years ago
What is the proportional relationship represented by this table?
77julia77 [94]

Answer:

the constant of proportionality is 0.2 and it is a proportional relationship

Step-by-step explanation:

2 divided by 10 is 0.2

1 divided by 5 is 0.2

they have the same answer so they are proportional

since they have the answer 0.2 it

is there constant of proportionality

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