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2 years ago

A triangle has side lenghts 10cm, 24cm, and 34cm. is the triangle acute, obtuse, or right?

Mathematics

1 answer:

Sergeu [11.5K]2 years ago

4 0

Imposible triangle, lagest side must be > sum other sides.

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What is 1/4 equal to?

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1/4 is equal to 25%, .25 , 2/8 on so on. Hope this helps

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Three things you use negative numbers for in real life

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Please solve this, will rate 5 stars and mark as STAR!

Nina [5.8K]

Answer:

$\boxed{5 \cdot \sqrt{2} \cdot \sqrt[6]{5} }$

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

Polygon ABCD goes through a sequence of rigid transformations to form polygon A′B′C′D′. The sequence of transformations involved

Valentin [98]

X axis because of its sequence that it is involved with a reflection which is a transformation

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3 years ago

HELP When 2((Three-fifths x + 2 and three-fourths y minus one-fourth x minus 1 and one-half y + 3)) is simplified, what is the r

lesya692 [45]

Answer:

<h2> StartFraction 7 over 10 EndFraction x + 2 and one-half y + 6</h2>

Step-by-step explanation:

Given the expression $2(\frac{3x}{5}+2\frac{3y}{4}-\frac{x}{4}-1 \frac{1}{2}y+3)$

To simplify the expression, we need to first collect the like terms of the functions in parentheses as shown;

$= 2(\frac{3x}{5}-\frac{x}{4}-1 \frac{1}{2}y+2\frac{3}{4}y+3)\\= 2(\frac{3x}{5}-\frac{x}{4}- \frac{3}{2}y+\frac{11}{4}y+3)\\$

Then we find the LCM of the resulting function

$= 2(\frac{3x}{5}-\frac{x}{4}- \frac{3}{2}y+\frac{11}{4}y+3)\\= 2(\frac{12x-5x}{20} - (\frac{6y-11y}{4})+3)\\= 2(\frac{7x}{20}- (\frac{-5y}{4})+3 )\\= 2(\frac{7x}{20}+ \frac{5y}{4}+3 )\\= \frac{7x}{10} + \frac{5y}{2} +6\\= \frac{7x}{10} + 2\frac{1}{2}y+6\\$

The final expression gives the required answer

7 0

2 years ago