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3 years ago

A triangle has side lengths measuring 20 cm, 5 cm, and n cm. Which describes the possible values of n?

Mathematics

2 answers:

Nuetrik [128]3 years ago

4 0

Answer:

D. 15 < n < 25 is the answer on edge

DENIUS [597]3 years ago

3 0

The answer is D. 15 < n < 25.

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Help me and u will get 100 points at the end

Lera25 [3.4K]

Answer:

x > 0 and y < 2

Step-by-step explanation:

By definition the inverse of a function is simply, when the domain and range are swapped. So if the domain of g(x) is x < 2 and y > 0, then the inverse will have these swapped. So x > 0 and y < 2

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1 year ago

Read 2 more answers

What's 2 and 5/10 in standard decimal form

Evgesh-ka [11]

2 5/10 = 2 1/2 = 2.5

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2 years ago

A mouse is traveling at a rate of 100 feet per hour. One half hour later, a snake smells the

Arlecino [84]

Let, snake catches the mouse in t hours. Till then mouse is travelling from

( t + 1.5 ) hours.

Distance, travelled by mouse in t hours is, $d = 100(t+ 1.5)$ .

To catch the mouse snake had to travel the same distance d in t hours.

So,

$200t = 100( t + 1.5 )\\\\2t = t + 1.5\\\\t = 1.5 \ hours$

Therefore, snake will catch mouse in 1.5 hours.

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3 years ago

The difference of five times a number and one is 24 find the number

dimaraw [331]

<h2>Answer:</h2><h2>let number be x</h2><h2>Step-by-step explanation:</h2><h2>5x - 1 = 24</h2><h2>5x = 25</h2><h2>x = 5</h2>

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2 years ago

One container is filled with mixture that is 25% acid. A second container is filled with a mixture that is 55% acid. The second

algol13

Answer:

Therefore the third container contains $\frac{735}{17}\%$ = 43.23 % acid.

Step-by-step explanation:

Given that, One container is filled with the mixture that 25% acid. 55% acid is contain by second container.

Let the volume of first container be x cubic unit.

Since the volume of second container is 55% larger than the first.

Then the volume of the second container is

$= (V\times \frac{100+55}{100})$ cubic unit.

$=\frac{155}{100}V$ cubic unit.

The amount of acid in first container is

$=V\times \frac{25}{100}$ cubic unit.

$=\frac{25}{100}V$ cubic unit.

The amount of acid in second container is

$=(\frac{155}{100}V \times \frac{55}{100})$ cubic unit.

$=\frac{8525}{10000}V$ cubic unit.

Total amount of acid $=(\frac{25}{100}V+\frac{8525}{10000}V)$ cubic unit.

$=(\frac{2500+8525}{10000}V)$ cubic unit.

$=(\frac{11025}{10000}V)$ cubic unit.

Total volume of mixture $=(V+\frac{155}{100}V)$ cubic unit.

$=\frac{100+155}{100}V$ cubic unit.

$=\frac{255}{100}V$ cubic unit.

The amount of acid in the mixture is

$=\frac{\textrm{The volume of acid}}{\textrm{The amount of mixture}}\times 100 \%$

$=\frac {\frac{11025}{10000}V}{\frac{255}{100}V}\times 100\%$

$=\frac{735}{17}\%$

Therefore the third container contains $\frac{735}{17}\%$ acid.

5 0

3 years ago