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2 years ago

5

Light travels about 180 million kilometers in 10 minutes. How far does it travel in one minute? How far does it travel in one se

cond?

2 answers:

astraxan [27]2 years ago

8 0

Answer:

18 million kilometers in one min. and 0.3 million kilometers in one second

Step-by-step explanation:

so the one min. is easy because all you have to do is take 180 and divide it by 10 and you get 18 million kilometers in one min. and then you would take 18 and divide it by 60 and you would get 0.3 million kilometers in one second

aliya0001 [1]2 years ago

7 0

Answer:

The light travels distance in 1 minute=18 million kilometers

The light travels distance in 1 second=0.3 million kilometers

Step-by-step explanation:

We are given that

Light travels total distance in 10 minutes=180 million kilometers

We have to find the distance travels by light in 1 minute and in 1 seconds.

By unitary method

Distance travel by light in 1 min= $\frac{180}{10}$ million kilometers=18 million kilometers

Distance travel by light in 1 min=18 million kilometers

We know that 1 minute=60 seconds

In 60 seconds, light travels distance =18 million kilometers

In 1 second, light travels distance = $\frac{18}{60}=0.3$ million kilometers

Hence, the light travels distance in 1 minute=18 million kilometers

The light travels distance in 1 second=0.3 million kilometers

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2 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

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Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

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2 years ago

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Answer:

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= 8.1 + 5

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I hope this was helpful, Please mark as brainliest

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2 years ago

Read 2 more answers

Question 1(Multiple Choice Worth 1 points)

tester [92]

The correct answer is d15

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2 years ago