Write the ratio as a fraction in simplest form, with whole numbers in the numerator and denominator.

One light-year equals 5.9 x 1012 miles. How many light-years are in 6.79

inna [77]

Option B

The number of light years in $6.79 \times 10^{16}$ miles is 11508 light years

<em><u>Solution:</u></em>

Given that,

One light-year equals 5.9 x 10^12 miles

Therefore,

$1 \text{ light year } = 5.9 \times 10^{12} \text{ miles }$

To find: Number of light years in $6.79 \times 10^{16}$ miles

Let "x" be the number of light years in $6.79 \times 10^{16}$ miles

Then number of light years in $6.79 \times 10^{16}$ miles can be found by dividing $6.79 \times 10^{16}$ miles by miles in 1 light year

$\text{Number of light years in } 6.79 \times 10^{16} miles = \frac{6.79 \times 10^{16}}{5.9 \times 10^{12}}\\\\\text{Use the law of exponent }\\\\\frac{a^m}{a^n} = a^{m-n}\\\\\text{Number of light years in } 6.79 \times 10^{16} miles = \frac{6.79}{5.9} \times 10^{16-12}\\\\\text{Number of light years in } 6.79 \times 10^{16} miles = 1.1508 \times 10^4\\\\\text{Number of light years in } 6.79 \times 10^{16} miles = 11508$

Thus number of light years in $6.79 \times 10^{16}$ miles is 11508 light years

3 0

2 years ago

I need to solve for b but I don't know how

lorasvet [3.4K]

It base times height

5 0

2 years ago

Finding product for 4 digit numbers

Amanda [17]

One answer would be15669012

7 0

2 years ago

Hello can someone help me simplify 4) and 5)? Thank you and please include steps :)

Leni [432]

Alrighty

remember some rules
$(ab)^c=(a^c)(b^c)$
and
$x^{-m}=\frac{1}{x^m}$
and
$x^0=1$ for all real values of x
and
$(a^b)^c=a^{bc}$
and
$(\frac{a}{b})^c=\frac{a^c}{b^c}$
and
(a/b)/(c/d)=(ad)/(bc)
and
$(a^b)(a^c)=a^{b+c}$
and
$\frac{a^m}{a^n}=a^{m-n}$
and don't forget pemdas
example: $-x^m=-1(x^m)$ but $(-x)^m=(-1)^m(x^m)$

so

4.
$(\frac{c^{-2}}{2})^{-2}=$

$\frac{(c^{-2})^{-2}}{2^{-2}}=$

$\frac{c^4}{\frac{1}{2^2}}=$

$\frac{c^4}{\frac{1}{4}}=$

$4c^4$

5.

$\frac{(-a)^4bc^5}{-a^2b^{-3}c^0}=$

$\frac{(-1)^4(a)^4bc^5}{-1(a^2)(\frac{1}{b^3}(1)}=$

$\frac{(1)a^4bc^5}{\frac{(-1)(a^2)}{b^3}}=$

$\frac{(a^4bc^5)b^3}{(-1)(a^2)}=$

$\frac{-a^4b^4c^5}{a^2}=$

$-a^2b^4c^5$