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3 years ago

How do I write 55555 in expanded form

Mathematics

2 answers:

marta [7]3 years ago

7 0

Answer:

FIFTY - FIVE THOUSAND FIVE HUNDRED AND FIFTY-FIVE

Step-by-step explanation:

WITCHER [35]3 years ago

6 0

5 ten thousands + 5 thousands + 5 hundreds + 5 tens + 5 ones

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5/15 = n/9 someone help me (proportions)

goblinko [34]

Answer:

Step-by-step explanation:

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2 years ago

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4. Rosa plays games at a fair. She can buy 8 game tokens for $ 1. Each game cost 2 tokens. How many games can she play with 120

bezimeni [28]

If each game costs 2 tokens, 120 tokens would equal 60 games.

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3 years ago

A sailboat travels a distance of 2 1/2 miles in 1/6 of an hour. Which complex fraction represents the unit rate in miles per hou

Vsevolod [243]

1/6 of an hour= 10 because 60mins=1hour & 60/6=10 so the sailboat travels 2 1/2 miles in 10 mins. So the complex fraction would be 2 &1/2 over 10. Hope it helps!

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3 years ago

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The population of a town has approximately doubled every 17 years since 1950. If the equation P=Po2k, where Po is the population

natka813 [3]

The population of a town has approximately doubled every 17 years since 1950.

the equation P= $P_02^k$ where Po is the population of the town in 1950, is used to model the population, P, of the town t years after 1950.

When t=17 yrs

P=2 $p_{0}$

for 1 year

The equation becomes

P = $P_02^\frac{t}{17}$ -------------(1)

Our original equation is

$P=P_02^k$ ----------------------------------(2)

equating expression 1 and 2

$P_02^\frac{t}{17}=P_{0}2^k$

Cancelling $P_{0}$ from both sides we get

$2^\frac{t}{17}=2^{k}$

t/17=k

⇒k=t/17 is the solution.

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3 years ago

Which expression is equivalent

gtnhenbr [62]

Answer:

Option B is correct.

$\frac{81m^2n^5}{8}$ is equivalent to $\frac{(3m^{-1}n^2)^4}{(2m^{-2}n)^3}$

Step-by-step explanation:

Given expression: $\frac{(3m^{-1}n^2)^4}{(2m^{-2}n)^3}$

Using exponents power:

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

$(a^n)^m = a^{nm}$

$a^m \cdot a^n = a^{m+n}$

Given: $\frac{(3m^{-1}n^2)^4}{(2m^{-2}n)^3}$

Apply exponent power :

⇒ $\frac{3^4 (m^{-1})^4(n^2)^4}{2^3(m^{-2})^3 n^3}$

⇒ $\frac{81 m^{-4}n^8}{8m^{-6}n^3} = \frac{81 m^{-4} \cdot m^6 n^8 \cdot n^{-3}}{8}$

⇒ $\frac{81 m^{-4+6} n^{8-3}}{8} = \frac{81 m^2 n^5}{8} = \frac{81m^2 n^5}{8}$

Therefore, the expression which is equivalent to $\frac{(3m^{-1}n^2)^4}{(2m^{-2}n)^3}$ is, $\frac{81m^2 n^5}{8}$

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3 years ago

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