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AfilCa [17]
3 years ago
14

the population of a town increases at the rate of 1% each year today the towns population is 8500 what will the population be in

five years
Mathematics
1 answer:
romanna [79]3 years ago
4 0

Answer:

8934 (8933.58542585)

Step-by-step explanation:

8,500 (population currently) * 1.01 (1 percent) ^ 5 (years) = 8933.58542585 ≅ 8934 people.

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4+2g+5g=-66
sergejj [24]
The answer is g=-10
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6 0
2 years ago
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I need help asap someone please help me i dont understand this question so can someone help me
eduard

Answer:

we conclude that:

\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}=\frac{1}{2p^3-p^2}

Step-by-step explanation:

Given the expression

\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}

\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{b}\div \frac{c}{d}=\frac{a}{b}\times \frac{d}{c}

=\frac{2p}{4p^2-1}\times \frac{6p+3}{6p^3}

=\frac{2p}{4p^2-1}\times \frac{2p+1}{2p^3}

\mathrm{Multiply\:fractions}:\quad \frac{a}{b}\times \frac{c}{d}=\frac{a\:\times \:c}{b\:\times \:d}

=\frac{2p\left(2p+1\right)}{\left(4p^2-1\right)\times \:2p^3}

cancel the common factor: 2

=\frac{p\left(2p+1\right)}{\left(4p^2-1\right)p^3}

cancel the common factor: p

=\frac{2p+1}{p^2\left(4p^2-1\right)}

=\frac{2p+1}{p^2\left(2p+1\right)\left(2p-1\right)}

cancel the common factor: 2p+1

=\frac{1}{p^2\left(2p-1\right)}

Expanding

=\frac{1}{2p^3-p^2}

Thus, we conclude that:

\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}=\frac{1}{2p^3-p^2}

7 0
3 years ago
H(x)=4log2(x−6)+1 <br><br> All real numbers greater than____
MariettaO [177]

\huge\underline\mathfrak\blue{Answer:-}

The demoin is all real numbers greater than 6

4 0
3 years ago
The oz of cheese goes with 8 oz of pasta. How much cheese is used for 3 oz of pasta?
brilliants [131]

Answer:

0.375

Step-by-step explanation:

1 > 8

x > 3

x=(3×1)÷8 = 0.375

3 0
3 years ago
A store can buy 8 cartons of milk for $24 or 7 cartons of milk for $28.
USPshnik [31]

Answer:

A. $1 less

Step-by-step explanation:

24 divided by 8 (unit rate) = $3

28 divided by 7 = $4

4-3=1

3 0
2 years ago
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