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Sergio [31]
3 years ago
15

Which statement about the expression 3(x + 2/5) is true ?

Mathematics
1 answer:
Jobisdone [24]3 years ago
5 0
A is the answer I think
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At the beginning of the year, Alyssa had $100 in savings and saved an additional $7 each week thereafter. Colton started the yea
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The equation that represents the amount saved by Alyssa after tt weeks is AA = 100 + $7tt

The equation that represents the amount saved by Colton after tt weeks is CC = $55 + $10tt

The amount they would have when they have same amount of money is $205.

<h3>What equation represents the total amount saved?</h3>

The total amount saved: inital amount + (amount saved per week x total number of weeks

AA = 100 + $7tt

CC = $55 + $10tt

<h3>What would be the amount when they have the same amount of money?</h3>

$55 + $10tt = 100 + $7tt

Combine similar terms and solve for tt

100 - 55 = 10tt - 7tt

45 = 3tt

tt = 15 weeks

100 + $7(15) = $205

To learn more about linear functions, please check: brainly.com/question/26434260

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What's 675890 in word form
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The population, P(t), of China, in billions, can be approximated by1 P(t)=1.394(1.006)t, where t is the number of years since th
vitfil [10]

Answer:

At the start of 2014, the population was growing at 8.34 million people per year.

At the start of 2015, the population was growing at 8.39 million people per year.

Step-by-step explanation:

To find how fast was the population growing at the start of 2014 and at the start of 2015 we need to take the derivative of the function with respect to t.

The derivative shows by how much the function (the population, in this case) is changing when the variable you're deriving with respect to (time) increases one unit (one year).

We know that the population, P(t), of China, in billions, can be approximated by P(t)=1.394(1.006)^t

To find the derivative you need to:

\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=\\\\\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'\\\\1.394\frac{d}{dt}\left(1.006^t\right)\\\\\mathrm{Apply\:the\:derivative\:exponent\:rule}:\quad \frac{d}{dx}\left(a^x\right)=a^x\ln \left(a\right)\\\\1.394\cdot \:1.006^t\ln \left(1.006\right)\\\\\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=(1.394\cdot \ln \left(1.006\right))\cdot 1.006^t

To find the population growing at the start of 2014 we say t = 0

P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(0)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^0\\P(0)' = 0.00833901 \:Billion/year

To find the population growing at the start of 2015 we say t = 1

P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(1)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^1\\P(1)' = 0.00838904 \:Billion/year

To convert billion to million you multiple by 1000

P(0)' = 0.00833901 \:Billion/year \cdot 1000 = 8.34 \:Million/year \\P(1)' = 0.00838904 \:Billion/year \cdot 1000 = 8.39 \:Million/year

6 0
3 years ago
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