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2 years ago

Kenneth has 22 math problems to do for homework. He

Mathematics

1 answer:

grin007 [14]2 years ago

5 0

Kenneth has 10 problems left because 22-12=10
If he does 1 problem every minute it will take him ten minutes

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A drawing contains 5 circles, 8 triangles, and 6 squares. What is the ratio of triangles to circles? Write the ratio using a fra

zavuch27 [327]

Answer:

8:5 or 8/5

Step-by-step explanation:

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2 years ago

Please help.

Luba_88 [7]

Answer:

6.5%

Step-by-step explanation:

The original price of the ticket was 100% of the price of the ticket.

The 15% coupe reduced the price to 85% of the price of the ticket.

The 10% fee on the discounted price adds 10% of 85% to 85%, so it adds 8.5% to 85%. By adding the fee he is at 93.5% of the original price.

100% - 93.5% = 6.5%

The overall percent off compared to the original price was 6.5%.

Answer: 6.5%

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2 years ago

After m

Whitepunk [10]

<span>length of finger nail in m month =10+3m
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3 years ago

Read 2 more answers

How many times do you need to multiply 271.96 by ten to get 2719.6

ivann1987 [24]

Every 10 you multipy by moves the decimla place to the right by 1 place
so it moves 1 place so therefor you need to multipy it by 10 ONE time
answer is 1 times

7 0

3 years ago

One model for the spread of a virusis that the rate of spread is proportional to the product of the fraction of the population P

Darya [45]

Answer:

The differential equation for the model is

$\frac{dP}{dt}=kP(1-P)$

The model for P is

$P(t)=\frac{1}{1-0.99e^{t/447}}$

At half day of the 4th day (t=4.488), the population infected reaches 90,000.

Step-by-step explanation:

We can write the rate of spread of the virus as:

$\frac{dP}{dt}=kP(1-P)$

We know that P(0)=100 and P(3)=100+200=300.

We have to calculate t so that P(t)=0.9*100,000=90,000.

Solving the diferential equation

$\frac{dP}{dt}=kP(1-P)\\\\ \int \frac{dP}{P-P^2} =k\int dt\\\\-ln(1-\frac{1}{P})+C_1=kt\\\\1-\frac{1}{P}=Ce^{-kt}\\\\\frac{1}{P}=1-Ce^{-kt}\\\\P=\frac{1}{1-Ce^{-kt}}$

$P(0)= \frac{1}{1-Ce^{-kt}}=\frac{1}{1-C}=100\\\\1-C=0.01\\\\C=0.99\\\\\\P(3)= \frac{1}{1-0.99e^{-3k}}=300\\\\1-0.99e^{-3k}=\frac{1}{300}=0.99e^{-3k}=1-1/300=0.997\\\\e^{-3k}=0.997/0.99=1.007\\\\-3k=ln(1.007)=0.007\\\\k=-0.007/3=-0.00224=-1/447$

Then the model for the population infected at time t is:

$P(t)=\frac{1}{1-0.99e^{t/447}}$

Now, we can calculate t for P(t)=90,000

$P(t)=\frac{1}{1-0.99e^{t/447}}=90,000\\\\1-0.99e^{t/447}=1/90,000 \\\\0.99e^{t/447}=1-1/90,000=0.999988889\\\\e^{t/447}=1.010089787\\\\ t/447=ln(1.010089787)\\\\t=447ln(1.010089787)=447*0.010039225=4.487533$

At half day of the 4th day (t=4.488), the population infected reaches 90,000.

8 0

3 years ago