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2 years ago

I need the answers please

Mathematics

1 answer:

lapo4ka [179]2 years ago

6 0

1 is parallel. 2. is skewed. 3. is perpendicular

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Question 1<br> Solve: 25.6 - 5y + 15. 3

QveST [7]

Answer:

40.9 - 5y

Step-by-step explanation:

25.6 - 5y + 15.3

40.9 - 5y ## ^ combine like-terms

8 0

3 years ago

One positive integer is 5 times another positive integer and their product is 320. What are the positive integers?

seropon [69]

Answer:

The integers are (x, y) = (40, 8).

Step-by-step explanation:

x = 5y

xy = 320

Substitute the first equation into the second equation.

(5y)(y) = 320

5y^2 = 320

y^2 = 64

y = 8 (y must be positive)

The integers are (x, y) = (40, 8).

6 0

2 years ago

Find -1/5 + (-3/5) in simplest form

guapka [62]

Answer: -0.8 or -4/5

Step-by-step explanation:

1/5 + 3/5 = 4/5 and u put the negatives so its -4/5. HOPE THIS HELPS :3 !!!!

6 0

3 years ago

I need these all written out on how they're done along with the answer please

taurus [48]

Well, these are quite simple really, especially 1,2,3 and 4. If you read the questions carefully, you can understand it well. You don’t need to know the answer, you just need to understand

8 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