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SpyIntel [72]
3 years ago
14

What is m = 2/3; (6, 1) in slope intercept form? :)

Mathematics
1 answer:
Marina86 [1]3 years ago
7 0

Answer:

y = 2/3x - 3

Step-by-step explanation:

First let's look at what slope-intercept form is.

y = mx + b

m is the slope, which in this case is 2/3

b is the y-intercept, which we will find later.

Now that we have identified the parts of the equation, let's plug in what we know. Use the point given for your x and y values.

y = mx + b

1 = 2/3(6) + b

Let's simplify this to find b.

[multiply] 1 = 4 + b

[subtract 4]

-3 = b

We now have our b-value, which is also our y-intercept. Plug in this value into the standard slope-intercept equation.

y = 2/3x - 3

This is your equation.

Check this by plugging in the point again.

1 = 2/3(6) - 3

[multiply] 1 = 4 - 3

[subtract]

1 = 1

This expression is true; therefore, your answer is correct.

Hope this helps!

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In the following expression, n is a rational number: Which expression below is equivalent to this expression?
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none of them

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1 year ago
A bacteria culture starts with 400 bacteria and grows at a rate proportional to its size. After 4 hours, there are 9000 bacteria
Kaylis [27]

Answer:

A) The expression for the number of bacteria is P(t) = 400e^{0.7783t}.

B) After 5 hours there will be 19593 bacteria.

C) After 5.55 hours the population of bacteria will reach 30000.

Step-by-step explanation:

A) Here we have a problem with differential equations. Recall that we can interpret the rate of change of a magnitude as its derivative. So, as the rate change proportionally to the size of the population, we have

P' = kP

where P stands for the population of bacteria.

Writing P' as \frac{dP}{dt}, we get

\frac{dP}{dt} = kP.

Notice that this is a separable equation, so

\frac{dP}{P} = kdt.

Then, integrating in both sides of the equality:

\int\frac{dP}{P} = \int kdt.

We have,

\ln P = kt+C.

Now, taking exponential

P(t) = Ce^{kt}.

The next step is to find the value for the constant C. We do this using the initial condition P(0)=400. Recall that this is the initial population of bacteria. So,

400 = P(0) = Ce^{k0}=C.

Hence, the expression becomes

P(t) = 400e^{kt}.

Now, we find the value for k. We are going to use that P(4)=9000. Notice that

9000 = 400e^{k4}.

Then,

\frac{90}{4} = e^{4k}.

Taking logarithm

\ln\frac{90}{4} = 4k, so \frac{1}{4}\ln\frac{90}{4} = k.

So, k=0.7783788273, and approximating to the fourth decimal place we can take k=0.7783. Hence,

P(t) = 400e^{0.7783t}.

B) To find the number of bacteria after 5 hours, we only need to evaluate the expression we have obtained in the previous exercise:

P(5) =400e^{0.7783*5} = 19593.723 \approx 19593.  

C) In this case we want to do the reverse operation: we want to find the value of t such that

30000 = 400e^{0.7783t}.

This expression is equivalent to

75 = e^{0.7783t}.

Now, taking logarithm we have

\ln 75 = 0.7783t.

Finally,

t = \frac{\ln 75}{0.7783} \approx 5.55.

So, after 5.55 hours the population of bacteria will reach 30000.

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Answer:

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