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irina1246 [14]
3 years ago
8

A line passes through point (8,-6) and has a slope of 3/4. How do you write an equation in slope-intercept form for the line?

Mathematics
1 answer:
satela [25.4K]3 years ago
4 0

Answer:

y=3/4x - 12

Step-by-step explanation:

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No Acute
No Right
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Write an algebraic expression for the perimeter of a rectangular garden with a width of 3 metres and a length of 2y metres. Simp
anyanavicka [17]

Answer:

P = 2(2y+3)

P = 30

Step-by-step explanation:

P = 2 (L+W) or P = L + L + W + W

P = 2 (2y+3)

P = 4y+6

Plug in 6 for y.

P = 24+6

P = 30

5 0
2 years ago
PLS HELP<br> Solve the equation for m. Show your work.<br> 19+m=30
Lena [83]

Answer:

11

Step-by-step explanation:

19 + m = 30

m = 30 - 19

m = 11

Hope it helps

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3 years ago
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6 red marbles out of 36 marbles
Andreyy89

Answer:

1/6

Step-by-step explanation:

P(red) = red/total

             =6/36

             =1/6

6 0
3 years 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.

6 0
3 years ago
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