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charle [14.2K]
3 years ago
8

PLEASE HELP THIS IS A PICTURE SO IF IT DONT LOAD RIGHT AWAY PLEASE BE PACTIENT THANK YOU.

Mathematics
2 answers:
amm18123 years ago
8 0
It would land on blue approximately 500 time C

vlabodo [156]3 years ago
4 0
I think.. 500, since blue is half of the wheel.
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Please help with this! It's due soon! Braileist or however you spell it will be given! I'm not the brightest and have other work
egoroff_w [7]

Answer:

Step-by-step explanation:

So, we se that there are 8 parts in total.

So, 50/8 is 6.25

So, 6.25*3 parts, which is how many there are for blue is 18.75

7 0
2 years ago
PLEASE HELP!!!!!! ILL GIVE BRAINLIEST *EXTRA 40 POINTS** !! DONT SKIP :((
Marina CMI [18]

Answer:

No. 1st is (0,2) 2nd is (0,5)

Step-by-step explanation:

3 0
3 years ago
Someone please help me with this !!!
a_sh-v [17]

Answer:

what does it say young buck

6 0
2 years ago
Hamster111 yes please
Rufina [12.5K]

Answer:

Indeed

Step-by-step explanation:

3 0
2 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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