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Rudiy27
2 years ago
8

I need help finding the value of n​

Mathematics
2 answers:
Degger [83]2 years ago
6 0
8n+3+9n+7=180
17n+10=180
17n=170
Answer
N=10
ivolga24 [154]2 years ago
4 0

Answer:

lakes 24

Step-by-step explanation:

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Which inequality matches the given situation if w represents weight in pounds:Baby tigers can be no larger than 4 pounds.
ozzi

Given:

w be the weight in pounds of a baby tigers.

To find:

The inequality if baby tigers can be no larger than 4 pounds.

Solution:

Baby tigers can be no larger than 4 pounds. It mean weight of baby tigers cannot be greater than 4. In other words, the weight of the baby tigers must be less than or equal to 4.

Let w represents weight in pounds and the weight of the baby tigers must be less than or equal to 4. So,

w\leq 4

Therefore, the required inequality is w\leq 4.

4 0
3 years ago
Please help me ASAP!!!
Marrrta [24]

Answer:

the range represents the number of users each month for 24 months

Step-by-step explanation:

this is because the graph shows a 24 month period.

4 0
3 years ago
Find the probability of spinning an A 3 times
Dima020 [189]
If the probability of spinning an A once is 1/12, then the probability of spinning an A 3 times is 3 × 1/12.

3 × 1/12 = 3/36

Simplify

3/36 = 1/12
6 0
3 years ago
Evaluate the expression ab for a = 10 and b = 4.
xz_007 [3.2K]

A x B

= 10 x 4

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Therefore answer is 40

7 0
3 years ago
Read 2 more answers
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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