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2 years ago

I need the expression

Mathematics

2 answers:

V125BC [204]2 years ago

7 0

Answer:

-7( 8n - m )

- 7 × 8n + 7× m

-56n + 7m

Oksi-84 [34.3K]2 years ago

3 0

What you need to do is the distributive property to start with this problem

-7(8n-m)

Multiply -7 inside the parentheses, which it will turn out to be -56n

Then you should put an addition operation on -7 and -m

-56n+7m

You shouldn’t really do anything else to these variables because they are different. So just leave it to that

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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

Mandy has a total of $2.00 in change in her purse. Complete each set of coins below to show amounts equivalent to $2.00.

Montano1993 [528]

Answer:

10 nickels

9 dimes

5 quarters

Step-by-step explanation:

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2 years ago

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If 1st and 4tg terms of G.p are 500 and 32 respectively it's second term is ?

bixtya [17]

Answer:

$T_{2} = 200$

Step-by-step explanation:

Given

Geometry Progression

$T_1 = 500$

$T_4 = 32$

Required

Calculate the second term

First, we need to write out the formula to calculate the nth term of a GP

$T_n = ar^{n-1}$

For first term: Tn = 500 and n = 1

$500 = ar^{1-1}$

$500 = ar^{0}$

$500 = a$

$a = 500$

For fought term: Tn = 32 and n = 4

$32 = ar^{4-1}$

$32 = ar^3$

Substitute 500 for a

$32 = 500 * r^3$

Make r^3 the subject

$r^3 = \frac{32}{500}$

$r^3 = 0.064$

Take cube roots

$\sqrt[3]{r^3} = \sqrt[3]{0.064}$

$r = \sqrt[3]{0.064}$

$r = 0.4$

Using: $T_n = ar^{n-1}$

$n = 2$ $r = 0.4$ and $a = 500$

$T_{2} = 500 * 0.4^{2-1}$

$T_{2} = 500 * 0.4^1$

$T_{2} = 500 * 0.4$

$T_{2} = 200$

<em>Hence, the second term is 200</em>

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Find all possible answers and check your solutions.Solve equation.

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A.
lxl=7
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Using the data set provided below, order the values from greatest to least.

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The answer will be E

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