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gtnhenbr [62]
3 years ago
14

HELP ASAP DUE AT 10:10

Mathematics
2 answers:
lozanna [386]3 years ago
5 0

Answer:

1) 128j

2)60k -32

adelina 88 [10]3 years ago
5 0
Sorry but I don’t know that
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12 points What’s 1+1
aliya0001 [1]

Answer:

2

Step-by-step explanation:

are you siruous

3 0
2 years ago
Read 2 more answers
17 13/18 divided by 2 7/9
erik [133]

let's firstly convert the mixed fractions to improper fractions and then divide.

\bf \stackrel{mixed}{17\frac{13}{18}}\implies \cfrac{17\cdot 18 +13}{18}\implies \stackrel{improper}{\cfrac{319}{18}}~\hfill \stackrel{mixed}{2\frac{7}{9}}\implies \cfrac{2\cdot 9+7}{9}\implies \stackrel{improper}{\cfrac{25}{9}} \\\\[-0.35em] ~\dotfill

\bf \cfrac{319}{18}\div \cfrac{25}{9}\implies \cfrac{319}{\underset{2}{~~\begin{matrix} 18 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}\cdot \cfrac{\stackrel{1}{~~\begin{matrix} 9 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}{25}\implies \cfrac{319}{50}\implies 6\frac{19}{50}

7 0
3 years ago
​_____ represent numerical data by varying the thickness and pattern of vertical lines.
Georgia [21]
The answer is 
Bar codes

4 0
3 years ago
What is each number in scientific notaion
ruslelena [56]

For the first one it is 5 x10^{8}  

and for the second one it is 0.0005 = 5 × 0.0001 = 5 × 10−4

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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