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3 years ago

8

Harry was assigned 75 problems for math homework. Harry did 1/5 of them correctly. How many problems did he get correct?

2 answers:

weeeeeb [17]3 years ago

6 0

Answer:

15

Step-by-step explanation:

75 divided by 5

svlad2 [7]3 years ago

5 0

Answer:

15 problems

Step-by-step explanation:

To find how many problems were correct, find 1/5 of 75:

Divide 75 by 5:

75/5

= 15

So, Harry got 15 problems correct.

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The rate of change in the number of bacteria in a culture is proportional to the number present. In a certain laboratory experim

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(a) $N(t)=10000e^{(\frac{ln5}{10})t }$

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(c) 4.3068 min.

Step-by-step explanation:

Rate of change in the number of bacteria is proportional to the number present.

Let N is the population of bacteria.

$\frac{dN}{dt}$ ∝ N ⇒ $\frac{dN}{dt}=kt$ { k = proportionality constant}

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$N(t_{1} )$ = 20,000

and $N(t_{1}+10 )=100,000$

(a) For population growth

$N(t)=N_{0}e^{kt}=10000e^{kt}$

$N(t_1)=10,000e^{kt_1}=20,000$

$e^{kt}=2$

$ln(e^{kt_1})=ln(2)$

$kt_1=ln(2)$

$t_{1}=\frac{ln2}{k}$ ----------(1)

$N(t_1+t_{10})=100,000$

$100,000=10,000e^{k(t_1+10)}$

$10=e^{k(t_1+10)}$

$ln10=ln[e^{k(t_1+10)}]$

$k(t_1+10)=ln10$

$k(t_1)=ln10-10k$

$t_1=\frac{ln10-10k}{k}$ ----------(2)

from equation (1) and (2)

$\frac{ln_2}{k}=\frac{ln10-10k}{k}$

$ln10-ln2=10k$

$k=\frac{ln5}{10}$

so expression will be

$N(t)=10000e^{(\frac{ln5}{10})t }$

(b) for t = 20

$N_{(20)}=10,000e\frac{ln5}{10}\times 20$

= $10,000\times e^{2ln5}$

= 10,000 × 25

= 25,000

(c) Since $t_1=\frac{ln2}{k}$ [from equation (1)]

$=\frac{ln2}{\frac{ln5}{10} }$

$=\frac{ln2}{ln5}\times 10$

= 4.3068

= 4.3068 min.

6 0

3 years ago