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2 years ago

Need help on this thank you if you help

Mathematics

1 answer:

MariettaO [177]2 years ago

4 0

Answer:

a. 2/5

b. 1/2

c. 100/9

Step-by-step explanation:

a: Multiply bottom and top by 100 to keep values the same and make the fraction a whole fraction:

.12 × 100 = 12

.3 × 100 = 30

Simplify: 12/30 = 2/5

b: Multiply bottom and top by 1,000 to keep values the same and make the fraction a whole fraction:

.125 × 1,000 = 125

.25 × 1,000 = 250

Simplify: 125/250 = 1/2

c: Multiply bottom and top by 100 to keep values the same and make the fraction a whole fraction:

4 × 100 = 400

.36 × 100 = 36

Simplify: 400/36 = 100/9

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Answer:

Step-by-step explanation:

12 divided by 1/2 = 24

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2 years ago

The orbit of a comet in a video game can be defined by the equation (x - 56)^2/150 + (y - 28)^2/256 = 1. Which of the following

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3 years ago

Can 0.2 0.6 3 and 9 make 24?

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3 years ago

Rafael received 5/6 of 180 volts how many votes did he receive

Kruka [31]

Answer: The answer would be 6

Here is an example:

If you divided 64 by 5/8, then you would get a number larger than 64.

To get the answer, multiply 64 * 5/8. So find what 64*5 is then divide that number by 8. 64*5=320. 8 goes into 32 4 times, so it goes into 320 40 times. So your answer is 40.

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3 years ago

The number of bacteria after t hours is given by N(t)=250 e^0.15t a) Find the initial number of bacteria and the rate of growth

Art [367]

Answer:

a) $N_0=250\; k=0.15$

b) 334,858 bacteria

c) 4.67 hours

d) 2 hours

Step-by-step explanation:

a) Initial number of bacteria is the coefficient, that is, 250. And the growth rate is the coefficient besides “t”: 0.15. It’s rate of growth because of its positive sign; when it’s negative, it’s taken as rate of decay.

Another way to see that is the following:

Initial number of bacteria is N(0), which implies $t=0$ . And $N(0)=N_0$ . The process is:

$N(t)=250 e^{0.15t}\\N(0)=250 e^{0.15(0)}\\ N_0=250e^{0}\\N_0=250\cdot1\\ N_0=250$

b) After 2 days means $t=48$ . So, we just replace and operate:

$N(t)=250 e^{0.15t}\\N(48)=250 e^{0.15(48)}\\ N(48)=250e^{7.2}\\N(48)=334,858\;\text{bacteria}$

c) $N(t_1)=4000; \;t_1=?$

$N(t)=250 e^{0.15t}\\4000=250 e^{0.15t_1}\\ \dfrac{4000}{250}= e^{0.15t_1}\\16= e^{0.15t_1}\\ \ln{16}= \ln{e^{0.15t_1}} \\ \ln{16}=0.15t_1 \\ \dfrac{\ln{16}}{0.15}=t_1=4.67\approx 5\;h$

d) $t_2=?\; (N_0→3N_0 \Longrightarrow 250 → 3\cdot250 =750)$

$N(t)=250 e^{0.15t}\\ 750=250 e^{0.15t_2} \\ \ln{3} =\ln{e^{0.15t_2}}\\ t_2=\dfrac{\ln{3}}{0.15} = 2.99 \approx 3\;h$

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3 years ago