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3 years ago

How to evaluate 3ac if a=4 and c=7

Mathematics

2 answers:

timofeeve [1]3 years ago

8 0

The answer is 1235.67 because I said so

Klio2033 [76]3 years ago

3 0

You Times 3 by 4 times 7=3x4x7=84

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Directions for (1-5): Plot the following on a number line 0.75 1.5 -2 ¾ -3 ⅕ 6.8

lana [24]

Answer:

|---|--------|---------|------|---------|-|

-2 ¾ -3 ⅕ 0.75 1.5 6.8

(This is a rough answer, it mainly just shows the order).

Step-by-step explanation:

1. -2 ¾

2. -3 ⅕

3. 0.75

4. 1.5

5. 6.8

7 0

2 years ago

Does arithmetic sequence demonstrate common difference or common ratio

makkiz [27]

Arithmetic sequences have a common difference (addition)
geometric sequences have a common ratio (multiplication)

4 0

3 years ago

Please helppp!!!!!!!

rjkz [21]

Answer:

Step-by-step explanation:

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

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