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3 years ago

What is the square root of 57? Help pleaseee

Mathematics

1 answer:

Free_Kalibri [48]3 years ago

8 0

Answer:

7.54983444

Step-by-step explanation:

hope this helps :)

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16^10 x 16^4 need the work

Alex787 [66]

Answer:

$16^{10}*16^{4} \\= 16^{14}$

Property used: $a^{m} * a^{n} = a^{m+n}$

3 0

3 years ago

Wang Yong owes the bank \$8500$8500dollar sign, 8500. To repay the debt, he paid a fixed amount back to the bank each month. Aft

34kurt

Complete question :

Wang Yong owes the bank $8500. To repay the debt, he paid a fixed amount back to the bank each month. After 12 months, his remaining debt was $6460. How much did Wang Yong pay each month?

Answer:

$170

Step-by-step explanation:

Given that:

Amount owed = $8500

Fixed amount replayed per month :

Amount left after 12 months = $6460

Amount paid per month :

Total amount paid over 12 months :

$(8500 - 6460) = $2040

Amount paid per month :

$2040 / 12

= $170

3 0

3 years ago

Read 2 more answers

Pls help question is on picture

NikAS [45]

Answer:

25 degrees

Step-by-step explanation:

5 * 5 = 25

6 0

3 years ago

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The lines on the graph below represent the cost of apples at four different stores.

katen-ka-za [31]

At Store A is the cost of apples the least. This is because it always has the lowest price for each pound compared to the other lines. Hope this helps!

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3 years ago

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

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

6 0

3 years ago