Which expression is equivalent to 63 + 27?

Mathematics

1 answer:

azamat2 years ago

5 0

The answer would be B.

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What is the sum of (2 + 6x) and (3 - x)?

Ahat [919]

Answer:

(2+6x)+(3-x)

5+5x is your answer

4 0

2 years ago

a baseball team paid eight times as much to their ace pitcher as they paid to their shortstop. if they paid 81 million for the t

GenaCL600 [577]

x +y = 81 million

let x be the short stop and y be the pitcher

so pitcher (y) gets 2 times what the the shortstop (x) gets so y = 2x

so x + 2x = 81 million

3x = 81

x = 27

the short stop makes 27 million

the pitcher makes 54 million

7 0

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

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