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3 years ago

Simplify the expression: 9a – 7a + 4 – 4 + 8

Mathematics

1 answer:

Ksju [112]3 years ago

8 0

Answer:

2a + 8

Step-by-step explanation:

9a - 7a + 4 - 4 + 8

=> 2a + 12 - 4

=> 2a + 8

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Simplify the ratio 9/12

Alex777 [14]

3:4 This is the answer

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3 years ago

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Which statement is correct?

fenix001 [56]

The second one that you did

5 0

3 years ago

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PLEASE HELP

kolbaska11 [484]

Answer:

a) P(x) = 32000*1.04^x
b) $37435
c) During year 7

Step-by-step explanation:

Given

Initial pay = $32000
Increase rate = 4%

a. Formula

P(x) = 32000*1.04^x

b. Year 5 is after 4 years, so we are looking for the value of P(4)

P(4) = $32000*1.04^4 = $37435

c. P(x) = 40000, x = ?

40000 = 32000*1.04^x
1.04^x = 40000/32000
1.04^x = 1.25
log 1.04^x = log 1.25
x = log 1.25 / log 1.04
x = 5.69, this is 6 years after

The required number of the years is 6 + 1 = 7

3 0

3 years ago

Can u pls help I don’t understand I’ll give u 15 points

Juliette [100K]

Answer: $\frac{4}{3}$

Step-by-step explanation:

This is a multiplication problem. You are multiplying $\frac{1}{3}$ by 4. This also means 4 divided by 3. They are both the same.

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

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

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