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3 years ago

How can these fractions be rewritten so they can be added together?

Mathematics

1 answer:

just olya [345]3 years ago

6 0

Answer:

8/12 + 3/12

Step-by-step explanation: multiply 2 and 3 by 4 to add these two fractions correctly.

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What is the value of x?

bonufazy [111]

Answer:

35 degrees

Step-by-step explanation:

7 0

4 years ago

Carly works less than 20 hours per week helping her grandfather on his farm.

horrorfan [7]

Answer:

h>20

Step-by-step explanation:

I'm not sure though

3 0

3 years ago

A baseball player swings and hits a pop fly straight up in the air to the catcher. The height of the baseball in meters t second

tresset_1 [31]

To solve for the time it reach the maximum height, you must solve the first derivative of the function and equate it to zero

h(t) = −4.9t^2 + 14.7t + 1
dh/ dt = -9.8t + 14.7
then equate to zero
-9.8t + 14.7 = 0
solve for t
t = 1.5 s

then the maximum height is when t = 1.5

h(t) = −4.9t^2 + 14.7t + 1
h(1.5) = −4.9(1.5)^2 + 14.7(1.5) + 1
h(1.5) = 12.025 m

4 0

3 years ago

What is the median of the data set 11,13,15,21,23,31,31,33? A. 31 B. 21 OC. 23 D. 22

SSSSS [86.1K]

Median:

The median in this set is the 5th term.

Therefore the median in this set is 23.

3 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