All categories

2 years ago

What is the operation of (ab)c = c(ab)

Mathematics

1 answer:

Kamila [148]2 years ago

6 0

Answer:

ican help you bat ican not uderstand

You might be interested in

How to write an inequality stating the restrictions of x

inessss [21]

For example
1>x>3
says x is less than 1 and greater than 3

6 0

3 years ago

Help plz i need it :(

mr_godi [17]

Answer:

Step-by-step explanation:

6 0

3 years ago

The value of x is 4 times the value of y.how to writ that in math

kakasveta [241]

I’m not exactly sure what you mean by “write it in math,” however I can tell you what I think.
4y=x
(It says that x is 4 times the value of y which means that y times 4 is x.)

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

Find the point on the sphere closest to the plane

Ratling [72]

Use the concept of Lagrange multipliers.

3 0

2 years ago