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3 years ago

How do you do this question?

Mathematics

1 answer:

lianna [129]3 years ago

3 0

Answer:

Solution : 126π/5

Step-by-step explanation:

Step 1 (Assign the 'values') : Using the cylinder method, given that x = 0 and x = 3, we will integrate along the x-axis from 0 to 3. The radius of each cylinder will be r = x, and the height should be the following,

h = y₁ - y₂ = x³ - 8

Step 2 (Substitute) : Remember that the volume of the graph will be in the following form,

$\mathrm{V\:=\:2\pi \int _a^b\:rhdx}$

If we substitute known values we should go in the following both. Remember that r = x, h = x³ - 8, b = 3, and a = 0 (our limits). Take a look at the attachment for those calculations. As you can see our solution is 126π/5.

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Write an equation for a line parallel to y=-3x-5 and passing through the point (4,-16)

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Answer:

Insert the numbers into the equation y=3x-5.

the rule states that in every equation the first number is x.so it will be -6=3(4)-5

-6=12-5

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3 years ago

Jerneii writes the set of ordered pairs below. The set represents a function.

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Step-by-step explanation:

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2 years ago

2.The number of tickets sold to a play for each showing is 78, 84, 87, 80, 91, 95, and 80.

Luden [163]

Answer:

$Min =78$

The maximum is :

$Max = 95$

Now we can calculate the median since the sample size os n =7 the median would be the middle value at the position 4 from the dataset ordered and we got:

$Median =84$

Now we can find the quartile 1 we analyze the first 4 values 78, 80,80, 84 and the first quartile would be:

$Q_1= \frac{80+80}{2}= 80$

Now we can find the quartile 3 we analyze the last 4 values 84,87, 91, 95 and the third quartile would be:

$Q_3= \frac{87+91}{2}=89$

And finally the 5 number summary would be:

$Min = 78, Q_1 = 80, Median=84, Q_3 = 89, Max=96$

Step-by-step explanation:

We have the following data given:

78, 84, 87, 80, 91, 95, and 80.

The first step is order the dataset on increasing way and we got:

78, 80,80, 84,87, 91, 95

We can begin finding the minimum value and for this case is:

$Min =78$

The maximum is :

$Max = 95$

Now we can calculate the median since the sample size os n =7 the median would be the middle value at the position 4 from the dataset ordered and we got:

$Median =84$

Now we can find the quartile 1 we analyze the first 4 values 78, 80,80, 84 and the first quartile would be:

$Q_1= \frac{80+80}{2}= 80$

Now we can find the quartile 3 we analyze the last 4 values 84,87, 91, 95 and the third quartile would be:

$Q_3= \frac{87+91}{2}=89$

And finally the 5 number summary would be:

$Min = 78, Q_1 = 80, Median=84, Q_3 = 89, Max=96$

5 0

3 years ago

The measure of the two sides of a triangle.

Phoenix [80]

Answer:

Step-by-step explanation:

The answer is A because well 3x would not change and 17x would not change but 7 and 5 would change because they are not variable numbers So when u do the mah you would get 3x2+17x -2

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

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

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3 years ago