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Varvara68 [4.7K]
2 years ago
6

(DO NOT GIVE ME AN WARNING FOR THIS IT IS PART OF MY SCHOOL PROJECT)

Mathematics
2 answers:
Contact [7]2 years ago
6 0

Answer:

47400000

Step-by-step explanation:

12 million * 79/20 = 47400000

Sliva [168]2 years ago
5 0

Answer:

12 million

Step-by-step explanation:

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Yasmeen bought 16 pizzas for her grade level classes.
torisob [31]

Answer:

183.20

Step-by-step explanation:

9.95x16=159.20

159.20+24=183.20

4 0
3 years ago
Find the general solution of the differential equation and check the result by differentiation. (Use C for the constant of integ
atroni [7]

Answer: y=Ce^(^3^t^{^9}^)

Step-by-step explanation:

Beginning with the first differential equation:

\frac{dy}{dt} =27t^8y

This differential equation is denoted as a separable differential equation due to us having the ability to separate the variables. Divide both sides by 'y' to get:

\frac{1}{y} \frac{dy}{dt} =27t^8

Multiply both sides by 'dt' to get:

\frac{1}{y}dy =27t^8dt

Integrate both sides. Both sides will produce an integration constant, but I will merge them together into a single integration constant on the right side:

\int\limits {\frac{1}{y} } \, dy=\int\limits {27t^8} \, dt

ln(y)=27(\frac{1}{9} t^9)+C

ln(y)=3t^9+C

We want to cancel the natural log in order to isolate our function 'y'. We can do this by using 'e' since it is the inverse of the natural log:

e^l^n^(^y^)=e^(^3^t^{^9} ^+^C^)

y=e^(^3^t^{^9} ^+^C^)

We can take out the 'C' of the exponential using a rule of exponents. Addition in an exponent can be broken up into a product of their bases:

y=e^(^3^t^{^9}^)e^C

The term e^C is just another constant, so with impunity, I can absorb everything into a single constant:

y=Ce^(^3^t^{^9}^)

To check the answer by differentiation, you require the chain rule. Differentiating an exponential gives back the exponential, but you must multiply by the derivative of the inside. We get:

\frac{d}{dx} (y)=\frac{d}{dx}(Ce^(^3^t^{^9}^))

\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*\frac{d}{dx}(3t^9)

\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*27t^8

Now check if the derivative equals the right side of the original differential equation:

(Ce^(^3^t^{^9}^))*27t^8=27t^8*y(t)

Ce^(^3^t^{^9}^)*27t^8=27t^8*Ce^(^3^t^{^9}^)

QED

I unfortunately do not have enough room for your second question. It is the exact same type of differential equation as the one solved above. The only difference is the fractional exponent, which would make the problem slightly more involved. If you ask your second question again on a different problem, I'd be glad to help you solve it.

7 0
2 years ago
you want to predict which movie will be the most popular among the students at your school next weekend. To do so, you ask a sam
Komok [63]
Because if you ask one student from each class you are getting a more accurate representation of the classes. If you were to ask one group you'd only get their opinion
6 0
2 years ago
Find the surface area of the triangular prism shown below.
Slav-nsk [51]

Answer:

i think it is 3380

multiply all the numbers and then divide by 5

6 0
3 years ago
Read 2 more answers
An important tool in archeological research is radiocarbon dating, developed by the American chemist Willard F. Libby.3 This is
Radda [10]

Answer:

Q(t) = Q_o*e^(-0.000120968*t)

Step-by-step explanation:

Given:

- The ODE of the life of Carbon-14:

                                       Q' = -r*Q

- The initial conditions Q(0) = Q_o

- Carbon isotope reaches its half life in t = 5730 yrs

Find:

The expression for Q(t).

Solution:

- Assuming Q(t) satisfies:

                                       Q' = -r*Q

- Separate variables:

                                      dQ / Q = -r .dt

- Integrate both sides:

                                       Ln(Q) = -r*t + C

- Make the relation for Q:

                                       Q = C*e^(-r*t)

- Using initial conditions given:

                                       Q(0) = Q_o

                                       Q_o = C*e^(-r*0)

                                      C = Q_o    

- The relation is:

                                       Q(t) = Q_o*e^(-r*t)

- We are also given that the half life of carbon is t = 5730 years:

                                       Q_o / 2 = Q_o*e^(-5730*r)

                                        -Ln(0.5) = 5730*r

                                        r = -Ln(0.5)/5730

                                        r = 0.000120968          

- Hence, our expression for Q(t) would be:

                                       Q(t) = Q_o*e^(-0.000120968*t)                                    

7 0
3 years ago
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