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3 years ago

Which equation is solved for s?

Mathematics

1 answer:

Allushta [10]3 years ago

6 0

Answer:

depends

Step-by-step explanation:

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ethan earned 5% of the maximum number of points possible . It ethan earned 40 points , what is the maximum number of points poss

Morgarella [4.7K]

Answer:

3840

Step-by-step explanation:

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3 years ago

Value 23.7 is 1/10 of

inna [77]

Is 19:17 because u have to value the number of the first like this 78:807y :997

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3 years ago

An author released his book on May 1. His daily sales steadily increased until they peaked and then steadily declined. He found

dem82 [27]

The domain of a certain function are all possible values of x that are applicable to the function or can make the function true. For this certain function, the domain is from the time that the book was released to the time that the sales was also zero due to steady decline in sales.

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3 years ago

Read 2 more answers

Cathy rolls a number cube that is labeled from 1 to 6.

MatroZZZ [7]

Answer: 2/3

Cathy rolls a number cube that is labeled from 1 to 6.

Total outcomes = {1,2,3,4,5,6}

she rolls an even number or a multiple of 3

Outcomes of an even number = {2,4,6}

Outcomes of multiple of 3 = {3,6}

so favorable outcomes of even number or a multiple of 3 = {2,3,4,6}

Probability ( any event) =number of favorable outcomes divide by total number of outcomes

probability(even number or a multiple of 3) = $\frac{4}{6}=\frac{2}{3}$

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3 years ago

Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview

mario62 [17]

Answer: The required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.$

So, the general solution of the given equation is

$y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.$

According to the given conditions, we have

$y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

and

$y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.$

From equation (ii), we get

$B=\dfrac{7}{3}.$

Thus, the required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

7 0

3 years ago