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4 years ago

6

6 is a factor of 133 84 116 118

1 answer:

tatiyna4 years ago

6 0

B.) 84
The reason why is because 6x14= 84

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In a writing competition, the first place winner receives 1/2 of the prize money. The second runner up receives 1/4 of what the

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Answer:

10,000 is the total amount of.money that was distributed

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3 years ago

2. A triangle has sides of lengths 15, 23, and 28. Is it a right triangle? Explain.

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Gvknj h executivesbsinsubs

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3 years ago

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At a diner, the lunch special gives you three choices. For the starter, you can pick from 3 choices (soup or salad or fries) For

MAXImum [283]

Answer:

1/12; 1/6

Step-by-step explanation:

1. the chance of getting soup is 1/3, the chance of getting a burger is 1/2, and the chance of getting coke is 1/2, multiply these chances to get (1/3) *(1/2) * (1/2) = 1/12

2. the chance of getting fries is 1/3, the chance of getting a burger is 1/2, and the chance of getting any drink is 1, multiply these chances to get (1/3) * (1/2) * 1 = 1/6

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2 years ago

Find the solution of the given initial value problem:<br><br> y''- y = 0, y(0) = 2, y'(0) = -1/2

igor_vitrenko [27]

Answer: The required solution of the given IVP is

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

Step-by-step explanation: We are given to find the solution of the following initial value problem :

$y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.$

Let $y=e^{mx}$ be an auxiliary solution of the given differential equation.

Then, we have

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

Substituting these values in the given differential equation, we have

$m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.$

So, the general solution of the given equation is

$y(x)=Ae^x+Be^{-x},$ where A and B are constants.

This gives, after differentiating with respect to x that

$y^\prime(x)=Ae^x-Be^{-x}.$

The given conditions implies that

$y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

and

$y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

Adding equations (i) and (ii), we get

$2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.$

From equation (i), we get

$\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.$

Substituting the values of A and B in the general solution, we get

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

Thus, the required solution of the given IVP is

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

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3 years ago

Simplify. Write your answer in standard form.<br><br><br>(4x^3+6x−7)+(3x^3−5x^2−5x+9)

UkoKoshka [18]

answer: 7x^ - 5x^2 + x + 2

step-by-step explanation: i think thats it

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3 years ago