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Korolek [52]
3 years ago
8

Which is an equation

Mathematics
2 answers:
Musya8 [376]3 years ago
8 0
Letter C is equation.
Montano1993 [528]3 years ago
4 0
The equation is option number c because someth = smth
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cestrela7 [59]

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P(X<=2) = 0.9736

Step-by-step explanation: Please find the attached file for the solution

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Dissimilar to similar fractions
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I NEED HELP ASAP!!!!!!! PLEASE HELP ME!!!!!!!!!
aniked [119]

Answer:

a = 33 degrees, b = 147 degrees

Step-by-step explanation:

The sum of all the angles in a triangle is 180 degrees, and to solve for the angle on the other side of 109 degrees is 71 degrees, so to solve for a, have 180 minus 71 and 76. Then the sum of 71 and 76 is angle b since its the result of a subtracting from 180.

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The temperature outside was -12 °. The temperature dropped 3 degrees. What is the new temperature?
tigry1 [53]

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3 years ago
Read 2 more answers
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}.

4 0
3 years ago
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