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Ostrovityanka [42]
3 years ago
13

Solve for k. k/2 +1/2=3

Mathematics
1 answer:
ikadub [295]3 years ago
7 0

Answer:

k = 5

Step-by-step explanation:

Solve for k :

\frac{k}{2} + \frac{1}{2} = 3

-Multiply both sides of the equation by the denominator 2 :

2 \times \frac{k}{2} + \frac{1}{2} = 3 \times 2

k + 1 = 6

-Subtract both sides by 1:

k + 1 - 1 = 6 - 1

k = 5

So, the value of k is 5.

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A differential equation is an equation involving an unknown function and its derivatives. Consider the differential equation 0.
navik [9.2K]

Answer: hi your question is poorly written below is the correct question

answer :

a) y1 = Asint,   y'1 = Acost  , y"1 = -Asint

b) y2 = Bcost,   y'2 = Bsint , y"2 = - Bcost

c) y = Asint + B cost satisfies the differential equation for any constant A and B

Step-by-step explanation:

y" + y = 0

Proves

a) y1 = Asint,   y'1 = Acost  , y"1 = -Asint

b) y2 = Bcost,   y'2 = Bsint , y"2 = - Bcost

c) y3 = y1 + y2 ,   y'3 = y'1 + y'2,  y"3 = y"1 + y"2

∴ y"1 + y1 = -Asint + Asint

  y"2 + y2 = -Bcost + Bcost

  y"3 - y3 = y"1 + y"2 - ( y1 + y2 )

               = y"1 - y1 + y"2 - y2  

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Hence we can conclude that y = Asint + B cost satisfies the equation for any constant A and B

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3 years ago
The product 72 and a number (please show work)
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That's as far as you can go.  You can't combine these terms. Stop there !


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4 years ago
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