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2 years ago

Help plz who ever helps they are nice

Mathematics

2 answers:

EleoNora [17]2 years ago

5 0

Answer:

i'll help

Step-by-step explanation:

slava [35]2 years ago

4 0

Im mean I’d be down to help

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ICE Princess25 [194]

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1 year ago

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Check whether the function yequalsStartFraction cosine 2 x Over x EndFraction is a solution of x y prime plus yequalsnegative 2

Jobisdone [24]

The question is:

Check whether the function:

y = [cos(2x)]/x

is a solution of

xy' + y = -2sin(2x)

with the initial condition y(π/4) = 0

Answer:

To check if the function y = [cos(2x)]/x is a solution of the differential equation xy' + y = -2sin(2x), we need to substitute the value of y and the value of the derivative of y on the left hand side of the differential equation and see if we obtain the right hand side of the equation.

Let us do that.

y = [cos(2x)]/x

y' = (-1/x²) [cos(2x)] - (2/x) [sin(2x)]

Now,

xy' + y = x{(-1/x²) [cos(2x)] - (2/x) [sin(2x)]} + ([cos(2x)]/x

= (-1/x)cos(2x) - 2sin(2x) + (1/x)cos(2x)

= -2sin(2x)

Which is the right hand side of the differential equation.

Hence, y is a solution to the differential equation.

6 0

3 years ago

Show me how you solve it

julia-pushkina [17]

Answer:

5^20

Step-by-step explanation:

Law of Exponent I

$\displaystyle \large{ \frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n} }$

Therefore:

$\displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4} = ({5}^{8 - 3})^{4} } \\ \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4} = ({5}^{5})^{4} }$

Law of Exponent II

$\displaystyle \large{( {a}^{m} ) ^{n} = {a}^{m \times n} }$

Thus:

$\displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4} = ({5}^{8 - 3})^{4} } \\ \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4} = ({5}^{5})^{4} } \\ \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4} = {5}^{5 \times 4} } \\ \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4} = {5}^{20} }$

7 0

2 years ago