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erma4kov [3.2K]
2 years ago
7

Solve for h in A = ½ ah - ½ bh. Solve for h in A = ½ ah - ½ bh .​

Mathematics
1 answer:
pantera1 [17]2 years ago
7 0

Answer:

h = A / ( 1/2 a - 1/2 b)

Step-by-step explanation:

A = ½ ah - ½ bh

Factor out an h

A = h( 1/2 a - 1/2 b)

Divide each side by ( 1/2 a - 1/2 b)

A / ( 1/2 a - 1/2 b) = h ( 1/2 a - 1/2 b)/ ( 1/2 a - 1/2 b)

A / ( 1/2 a - 1/2 b) = h

h = A / ( 1/2 a - 1/2 b)

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Obtain the general solution to the equation. (x^2+10) + xy = 4x=0 The general solution is y(x) = ignoring lost solutions, if any
alukav5142 [94]

Answer:

y(x)=4+\frac{C}{\sqrt{x^2+10}}

Step-by-step explanation:

We are given that a differential equation

(x^2+10)y'+xy-4x=0

We have to find the general solution of given differential equation

y'+\frac{x}{x^2+10}y-\frac{4x}{x^2+10}=0

y'+\frac{x}{x^2+10}y=4\frac{x}{x^2+10}

Compare with

y'+P(x) y=Q(x)

We get

P(x)=\frac{x}{x^2+10}

Q(x)=\frac{4x}{x^2+10}

I.F=e^{\int\frac{x}{x^2+10} dx}=e^{\frac{1}{2}ln(x^2+10)}

e^{ln\sqrt(x^2+10)}=\sqrt{x^2+10}

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y\cdot \sqrt{x^2+10}=\int \frac{4x}{\sqrt{x^2+10}}+C

y\cdot \sqrt{x^2+10}=4\sqrt{x^2+10}+C

y(x)=4+\frac{C}{\sqrt{x^2+10}}

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3 years ago
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3 years ago
In the inequality 6a+4b>10, what could be the possible value of a if b=2?
Arlecino [84]

We are given the following inequality:

6a+4b>10

If we replace b = 2, we get:

\begin{gathered} 6a+4(2)>10 \\ 6a+8>10 \end{gathered}

Now we solve for "a" first by subtracting 8 on both sides:

\begin{gathered} 6a+8-8>10-8 \\ 6a>2 \end{gathered}

Now we divide both sides by 6

\frac{6a}{6}>\frac{2}{6}

Simplifying:

a>\frac{1}{3}

Therefore, for b = 2, the possible values of "a" are those that are greater than 1/3

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