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3 years ago

V=s^2+ 1/2sh, solve for H.

1 answer:

vladimir1956 [14]3 years ago

3 0

V = s^2 +1/2sh
Reorder so common terms are on one side.
s^2 +(sh)/2 = v
Subtract s^2 from both sides.
(sh)/2 = -s^2 +v
Multiply both sides of the equation by 2.
sh = -s^2 *2 +v *2
Simplify each term.
sh = -2s^2 +2v
Divide each term by s and simplify.

h = -2s + (2v)/s

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|1/2b-8|=|1/4b-1|<br> b=____ and ____

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b = 12 and 28

Step-by-step explanation:

The absolute value equation |1/2b-8| = |1/4b-1| resolves to a piecewise linear function with three pieces. There are two solutions.

<h3>Domains</h3>

The absolute value function on the left has a turning point where its value is zero:

1/2b -8 = 0

b -16 = 0

b = 16

The absolute value function on the right has a turning point where its value is zero:

1/4b -1 = 0

b -4 = 0

b = 4

For b > 16, both absolute value functions are identity functions. In this domain, the equation is ...

1/2b -8 = 1/4b -1

For 4 < b < 16, the function on the left negates its argument, so the equation in this domain is ...

-(1/2b -8) = 1/4b -1

For b < 4, both functions negate their arguments, so the equation in this domain is ...

-(1/2b -8) = -(1/4b -1)

For the purpose of finding the value of b, this is effectively identical to the equation for b > 16. (The value of b does not change if we multiply both sides of the equation by -1.)

<h3>Solutions</h3>

<u>Domain b < 4 ∪ 16 < b</u>

1/2b -8 = 1/4b -1

2b -32 = b -4 . . . . . . . . multiply by 4

b = 28 . . . . . . . . . . . . add 32-b to both sides

This solution is in the domain of the equation, so is one of the solutions to the original equation.

<u>Domain 4 < b < 16</u>

-(1/2b -8) = 1/4b -1 . . . . equation in this domain

-2b +32 = b -4 . . . . . . multiply by 4

36 = 3b . . . . . . . . . . . add 2b+4 to both sides

12 = b . . . . . . . . . . . . divide by 3

This solution is in the domain of the equation, so is the other solution to the original equation.

<h3>Graph</h3>

For the purposes of the graph, we have define the function g(b) to be the difference of the two absolute value functions. The solutions are found where g(x) = 0, the x-intercepts. The graph shows those to be ...

b = 12 and b = 28

<em>Additional comment</em>

Defining g(b) = |1/2b-8| -|1/4b-1|, we can rewrite it as ...

$g(b)=\begin{cases}7-\dfrac{1}{4}b&\text{for }b < 4\\-\dfrac{3}{4}b+9&\text{for }4\le b < 16\\\dfrac{1}{4}b-7&\text{for }16\le b\end{cases}$