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Mathematics 19 Online
OpenStudy (kewlgeek555):

Just ONE Question Please?!

OpenStudy (kewlgeek555):

IV. Marcie wants to enclose her yard with a fence. Her yard is in the shape of a triangle attached to a rectangle. See the figure below.

OpenStudy (kewlgeek555):

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OpenStudy (kewlgeek555):

The area of this figure can be found by the formula A = (wh) + 0.5(bh). If Marcie wants the total area to be larger than a specified value, she can use the formula A > (wh)+ 0.5(bh). Rewrite this formula to solve for b. Show all steps in your work.

OpenStudy (kewlgeek555):

I really don't know how to start this - so if you could guide me.

OpenStudy (kewlgeek555):

@mathstudent55 Is this correct how I solved this problem? A > (wh) + 0.5(bh) Okay, so first, this is how I do it. I ignore the inequality symbol and I just think of it as an equal sign for now. But I already have in mind that if I multiply or divide, I have to flip the sign. A = (wh) + 0.5(bh) A ÷ (wh) = 0.5bh A ÷ (wh) ÷ 0.5 = bh A ÷ (wh) ÷ 0.5 ÷ bh = b OR \[\frac{ a }{ wh }\div \frac{ 0.5 }{ bh }=b\]

OpenStudy (kewlgeek555):

Or anyone looking.

OpenStudy (anonymous):

\(0.5=\dfrac{1}{2}\) \(\dfrac{1}{2} \times bh = \dfrac{bh}{2}\) \(A=wh+\dfrac{bh}{2}\) Then solve for b.

OpenStudy (kewlgeek555):

I don't really get this. Are these steps?

OpenStudy (anonymous):

Just demonstrating equivalencies. You could have kept the decimal, I suppose, but I find it easier to use fractions. I've just rewritten the equation. The first step would be to subtract wh from both sides.

OpenStudy (anonymous):

\(A-wh>\cancel{wh-wh}+\dfrac{bh}{2}\) \(A-wh>\dfrac{bh}{2}\)

OpenStudy (anonymous):

The goal is to isolate b, so multiply both sides by 2, then divide by h.

OpenStudy (kewlgeek555):

Oh okay. Well, we are opposite. I hate fractions in my questions - but it is bearable. Go on.

OpenStudy (anonymous):

OK - sticking with decimals then... \(A > (wh) + 0.5(bh)\) The process is very similar. 1. subtract wh from both sides. 2. divide both sides by 0.5 3. divide both sides by h.

OpenStudy (anonymous):

\(A-wh>\cancel{wh-wh}+0.5bh\) \(A-wh = 0.5bh\)

OpenStudy (kewlgeek555):

Okay. I am getting it now. Sorry for the decimal fraction thing.

OpenStudy (kewlgeek555):

Now I divide 0.5, right?

OpenStudy (anonymous):

No need to apologize. It's your homework, do it the way you feel comfortable. Do you have the answer?

OpenStudy (kewlgeek555):

Then divide h?

OpenStudy (anonymous):

Exactly.

OpenStudy (kewlgeek555):

Awesome, so it would be \[a-wh \div \frac{ 0.5 }{ h }=b\] P

OpenStudy (kewlgeek555):

ut it in fraction form. :3

OpenStudy (anonymous):

Working with the decimal makes it more complicated. Let me show you:

OpenStudy (anonymous):

\(A-wh>\cancel{wh-wh}+\dfrac{bh}{2}\) 1. Subtract wh from both sides: \(A-wh>\dfrac{bh}{2}\) 2. Multiply both sides by 2: \(2A-2wh>bh\) or \(2(A-wh)>bh\) 3. Divide both sides by h: \(\dfrac{2A-2wh}{h} >b\) or \(\dfrac{2(A-wh)}{h}>b\)

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