Just ONE Question Please?!
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.
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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.
I really don't know how to start this - so if you could guide me.
@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 \[\frac{ a }{ wh }\div \frac{ 0.5 }{ bh }=b\]
Or anyone looking.
\(0.5=\dfrac{1}{2}\) \(\dfrac{1}{2} \times bh = \dfrac{bh}{2}\) \(A=wh+\dfrac{bh}{2}\) Then solve for b.
I don't really get this. Are these steps?
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.
\(A-wh>\cancel{wh-wh}+\dfrac{bh}{2}\) \(A-wh>\dfrac{bh}{2}\)
The goal is to isolate b, so multiply both sides by 2, then divide by h.
Oh okay. Well, we are opposite. I hate fractions in my questions - but it is bearable. Go on.
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.
\(A-wh>\cancel{wh-wh}+0.5bh\) \(A-wh = 0.5bh\)
Okay. I am getting it now. Sorry for the decimal fraction thing.
Now I divide 0.5, right?
No need to apologize. It's your homework, do it the way you feel comfortable. Do you have the answer?
Then divide h?
Exactly.
Awesome, so it would be \[a-wh \div \frac{ 0.5 }{ h }=b\] P
ut it in fraction form. :3
Working with the decimal makes it more complicated. Let me show you:
\(A-wh>\cancel{wh-wh}+\dfrac{bh}{2}\) 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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