Question

the area of a trapezoid a can be found using the equation $a = \frac{1}{2}h(x + y)$ where x and y are the two bases and h is the height. solve the equation for x.
farming a farmer needs to enclose several cows in a field. the farmer knows she wants a rectangular enclosure with a width of 10 meters. she has 78 meters of fencing. if she uses all her fencing, how long can the fenced enclosure be? the formula for the perimeter of a rectangle is $p = 2l + 2w$, where p is the perimeter, w is the width, and l is the length. solve the perimeter formula for l. find the length of the fence.

Explanation:

First Sub - Question (Solving \(A=\frac{1}{2}h(x + y)\) for \(x\))

Step 1: Eliminate the fraction

Multiply both sides of the equation \(A=\frac{1}{2}h(x + y)\) by 2 to get rid of the fraction.
\(2A=h(x + y)\)

Step 2: Isolate the term with \(x\)

Divide both sides of the equation \(2A = h(x + y)\) by \(h\) (assuming \(h
eq0\)).
\(\frac{2A}{h}=x + y\)

Step 3: Solve for \(x\)

Subtract \(y\) from both sides of the equation \(\frac{2A}{h}=x + y\).
\(x=\frac{2A}{h}-y\)

Step 1: Solve \(P = 2l+2w\) for \(l\)

Subtract \(2w\) from both sides of the equation \(P = 2l+2w\).
\(P-2w = 2l\)
Then divide both sides by 2.
\(l=\frac{P - 2w}{2}\)

Step 2: Substitute \(P = 78\) and \(w = 10\) into the formula for \(l\)

Substitute the values into \(l=\frac{P - 2w}{2}\), we get \(l=\frac{78-2\times10}{2}\)
First, calculate the numerator: \(78 - 2\times10=78 - 20 = 58\)
Then, divide by 2: \(l=\frac{58}{2}=29\)

Answer:

\(x = \frac{2A}{h}-y\)

Second Sub - Question (Solving \(P = 2l+2w\) for \(l\) and finding \(l\) when \(P = 78\) and \(w = 10\))