Let:

\(\displaystyle Length=x\Rightarrow Width=x-3\)

The perimeter of a rectangle is \(\displaystyle P=2L+2W\), where \(\displaystyle L\) is the length and \(\displaystyle W\) is the width of the rectangle. The perimeter is known so we can set up an equation in terms of \(\displaystyle x\) and solve it:

\(\displaystyle P=2L+2W=2(x)+2(x-3)=36\)

\(\displaystyle \Rightarrow2x+2x-6=36\)

\(\displaystyle \Rightarrow 4x-6=36\)

\(\displaystyle \Rightarrow 4x=42\)

\(\displaystyle \Rightarrow x=10.5\)

So we can get:

\(\displaystyle Length=x=10.5\) inches

\(\displaystyle Width=x-3=10.5-3=7.5\) inches

The perimeter of a rectangle is \(\displaystyle P=2L+2W\), where \(\displaystyle L\) is the length and \(\displaystyle W\) is the width of the rectangle. In order to find the perimeter we can substitute the \(\displaystyle L=t+1\) and \(\displaystyle W=t-1\) in the perimeter formula:

\(\displaystyle P=2L+2W\)

\(\displaystyle =2(t+1)+2(t-1)\)

\(\displaystyle =2t+2+2t-2\)

\(\displaystyle =4t\)

The length of the rectangle is known, so we can find the width in terms of \(\displaystyle x\):

 \(\displaystyle Width=Length-3=(2x+4)-3=2x+1\)

The perimeter of a rectangle is \(\displaystyle P=2L+2W\), where \(\displaystyle L\) is the length and \(\displaystyle W\) is the width of the rectangle.

In order to find the perimeter we can substitute the \(\displaystyle L=2x+4\) and \(\displaystyle W=2x+1\) in the perimeter formula:

\(\displaystyle P=2L+2W\)

\(\displaystyle =2(2x+4)+2(2x+1)\)

\(\displaystyle =4x+8+4x+2\)

\(\displaystyle =8x+10\)

Substitute \(\displaystyle t=5\) to get \(\displaystyle L\) and \(\displaystyle W\):

\(\displaystyle L=t^2=5^2=25\)

\(\displaystyle W=5t=5\times 4=20\)

The perimeter of a rectangle is \(\displaystyle P=2L+2W\) , where \(\displaystyle L\) is the length and \(\displaystyle W\) is the width of the rectangle. So we have:

\(\displaystyle P=2L+2W=2\times 25+2\times20=50+40=90\) inches

Now we should divide the perimeter by 12 in order to convert to feet:

\(\displaystyle Perimeter=90\div 12=7.5\) feet

So the perimeter is 7 feet and 6 inches, which is between 7 and 8 feet.

The perimeter of a rectangle is twice the sum of its length and its width; a rectangle with dimensions 55 inches and 15 inches has perimeter 

\(\displaystyle 2 (55+15) = 140\) inches.

All of the polygons in the choices are regular - that is, all have congruent sides - and all have sidelength two feet, or 24 inches, so we divide 140 by 24 to determine how many sides such a polygon would need to have a perimeter equal to the rectangle. However, 

\(\displaystyle 140 \div 24 = 5 \frac{5}{6}\),

so there cannot be a regular polygon with these characteristics. All of the choices fail, so the correct response is that none are correct.

The area of Mark's lawn is \(\displaystyle 225 \cdot245 = 55,125\). The amount of grass seed he needs is \(\displaystyle 55,125 \div 400 \approx 137.8\) pounds.

He has two options.

Option 1: he can buy three fifty-pound bags for \(\displaystyle \$45 \cdot3 = \$135\)

Option 2: he can buy two fifty-pound bags and four ten-pound bags for \(\displaystyle \$45 \cdot 2 + \$13 \cdot 4 = \$142\)

The first option is the more economical. 

The area of a rectangle is given by multiplying the width times the height. As a formula:

\(\displaystyle Area=wh\)

Where:

\(\displaystyle w\) is the width and \(\displaystyle h\) is the height. So we can get:

\(\displaystyle Area=wh=(t+1)(t-1)=t^2-1\)

The area of a parallelogram is given by:

\(\displaystyle Area=ba\)

Where \(\displaystyle b\) is the base length and \(\displaystyle a\) is the corresponding altitude. In this problem we have:

\(\displaystyle b=2a\)

or

\(\displaystyle a=\frac{b}{2}\)

So the area of the parallelogram would be:

\(\displaystyle Area_{(Parallelogram)}=ba=b\times \frac{b}{2}=\frac{b^2}{2}\)

The area of a square is given by:

\(\displaystyle Area=x^2\)

weher \(\displaystyle x\) is the side length of a square. In this problem we have \(\displaystyle b=x\), so we can write:

\(\displaystyle Area _{(Square)}=x^2=b^2\)

Then:

\(\displaystyle \frac{Area_{(Parallelogram)}}{Area_{(Square)}}=\frac{\frac{b^2}{2}}{b^2}=\frac{1}{2}\)

or:

\(\displaystyle {Area_{(Square)}}=2\cdot {Area_{(Parallelogram)}}\)

Solution 1:

We can divide the rectangle width and height by the square side length and multiply the results:

rectangle width \(\displaystyle \div\) square length = \(\displaystyle 10\div 2=5\)

rectangle height\(\displaystyle \div\)square length = \(\displaystyle 4\div 2=2\)

\(\displaystyle 5\times 2=10\)

Solution 2:

As the results of the division of rectangle width and height by the square length are integers and do not have a residual, we can say that the squares can be perfectly fitted in the rectangle. Now in order to find the number of squares we can divide the rectangle area by the square area:

Rectangle area = \(\displaystyle 4\times 10=40\) square inches

Square area = \(\displaystyle 2\times 2=4\) square inches

So we can get:

\(\displaystyle 40\div 4=10\)

The area of a rectangle is given by multiplying the width times the height. That means:

\(\displaystyle Area=wh\)

where:

\(\displaystyle w=\) width and \(\displaystyle h=\) height.

We know that: \(\displaystyle w=h+2\Rightarrow h=w-2\). Substitube the \(\displaystyle w\) in the area formula:

\(\displaystyle Area=80=wh=(h+2)\times h\Rightarrow 80=h^2+2h\Rightarrow h^2+2h-80=0\)

Now we should solve the equation for \(\displaystyle h\):

\(\displaystyle h^2+2h-80=0\Rightarrow h=-1\pm \sqrt{1^2+1\times 80}=-1\pm \sqrt{81}=-1\pm 9\)

The equation has two answers, one positive \(\displaystyle (h=8)\) and one negative \(\displaystyle (h=-10)\). As the length is always positive, the correct answer is \(\displaystyle h=8\) inches.