4 2/7 30/7=4 with a remainder of two. The 2 goes as the numerator over the denominator of 7.
To get which design would have maximum area we need to evaluate the area for Tyler's design. Given that the design is square, let the length= xft, width=(120-x)
thus:
area will be:
P(x)=x(120-x)
P(x)=120x-x²
For maximum area P'(x)=0
P'(x)=120-2x=0
thus
x=60 ft
thus for maximum area x=60 ft
thus the area will be:
Area=60×60=3600 ft²
Thus we conclude that Tyler's design is the largest. Thus:
the answer is:
<span>Tyler’s design would give the larger garden because the area would be 3,600 ft2. </span>
I am pretty sure you can only make one.
If two sides are known, and one of the angles, then the other bits can be deduced and are fixed.
The radius of the circle is 15 cm,
The diameter of the circle is 30 cm,
The circumference of the circle is 94.248 cm,
The area of the circle is 706.86 cm^2
The radius is given in the diagram as half the circle, which is 15 cm.
The diameter is double the radius because the diameter measures the circle from edge to edge, so 15•2=30 cm.
The circumference of the circle is 2•3.14•r=C,
2•3.14=6.28, 6.28•15= 94.248 cm.
The area of the circle is 3.14•r^2, so the radius squared is 225 (15•15) and 225•3.14=706.86 cm squared :)
I don't think it is. I may be wrong though
