Answer:
Solve by substitution
[y=5/2x-4]
[y= -x+3]
Substitute y = -x +3
[-x+3=5/2x-4]
Isolate x for -x+3=5/2x-4: x=2
For y = -x+3
Substitute x = 2
y= -2+3
Simplify
y=1
The solutions to the system of equations are:
y=1 x=2
How many times larger is a 10 pound dog than a hamster weighing five eighth pounds
2 answers:
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a)The total inside length of the curved ends of the track will be 25.13 meters
b)The total distance around the inside of the track will be 275.13 meters
c)The area inside the track is 10050 square meters.
<h3>What is an area?</h3>The space occupied by any two-dimensional figure in a plane is called the area. The space occupied by the circle in a two-dimensional plane is called the area of the circle.
Given that:-
A stadium has a central rectangular area 125m long by 80m wide. there are two semi-circular ends. a running track goes all the way around.
a) what is the total inside length of the curved ends of the track?
The length of the curved track means it is a perimeter of the circle so the radius of the semicircle will be 4 meters.
So the length will be = π r + πr = 2 πr
= 2 x π x 4 = 25.13
b) what is the total distance around the inside of the track?
The total distance of the track will be the sum of the length of the two semicircles and the longer side of the rectangle.
Total length = 125 + 125 + 25.13
= 275.13 meters
c) what is the area inside of the track?
The area will be equal to the area of the two semicircular tracks and the rectangular track.
Total Area = Area of semicircle + Area of semicircle + Area of rectangle
= ( π/2 ) r² + (π/2) r² + ( L x W )
= π r² + ( L x W)
= π ( 4 )² + ( 125 x 8 )
= 10050.26 square meters.
Therefore the total inside length of the curved ends of the track will be 25.13 meters. The total distance around the inside of the track will be 275.13 meters. The area inside the track is 10050 square meters.
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