4. a horse is tethered to one corner of a square plot of side 42 m, by a rope 35 m long. find i) what area it can graze ii) what area will be left ungrazed.

Answers:

i) It can graze an area of approximately 961.625 square meters

ii) The ungrazed area is roughly 802.375 square meters

note: I used pi = 3.14 which is approximate; if you want more accuracy, use more decimal digits in pi

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Explanation:

Check out the attached image below for a visual diagram.

The area of a circle is the formula

A = pi*r^2

Use r = 35 as the radius and pi = 3.14 as the approximation for pi, to get,

A = pi*r^2

A = 3.14*35^2

A = 3846.5

Then divide this by 4 to get the area of a quarter circle

A/4 = 3846.5/4 = 961.625

This is the approximate area the horse can graze

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Subtract this from 42^2 = 42*42 = 1764 because this is the area of the entire square. Doing this will get you the ungrazed area

Ungrazed area = (entire plot) - (grazed area)

Ungrazed area = (entire area of square) - (area of quarter circle)

Ungrazed area = 1764 - 961.625

Ungrazed area = 802.375

the answer is 200

step-by-step explanation:

the way the question is worded, it seems to imply the $264 interest is simple interest, not compound interest.

the simple interest for 2 years is $264; the simple interest for 1 year is $132.

$132 interest on a $600 investment means an interest rate of (132/600)*100 = 22%.

the value of $200 at 22% interest compounded annually for three years is

200%281.22%29%5e3

the amount of interest is that value, minus the original $200.

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."