I assume it would be y=65,536x
0=65,536(0)
65,536=65,536(1)


- <u>A </u><u>triangle </u><u>with </u><u>sides </u><u>11m</u><u>, </u><u> </u><u>13m </u><u>and </u><u>18m</u>

- <u>We</u><u> </u><u>have </u><u>to </u><u>check </u><u>it </u><u>whether </u><u>it </u><u>is </u><u>right </u><u>angled </u><u>triangle </u><u>or </u><u>not</u><u>? </u>


According to the Pythagoras theorem, The sum of the squares of perpendicular height and the square of the base of the triangle is equal to the square of hypotenuse that is sum of the squares of two small sides equal to the square of longest side of the triangle.
<u>We </u><u>imply</u><u> </u><u>it </u><u>in </u><u>the </u><u>given </u><u>triangle </u><u>,</u>





<u>From </u><u>Above </u><u>we </u><u>can </u><u>conclude </u><u>that</u><u>, </u>
The sum of the squares of two small sides that is perpendicular height and base is not equal to the square of longest side that is Hypotenuse

First, we should figure out the area of the entire face of the clock because we need that information to solve the problem. The formula for the area of a circle is A=pi*r^2. Since we know that r (radius) is equal to 11.25 feet, we can plug this in for r and solve for A: A=pi*11.25^2 which equals A=397.61 ft^2 rounded to the nearest hundredth.
Now, to find the area the hand sweeps over in 5 minutes, we should determine how much of the clock the hand sweeps over in 5 minutes. Think about it like this: since 5 minutes goes into 60 minutes 12 times (60/5=12), then 5 minutes is one twelfth of the clock's face. Therefore, we are going to divide the total area by 12 (397.61/12) to get 33.13 ft^2, so the answer is C.
I hope this helps.
X/7-3=6
X/7=6+3
X/7=9
7x9=63
63/7-3=6
9-3=6
I would say choice “C”