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Question

My teacher asked us some question for fun but it seem hard...

Write a formula giving the area of a circle in terms of its circumference.

Solve the equation ax+b= cx+d for x in term of a.b.c an d. iner what conditions is there no solution? under what condition are all real numbers solutions?

You are hanging fliers around a cylindrical kiosk that has a diameter of 5 feet. you want to hang 15 fliers that are 8.5 inches wide to they are evenly placed. how far apart should the fliers be placed?

Answer 1

1)

We know A=pi*r^2

We know C=2pi*r

Let's write area in terms of circumference

C=2pi*r
r= C/2pi

This is radius in terms of circumference so

A=pi*r^2
A=pi * (C/2pi)^2
A= pi * C^2/(4pi^2)
A= C^2/(4pi)


2)

ax+b = cx+ d
ax - cx = d - b
x(a-c) = d-b
x = (d-b)/(a-c)

We have no solution when a = c and b and d are different.
We all all real number solutions if a = c and b=d because it would give you the same line.

c)
The cylinder has a circular base so the distance around the kiosk is a circle.

C=2pi*r
C=pi*d
C=5pi ft

convert this to inches
you will get 60pi inches

If each flyer is 8.5 inches, 15 of them will take up 127.5 inches. The rest will be the space between them. Take 60pi - 127.5.

You will get 60.99555 inches. Divide this by 15 spaces...one in between two fliers. You'll get about 4 inches apart.

Answer 2

Write a formula giving the area of a circle in terms of its circumference.

You know that Area A = pi(r^2) and that the Circumference C=(2pi)r. Solving the second equation for r we have C/(2pi)=r. Substituting this value into the area equation we have A=pi( C/(2pi))^2. Simplifying this result we have A=(C^2)/(4pi).





Solve the equation ax+b= cx+d for x in term of a.b.c an d d. Under what conditions is there no solution? under what condition are all real numbers solutions?

ax+b= cx+d

Subtract cx and b from both sides to get
ax-cx= d-b

Factor the left side to get
x(a-c)=d-b

Divided both sides by (a-c) to get
x=(d-b)/(a-c)

If a=c AND b does NOT equal d, then you will have no solutions.

If a=c AND b=d then all real numbers will be solutions.



You are hanging fliers around a cylindrical kiosk that has a diameter of 5 feet. you want to hang 15 fliers that are 8.5 inches wide to they are evenly placed. how far apart should the fliers be placed?

The distance around the cylinder is pi(d)=5pi approximately 15.708 ft. The width of all 15 fliers is 15(8.5inches)=127.5in=10.625 ft.

Subtracting these two values we have15.708 ft -10.625 ft.=5.083ft. This is the total length of all the space between the fliers. Thus we divide this value by 15 to determine the space needed between each flier. So 5.083ft/15= approximately .3389 feet which is a little over 4 inches between each flier.

Answer 3

Well, A=Pi*r^2, and C=2*Pi*r
So, r=C/(2*Pi)
And, A=Pi*C^2/(2*Pi)^2
So, A=(Pi*C^2)/(4*Pi*Pi)
A=C^2/(4*Pi)

ax+b=cx+d
ax-cx=d-b
(a-c)x=d-b
x=(d-b)/(a-c)

If a=c, you have division by 0, which is impossible, and thus no solution. Some people would say the answer is infinity, but what if c is just a tiny bit bigger than a? The answer would be close to negative infinity. So, what's the answer? Positive or negative infinity? The answer is neither...there is no solution.

Likewise, when d=b and a=c, you get 0/0. In that case, so long as d=b and a=c, you could use any value you like for a and d, so long as a=d as well. And THAT means that so long as a=b=c=d, any real number can be substituted for a.

d=5
C=Pi*d
C=(~)3.141*5=15.705 feet, or 157.05+31.41 inches.
157.04+31.41 = 188.45 inches.

188.45/15=12.56 inches per page.
Since each page is 8.5 inches wide, 12.56-8.4 = 4.06 inches is the spacing between fliers.

Answer 4

c= 2π r = dπ
r= c/2π

A = π r²
A = π (c/2π)²
A = c² / 4π


ax+b= cx+d
ax - cx = d - b
x(a-c) = d - b
x = (d-b)/(a-c)

c= 2π r = dπ = 5π ≈ 15.707 ft OR 188.5 inches after you multiply by 12 inches/foot.

Now 188.5 inches divided by 15 to get 12.56 inches available per flier.
If each flier occupies 8.5 inches of that space you are left with 4.06 inches of space in between fliers.

12.56- 8.5 = 4.06 inches

Answer 5

area of a circle =pi r^2=1/2r(circumference)
ax+b= cx+d
ax-cx=d-b
x(a-c)=d-b
x=(d-b)/(a-c)

Answer 6

Challenge??? ........No
Homework??? .......Yes