CPhill

Here's another way to see this.....if the perimeter = 66, then a side must = 11.

And we have 12 congruent right triangles each having a hypoteneuse of 11 (the radius) and a leg (1/2) the length of a side = 5.5.   And the apothem would form the other leg.

So we have...

apothem^2 = 11^2 - (5.5)^2

apothem^2 = (2^2)(5.5)^2 - (5.5)^2 = (4)(5.5)^2 - (5.5)^2    ....factor out the (5.5)^2......

apothem^2 = (5.5)^2 * (4 -1)

apothem^2 = (5.5)^2 * (3)  ......      take the square root of each side  .....

apothem = 5.5√3  

Just as jboy314 found !!!!

Have you tried the on-site calculator?? I think it's still working OK.....last time I checked......

Thanks, Anonymous....genius in simplicity...!!!!

Love the "Scary Homework" one....that's how I used to feel about it, too!! (Hey, doesn't everyone???)

"Points and Thumbs" from me......

This might seem a little more daunting than it really is....what we're basically saying is that when we take some number "3d" and we divide it by 1288, we get a remainder of "1."

Or, put another way..........3d = 1288n + 1

So we're really looking for some "n" such that 1288n + 1 is divisible by 3.

Note that 1288(1) + 1  = 1289   .... And when I can add the digits of a number and that sum is divisible by 3, then that number is divisible by 3. So when we add the digits of 1289, we get 20,and that's not divisible by 3, so 1289 is not divisible by three.

Now, note that 1288(2) + 1  = 2577. And when we add the digits of 2577, we get 24. That's divisible by 3, so 2577 is divible by 3, and that's what we need!!

So 2577/3 = 859 = d....just as Alan and heureka found!!!

And, as noted by Alan, 3d = 1288n + 1 will hold true for positive integers d and n whenever n = 3k + 2 for k = 0, 1, 2.......

heureka's answer is a good one....how might we arrive at that????

We note that:

tan(41) = h/x     and that   tan(48) = h/(140-x)   where "x" is the horizontal distance from the side of the bridge where the 41 degree angle of depression is observed to the point where the perpendicular line representing the height intersects the bridge and  "h" is the height.

Using the second  equation, we can write h = (140-x)tan(48) and substituting this into the first equation for h, we have

tan(41) = (140-x)tan(48)/x    multiply both sides by x

xtan(41) = (140-x)tan(48)     simplifying the right, we have

xtan(41) = 140tan(48) - xian(48)   add xtan(48) to both sides

xtan(41) + xtan(48) = 140tan(48)   simplify the left side

x [tan(41) + tan(48)] = 140tan(48)   divide both sides by [tan(41) + tan(48)]

x = 140tan(48) / [tan(41) + tan(48)] 

And using the frist equation, we can write

h = xtan(41)

And substituting in for "x"  we have

h = 140tan(48) / [tan(41) + tan(48)] tan(41)  =

[tan(48)tan(41)] / [tan(41) + tan(48)] * 140 = about 68.2669m or 68.27m or just 68.3m  (rounded to nearest tenth)......which is pretty much what heureka found!!!

Here's a picture to demonstrate the situation:

When I see the term "rational function" I think of something like this .... r(x) = f(x) / g(x)

Do you mean this???..........

r(x) = (x^2 -16) / x^2

If so, the x intercepts that jboy314 gave are correct.......there are no y intercepts because the function does not exist when x = 0......here's the graph

I'm taking that you mean "arc length"

The arc length (S) is given by.....

S = rθ     where r is the radius and "theta" is in radians

So 60 ° = pi/3 rads   and we have

S = 32(pi/3) = about 33.51 inches

And 135 °  = (3/4)pi rads

So we have

S = 32(3/4)pi = about 75.4 inches

And for 5/4pi rads, we have

S = 32(5/4)pi = about 125.7 inches

Let's do "b" first.....since this is an octagon. the measure of <AOB = 360/8 = 45

And m<AOX is 1/2 of that = 22.5

And that's a  and b

c) Note that the measure of <OAX = 90- 22.5 = 67.5

And we can find OX by the Law of Sines

OX/sin67.5 = OA/sin90 ......  (sin 90 = 1)

OX/sin67.5 = 9

OX = 9sin67.5 = about 8.3m

d) We can find AB with the Law of Cosines

AB^2 = OA^2 + OB^2 - 2(OA)(OB)cos45

AB^2 = 9^2 + 9^2 - 2(9)(9)cos45

AB^2 = 162 - 162cos45 = 47.448701447706    take the square root of both sides

AB =  about 6.9m

e) The perimeter = 8(6.9)= about 55.2 m

f) Area...we have 8 congruent triangles....the area of each is given by (1/2)(9^2)sin45 = 28.6378246380735m^2

So.... 8*28.6378246380735 = about 229m^2

Hope I haven't made any serious errors here.....!!!

The radio would be located at (-5, -3) in the Cartesian Plane and would have a range of 13 units.

Basically, the situation is modeled by a circle with a center at (-5, -3) and a radius of 13.