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One thing you might try is to compare evaporation rates for various water scenarios.

For example pour a gallon of water into buckets with

a) nothing

b) mostly gravel

c) mostly sand

d) a 50/50 mix

e) a bunch of ordinary sponges

weigh the buckets after you've done this.

Put these buckets into the same area and record the weight of the buckets vs. time

Since the gravel/sand/sponges don't change weight you are essentially measuring the evaporation rate.

Do this all carefully including measurement error.  Grind your data through standard analysis routines to 

come up with the rate +/- your final measurement error.

Present it all cleanly and clearly.

Cheap and easy experiment.  Gives you a chance to do some real data analysis.  This is actually a fairly important real life problem.

First...realize that 180-150=30 degrees, since it is a straight line. Since the angles in a triangle add up to 180, the angle in the topmost triangle is 180-30-45=105. Again, there is a straight line, so 180-105=95 degrees(one of the angles in the quadrilateral.)

Next, by opposite angle, the triangle to the leftmost has angles 30 degrees and 50 degrees, so the remaining measure is 180-30-50=100 degrees. And, 180-100=80 degrees.

Now, that we filled up the angles in the quadrilateral and trying to solve for x, we get 360-130-95-80=55 degrees.

The angle supplementary to 130° = 50°

So...the missing angle in the top triangle = 180 - 45 - 50 = 85°

And the angle supplemental to this is 95°

So....x = 180 - 95 - 30  =  180 - 125  =  55°

I sometimes forget that, too! 

Thank you! 

Ah....thanks, Tetre.....I always forget that they want the perimeter of the whole sector....not just the "curved" part

Best Answer !!!

Supplementary angles add to 180deg

Interior angles of a triangle add to 180deg

Interior angles of a quadrilateral add to 360deg

Rearranging, we have 3x + 4y = x - 8y + 24= 3x-x=-12y+24.

Therefore, 2x=-12y+24 and \(x=-6y+12.\)

50 bucks

The perimeter of the semi-circle is half the circumference plus the diameter.

In this case, we have three as the radius, so the diameter is 6 inches long, and the perimeter around the arc is \(\frac{2\pi*3}{2}=3\pi.\) Thus, the answer is \(\boxed{3\pi+6}.\)

what's your budget?

Semi-circle perimeter =   pi * radius   =

pi * 3 =

3 pi  cm

EDIT....see tertre's answer below.....mine is flawed!!!

Area of a semi-circle  =  (1/2)pi *(radius)^2.....so ...

(1/2) (3.142) * (8.2)^2 =  

105.6 cm^2

LOL

i would like something a bit more non-indecent

Design an Orgasmatron. These always impress the judges.

If you make one that’s WIFI enabled for virtual orgies, you’ll not only impress the judges you’ll make a fortune!

Nvm figured it out the slope is 2 and the y intercept is -4 so the sum is -2.

Nevermind, I got the answer

Exterior angle ad to 360    2+3+4  parts =   360/9  = 40 degrees per part

the largest INTERIOR angle will have the SMALLEST exterior angle (2 parts)     the int and ext angles add to 180

180  -  2 parts x 40 degress/part  = 100 degrees

I found only one:8/5. What did I do wrong?

200 x 8 x 5 x 4 x 10 =  

The answer, of course, is 320,000

You can multiply the numbers one after another manually,

Or you can do it the easy way, with a calculator,

Or you can do it in your head:

look at the whole thing at once,

and recognize that 5 times 4 is 20,

then times the 10 takes it up to 200 

which gives you 200 x 8 x 200. 

You square 200 for 40,000,

then that times 8 gives you 320,000.

The hardest part is keeping up with the number of zeros.

Excellent finding tanmai79. I am impressed!

Have you tried drawing it?

It seems to me that both circles are on the same side of the line or

They are on opposite sides of the line.

SO I think there is only 2 possible lengths of OP.  You can sketch the diagram and find these 2 lengths all by yourself.

You can let me/us know how you get on or you can ask questions if you like. :)

PLEASE NO ONE ANSWER OVER ME.

Hmm!   -arcsin(x) is the same as arccos(x) to within a constant, so either is acceptable in a solution to the indefinite integral.

I have a picture with dimensions x and y (in inches), such that x and y are both integers greater than one. I would like to place this picture in an elongated frame of dimensions (2x + 3) and (y+2). If I measured the area of the frame to be 34 square inches, what is the area of the picture in square inches?