CPhill

Call the number of 20 cent coins "x"

Let the number of 10 cent coins be 38 - x

So, the number of coins times their value each gives us the total value....so we have

20x + 10(38 - x)  = 530   simplify

20x + 380 - 10x = 530

10x + 380 = 530    subtract 380 from each side

10x = 150    divide both sides by 10

x = 15   ....and that's the number of 20 cent coins

If you know the base and the height...it is

Area  =  (1/2)base x height

If you know two sides and the included angle between them, it is

Area = (1/2) ab sin θ    where a,b are the known sides and θ is the angle between them

There is also a formula if you only know the sides and nothing else....but....it's a little complicated......if this is the case....post back and I'll run through it for you.... 

Like Melody, I'm on shaky ground here....so.....you might have to take this with a grain of salt....

By the Integral Test, the function in the integrand must be eventually positive and decreasing on the interval of interest - in this case, 0 to ∞.

Here's the graph of  y = cos(x)*e^(-sin(x))

I think Anonymous forgot to check his/her answer here

Taking it from this point, we have

x + ( x - 1)  = 7

2x - 1  = 7    add 1 to both sides

2x = 8     divide both sides by 2

x = 4

And using x + y = 7

4 + y = 7   ....  so.....y = 3

And the answer is  (x, y) = (4, 3)

The semi-annual answer given by Anonymous is correct

For the annual compounding, we have

1500(1.03)^4 = $ 1688.26

This will occur at 45° ± n180°  where n is an integer

Let's tnink about this for just a second.....how could subtratcing 5.6 from 13.48 result in something a little more than 1 ???

That doesn't even pass the "reasonable" test......!!!

I'd probably trust the calculator on this one........

That must be the second oldest "gag" question in history......!!!!

Right after..."Who is buried in Grant's Tomb ??"

It he dimensions are 1/2 the original size, the new area is 1/4 as much - the square of the scale factor (1/2).   So the new area is 16 sq units.

Yeah....I checked this with WolframAlpha.....3.9 is correct.....!!!