CPhill

4 ³√54 +³√250 =

4 [³√(27*2)] +³√(125*2) =

4 [3 ³√(2)] + 5 [³√(2)] =

12 [ ³√(2)] + 5 [³√(2)]  =

17 ³√(2)

Here's my (belated) take on this one:

We can write:

4^(0.5log₄(9) - 0.25log₂(25))  ...as.....

   [4^log₄(3)] / [ 4^ log₂(5)^(.5)]

4^log₄(3)/ [4^{(log₂(5)/2)]}    ......   the numerator simplifies to 3

Note that log₂(5) is just a number.....call it "a'  ....so we have

4^(a/2) = [4^(1/2)]^a = 2^(a) =

2^{log₂ 5} = 5

So our answer is just  .....  3/5

There is a "formula" to solve this, but let's use some logic, instead. If his average was 81 after 6 rounds, he must have had 6*81 = 486 total strokes.

So, totaling his first 5 rounds gives.......409 strokes. So, his last round must have been 486- 409 = 77.

(r¼s ²⁄5)²º=

———————

       r²

OK , using the rule of exponents that says (a^m) ⁿ =  a ^m*n  the numerator becomes

(r⁵ * s⁸ )     and we have

(r⁵ * s⁸ ) / r²    which simplifies to

r³ * s⁸

A "semi" truck.......!!!!!

HAHAHA!!!  Loved the "Stop Sign" one !!!!

I love these daily laughs, rosala.......points from me......!!!!

Find and equivalent algebraic expression for the composition, cot(arcsin(x))

So we're looking for the cotangent of an angle whose sine = x  =  x/1

So using the Pythagorean Theorem, the side adjacent to the angle is √(1 - x²)

So the cotangent  is   √(1 - x²) / x

cos(x)/(1+sin(x)) =sec(x)-tan(x)

OK, on the right side we have

1/cos(x) - sin(x)/cos(x)

(1 - sin(x)) /  cos(x)    Multiply numerator and denominator by cos(x)

[(cos(x)* (1- sin(x)] / [cos²(x)]

[(cos(x)* (1- sin(x)]/ (1 - sin²(x))

[(cos(x)* (1- sin(x)]/ [(1 + sin(x)) * (1 - sin(x)]

cos(x)/(1+sin(x))   which equals the left hand side

p ‾²⁄7

—————

p¹⁄7 *p¯4⁄7

So the denominator becomes p ^(-3/7)

So we have

p ^(-2/7) / p ^(-3/7)

And by the laws of exponents we can write

p ^{(-2/7) - (-3/7)} =

p ^(1/7)

a 7⁄8 *a‾³⁄4

OK, in this situation, we want to "keep" the base "a" and add the exponents.

So...

a ^{(7/8) + (-3/4)} =

a ^{(7/8) + (-6/8)} =

a ^1/8