The only 5-digit numbers that do the job are 21,143 and 28,847 .
So, either
a = 1 and
b = 3
or
a = 8 and
b = 7 .
The sum of the values that 'a' can have is (1 + 7) = <em>8</em> .
I shall cherish every one of your generous 5 points.
Rewrite 512x^3 as (8x)^3
(8x)^3 + 2197y^3
Rewrite 2197y^3 as (13y)^3
(8x)^3 + (13y)^3
Since both terms are perfect cubes, factor using the sum of cubes formula.
a^3 + b^3 = (a + b)(a^2 - ab + b^2) where a = 8x and b = 13y.
(8x + 13y)((8x)^2 - (8x)(13y) + (13y)^2)
Simplify
(8x + 13y)(64x^2 - 104xy + 169y^2)
I believe an irregular galaxy is what it's called
Answer:
a. x > 0.2
Step-by-step explanation:
In either order, subtract the left-side constant and divide by the x-coefficient.
3x +2.4 > 3
3x > 0.6 . . . . . . subtract 2.4
x > 0.2 . . . . . . . divide by 3
__
or
x + 0.8 > 1 . . . . divide by 3
x > 0.2 . . . . . . . subtract 0.8
First off, your chances of red are not really 50-50. You are overlooking the 0 slot or the 00 slot which are green. So, chances of red are 18 in 37 (0 slot) or 38 (0 and 00 slots). With a betting machine, the odds does not change no trouble what has occurred before. Think through the simplest circumstance, a coin toss. If I toss heads 10 times one after the other, the chances of tails about to happen on the next toss are still on a 50-50. A betting machine has no ability, no plan, and no past.
Chances (0 slot) that you success on red are 18 out of 37 (18 red slots), but likelihoods of losing are 19 out of 37 (18 black plus 0). For the wheel with both a 0 and 0-0 slot, the odds are poorer. You chances of red are 18 out of 38 (18 red slots win), and down are 20 out of 38 (18 black plus 0 and 00). It does not really matter on how long you play there, the probabilities would always continue the same on every spin. The lengthier you play, the more thoroughly you will tie the chances with a total net loss of that portion of a percent in accord of the house. 18 winning red slots and either 19 or 20 losing slots.
