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What is 5 x3?

Updated: 4/28/2022
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Q: What is 5 x3?
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Simplify 9x 5 - x 3?

I am trying to read this right: (9 x 5) - x3 = ? or 9 x (5 - x3) = ? Which one is it?


Factor the polynomial of x3 plus 5x2 - x - 5?

x3 + 5x2 - x - 5 = (x2 - 1)(x + 5) = (x + 1)(x - 1)(x + 5)


X3 plus x2 plus 5x plus 5?

10x + 5 unless you know what x & y are * * * * * Perhaps this answer will be of help: x3 + x2 + 5x + 5 = (x + 1)(x2 + 5). If you are willing to go to complex roots, then, x3 + x2 + 5x + 5 = (x + 1)(x2 + 5). = (x + 1)(x + i√5)(x - i√5); in which case, x = -1 or ±i√5.


How do you write an expression using a single exponent?

For example, (x3)(x4) = (x3+4) = x7 Also, (x5)2 = x(5)(2) = x10


How many times in a twelve hour period does the sum of the digits on a digital clock equal six and prove it?

notation: natural numbers = 0,1, 2, 3, 4, 5, ....., (some define it without the zero, though) <= means smaller than or equal to, {} is set notation and means a set of numbers : (such that) then some condition. For example {x: x is not a duck} is the set of all things not a duck. Our goal is to prove that there are 21 different times. let x1 = hours, x2 = tens of minutes, x3 = minutes. We are going to prove the statement about the set {x1, x2,x3: 1<=x1 <= 12, 0<= x2<=5, 0<=x3 <= 59, x1 + x2 + x3 = 6}. It will be taken by assumption that this set is the set of digital clock combinations that add up to 6. So then, we must prove that there are unique 21 elements in the set {x1 + x2 + x3 : 1<= x1 <= 12, 0<= x2<=5, 0<=x3 <= 59, x1 + x2 + x3 = 6}. {x1 , x2 , x3 : 1<= x1 <= 12, 0<= x2<=5, 0<=x3 <= 59, x1 + x2 + x3 = 6} = {x1 , x2 , x3 : 1<= x1 <= 6, 0<= x2<=5, 0<=x3 <= 5, x1 + x2 + x3 = 6} because x3<=6, and because if x1 >=1, then x2 + x3 <=5, and x3, x2 >= 0 , so surely x3, x2 <= x5. Either x1 = 1, 2, 3, 4, 5, or 6. Next, x1 + x2 + x3 = 6, so x2 + x3 = 6 - x1. There are n+1 natural numbers between 0 and n (I'm being lazy and not proving this, but the proof would be so much longer if I proved it), and since 0 <= x2 <= 5 <= 6-x1, there are at most 6-x1+1 values of x2 for each value of x1. When x1 = 1, there are a maximum of 6, when x1 = 2, there are 6-2+1 = 5, when x1 = 3, there are 6-3+1 = 4, when x1 = 3, there are 3, then 2, and then 1. Summing this up gives us a maximum of 21. So it is at most 21 and at least 21, so exactly 21.