Asking the Math Gods...
The factors of 15 Answer : 1,3,5,15
so 3 and 5 are the only that would get 15
therefore this problem isnt correctly written.. You can't get 6 by adding any of the factored numbers
<span>When flipping two standard American quarters, there are four independent possible outcomes:
-Tails, tails
-Heads, heads
-Heads, tails
-Tails, heads
Looking, then, at these four outcomes, there are three of those that include at least one head. As such, the answer to this question is three possibly different ways for her to achieve the desired outcome.</span>
18 and 35. The numbers whose sum 53 are 18 and 35.
The key to solve this problem is using a system of equations.
There are two numbers whose sum is 53. This number can be represented as x and y. So:
x + y = 53
Three times the smaller number is equal to 19 more than the larger. Let's set x as the smaller number and y the larger number. So:
3x = 19 + y
Clear y in both equations and let's use the equalization method to solve for x:
y = 53 - x and y = 3x - 19
Then,
53 - x = 3x - 19
53 + 19 = 3x + x ---------> 3x + x = 53 + 19 -------> 4x = 72
x = 72/4 = 18
To find y, let's substitute x = 18 in the equation x + y = 53
18 + y = 53 --------> y = 53 - 18
y = 35
X² + 2x +3
If the discriminent Δ = b²-4.a.c <0 , that means the roots (or the zero value) are nit REAL Number (they are imaginary or complex)
Δ = 2² - 4(1)3) = 4 - 12 = -8
since Δ = - 8 < 0 there are no real roots
