n, n + 2, n + 4, n + 6 - four consecutive odd integers
3 times the largest decreased by 19 is twice the second integer
3(n + 6) - 19 = 2(n + 2) <em>use distributive property</em>
(3)(n) + (3)(6) - 19 = (2)(n) + (2)(2)
3n + 18 - 19 = 2n + 4
3n - 1 = 2n + 4 <em>add 1 to both sides</em>
3n = 2n + 5 <em>subtract 2n from both sides</em>
n = 5
n + 2 = 5 + 2 = 7
n + 4 = 5 + 4 = 9
n + 6 = 5 + 6 = 11
<h3>Answer: 5, 7, 9, 11</h3>
There are two ways to solve this question.
1) To solve this question, we need to substitute a = 6 and b = -3 into the given expression and then evaluate:
(-a)(b)(-a + b)
= (-6)(-3)(-6 + (-3))
= 18(-9)
= -162
2) An alternative method is to simplify (-a)(b)(-a + b) into an expression without brackets and then substitute a = 6 and b = -3:
1. (-a)(b)(-a + b)
= (-ab)(-a + b)
= -ab*(-a) + (-ab)*b
= a^(2)b+ (-ab^(2))
= a^(2)b - ab^(2)
2. a^(2)b - ab^(2)
= 6^(2)*(-3) - 6*(-3)^2
= 36*(-3) - 6*9
= -108 - 54
= -162
The key to either method is to be careful with placement of brackets, especially where there are negative values involved.
Ok........
How many were severe
64.47
How many total injuries in total
(I think) 783.154
H0P3 It H3LPS :)