Answer:
green 5 yellow 3
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given the complex notations a = 5i + j, b = i − 2j, we are to evaluate the following:
1) a + b
= 5i+j + (i-2j)
= 5i+j+i-2j
collect like terms
= 5i+i+j-2j
= 6i-j
<em>Hence a+b = 6i-j</em>
2) 2a+3b
= 2(5i+j) + 3(i-2j)
open the parenthesis
= 10i+2j+3i-6j
collect like terms
= 10i+3i+2j-6j
= 13i-4j
<em>Hence 2a+3b = 13i-4j</em>
3) |a| = √x²+y²
Given a = 5i+j; x = 5, y = 1
|a| = √5²+1²
|a| = √25+1
<em>|a| = √26</em>
4) |a-b|
First we need to calculate a-b
= a - b
= 5i+j - (i-2j)
open the parenthesis
= 5i+j-i+2j
collect like terms
= 5i-i+j+2j
= 4i-3j
|a-b| = √4²+(-3)²
|a-b| = √16+9
|a-b| = √25
|a-b| = 5
The sum of two consecutive even integers is at most seven more than half the sum of the next two consecutive even integers
two consecutive even integers: n, n+2
next two consecutive even integers n+4, n+6
n+(n+2) >= 7+1/2(n+4 +n+6)
combine like terms
2n+2 >= 7 +1/2(2n+10)
distribute 1/2
2n+2 >= 7+n+5
2n+2>= n+12
subtract n from each side
n+2>= 12
subtract 2 from each side
n>10
4 integers : 10, 12 ,14, 16
10 cups and I'm sure hope it helps !