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
Answer:
27.5 cm
Step-by-step explanation:
because 0.25 x 110 = 27.5
Answer:
Step-by-step explanation:
23x-12-4x+5x
24x-12 combine the 23x-4x+5x to get 24x
19-14+3x+12x
15x+5 19-14=5 3x+12x=15x
6x+19-14x+9
-8x+28 -14x+6x= -8x 19+9=28
Step-by-step explanation:
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Answer: 529 students
Step-by-step explanation:
To solve this, we will be using the formula for finding averages:
> (total sum of student weights) ÷ (total sum of students) = (average of student weights)
Plugging in the values that are given in the problem, you get:
> 16,293.2 ÷ x = 30.8
Where x is equal to the total sum of students. From here, it's a simple equation to isolate x, and find out what numerical value x is equal to. In case you are unsure how to do this, I've created a clear step by step process for how to do this (NOTE: The slash represents division, while the asterisk represents multiplication.):
> 16,293.2 / x = 30.8
> (16,293.2 / x) * x = (30.8) * x
> 16,293.2 = 30.8 * x
> (16,293.2) / 30.8 = (30.8 * x) / 30.8
> 529 = x
In total, there are 529 students in the school.
