First, find what percentage of students had 3 or more by adding up your known percents:
45% + 23 % + 21% + x% = 100%
x = 11%
Since you're given that 96 students had 2 or more, you add up the percentages of 2 and 3 or more:
11 + 21 = 32%
Now set up a proportion that relates it to the whole:

This will allow you to find the total number of students at the school.
Cross multiplying and solving for x results in 300 total students.
Question 1:
45% had one or more absences. 45% of 300 students is
135 students.
Question 2:
As we found before, 11% of students had three or more. 11% of 300 is
33 students.
Factor by grouping. Group up the terms into pairs, factor each pair, then factor out the overall GCF.
x^3 + 2x^2 - 16x - 32
(x^3 + 2x^2) + (-16x-32) ... pair up terms
x^2(x + 2) + (-16x - 32) ... factor x^2 from the first group
x^2(x + 2) - 16(x + 2) ... factor -16 from the second group
(x^2 - 16)(x + 2) .... factor out (x+2)
(x - 4)(x + 4)(x + 2) .... Use the difference of squares to factor x^2-16
---------------------------
The original expression completely factors to (x - 4)(x + 4)(x + 2)
The three factors are x - 4 and x + 4 and x + 2
Answer:
<em>2 solutions</em>
Step-by-step explanation:
Given the expression
2m/2m+3 - 2m/2m-3 = 1
Find the LCM of the expression at the left hand side:
2m(2m-3)-2m(2m+3)/(2m+3)(2m-3) = 1
open the bracket
4m²-6m-4m²-6m/(4m²-9) = 1
Cross multiply
4m²-6m-4m²-6m = 4m² - 9
-12m = 4m² - 9
4m² - 9+12m = 0
4m² +12m-9 = 0
<em>Since the resulting equation is a quadratic equation, it will have 2 solutions since the degree of the equation is 2</em>
Answer:
y=30x+50 and y=40
Step-by-step explanation:
Answer:
The answer is
<h2>( 2 , 1)</h2>
Step-by-step explanation:
The midpoint M of two endpoints of a line segment can be found by using the formula

where
(x1 , y1) and (x2 , y2) are the points
From the question the points are
(-4,-8) and (8, 10)
The midpoint is

We have the final answer as
<h3>( 2 , 1)</h3>
Hope this helps you