Before we go into this, here's the formula: (y2-y1)/(x2-x1)
Pair 1: RS TU
> Slope of RS = 2/3
> Slope of TU = 2/3
Since both slopes are they same, they are parallel.
Pair 2: RS ST
> Slope of RS = 2/3
> Slope of ST = -3/2
Since the slopes are negative reciprocals of each other, they are perpendicular.
Same goes for the last pair, which is TU ST.
> Slope of TU = 2/3
> Slope of ST = -3/2
They're negative reciprocals, hence they are perpendicular.
Hope this helps! :)
It’s C because if you look at the problem and solve it the right way it’ll be 10.50
Answer: 954
Step-by-step explanation:
Using the condition given to build an inequality, it is found that the maximum number of junior high school student he can still recruit is of 17.
<h3>Inequality:</h3>
Considering s the number of senior students and j the number of junior students, and that he cannot recruit more than 50 people, the inequality that models the number of students he can still recruit is:
In this problem:
- Already recruited 28 senior high students, hence
.
- Already recruited 5 junior high students, want to recruit more, hence
.
Then:
The maximum number of junior high school student he can still recruit is of 17.
You can learn more about inequalities at brainly.com/question/25953350
To solve this problem you must apply the proccedure shown below:
1. You have the following quadratic equation given in the exercise above:
2x²−6x+1=0
2. The quadratic equation is:
x=[-b±√(b²-4ac)]/2a
Where:
a=2
b=-6
c=1
3. When you substitute these values into the quadratic equation, you obtain:
x=3/2-√(7)/2
x=3/2+√(7)/2
