Answer:
Quadrilateral ABCD is not a square. The product of slopes of its diagonals is not -1.
Step-by-step explanation:
Point A is (-4,6)
Point B is (-12,-12)
Point C is (6,-18)
Point D is (13,-1)
Given that the diagonals of a square are perpendicular to each other;
We know that the product of slopes of two perpendicular lines is -1.
So, slope(m) of AC × slope(m) of BD should be equal to -1.
Slope of AC = (Change in y-axis) ÷ (Change in x-axis) = (-18 - 6) ÷ (6 - -4) = -24/10 = -2.4
Slope of BD = (Change in y-axis) ÷ (Change in x-axis) = (-1 - -12) ÷ (13 - -12) = 11/25 = 0.44
The product of slope of AC and slope of BD = -2.4 × 0.44 = -1.056
Since the product of slope of AC and slope of BD is not -1 hence AC is not perpendicular to BD thus quadrilateral ABCD is not a square.
Because the numerater and denomenater are = hence 1/1, 2/2, 3/3, 4/4 5/5, 6/6, 7/7, 8/8, etc. are all = to one
hope this helps :)
Answer:
1/15
Step-by-step explanation:
First you must figure out your total outcomes, int the problem it said ten. Than take the number of students and put the number of students over the number of outcomes. This will tell you the probability of drawing one student. Next you must subtract 1 student and one member from your original equation. This is because you have assumed that the first drawing had a favorable outcome. So that leaves you with 2/9. Next you must multiply both the first fraction (3/10) and the second fraction (2/9) to come up with the total probability of the event occurring.
Hope that helped.
