Answer:
perpendicular
Step-by-step explanation:
To determine if AB and CD are parallel, perpendicular, or neither, we need to get the slope of AB and CD first
Given A (−1, 3), B (0, 5),
Slope Mab = 5-3/0-(-1)
Mab = 2/1
Mab = 2
Slope of AB is 2
Given C (2, 1), D (6, −1)
Slope Mcd = -1-1/6-2
Mcd = -2/4
Mcd = -1/2
Slope of CD is -1/2
Take their product
Mab * Mcd = 2 * -1/2
Mab * Mcd = -1
Since the product of their slope is -1, hence AB and CD are perpendicular
Let's convert the equation in point-slope form to slope-intercept form.
y - 4 = -(x - 6)
Multiply everything out.
y - 4 = -x + 6
Add 4 to both sides.
y = -x + 10
To be perpendicular, the slope of the new has to be the reciprocal of the other line. To slope in this equation is -1/1. To get the reciprocal, flip the numerator and denominator and change the sign too.
-1/1 becomes 1/-1, and changes to 1/+1, or 1.
y = x + b
Input the coordinate point and solve for b.
-2 = -2 + b
Add 2 to both sides.
0 = b
y = x
Answer:
P(3) is true since 2(3) - 1 = 5 < 3! = 6.
Step-by-step explanation:
Let P(n) be the proposition that 2n-1 ≤ n!. for n ≥ 3
Basis: P(3) is true since 2(3) - 1 = 5 < 3! = 6.
Inductive Step: Assume P(k) holds, i.e., 2k - 1 ≤ k! for an arbitrary integer k ≥ 3. To show that P(k + 1) holds:
2(k+1) - 1 = 2k + 2 - 1
≤ 2 + k! (by the inductive hypothesis)
= (k + 1)! Therefore,2n-1 ≤ n! holds, for every integer n ≥ 3.
