Point Mis the midpoint of AB. AM = 3x + 3, and AB= 83 – 6.
2 answers:
Answer:
Step-by-step explanation:
AM = 3x + 3
AB = 8x - 6
Point M is the midpoint of AB.
So, AM = AB/2
3x + 3 = (8x - 6)/2
Multiplying both sides by 2.
2(3x + 3) = 8x - 6
Expanding brackets.
6x + 6 = 8x - 6
Subtracting 6x from both sides.
6 = 2x - 6
Adding 6 to both sides.
12 = 2x
Dividing both sides by 2.
6 = x
Let x = 6 for the length of AM.
3(6) + 3
18 + 3
21
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The given point (-3/5 , y) lies in the third quadrant.
It is also given that the point lies on a unit circle.
For a point (x,y) lying on a unit circle a and y are defined as:
x = cos θ
y = sin θ
So, we can say for the point (-3/5 , y) the value -3/5 is equal to cos θ
sec θ is the reciprocal of cos θ.
So, sec θ = -5/3
Using Pythagorean identity we can first find sin θ.
Since the point lies in 3rd quadrant, both sin and cos will be negative.
So, now we can write:
Answers:
sec θ = -5/3
cot θ = 3/4
It is also given that the point lies on a unit circle.
For a point (x,y) lying on a unit circle a and y are defined as:
x = cos θ
y = sin θ
So, we can say for the point (-3/5 , y) the value -3/5 is equal to cos θ
sec θ is the reciprocal of cos θ.
So, sec θ = -5/3
Using Pythagorean identity we can first find sin θ.
Since the point lies in 3rd quadrant, both sin and cos will be negative.
So, now we can write:
Answers:
sec θ = -5/3
cot θ = 3/4
Answer:
(A)-494
Step-by-step explanation:
Given the arithmetic series
The terms in the sequence are:
- When k=1, 4-3k=4-3(1)=1
- When k=2, 4-3k=4-3(2)=-2
- When k=3, 4-3k=4-3(3)=-5
Therefore, the terms in the sequence are: 1, -2, -5, ...
First term, a =1
Common difference, d=-2-1=-3
The sum of an arithmetic series,
Therefore:
The correct option is A.