Answer:
P and Q are two points on the line x-y+1=0 and are at a distant of 5 units from the origin. Find the area of triangle POQ.
Step-by-step explanation:
P and Q are the intersection points of
x-y+1 = 0 and the circle x^2 + y^2 = 25
sub y = x+1 into the circle
x^2 + (x+1)^2 = 25
x^2 + x^2 + 2x + 1 - 25 = 0
x^2 + x - 12 = 0
(x+4)(x-3) = 0
x = 3 or x = -4
y = 4 or y = -3
so P(3,4) and Q(-4,3) are our two points
Height of triangle.
h = |0 - 0 + 1|/√2 = 1/√2
PQ = √( (-7)^2 + 1^2) = √50 = 5√2
area POQ = (1/2)(1/√2)(5√2) = 5/2 square units
hope this helped
Answer:
3: 2q − 4 = 20
Step-by-step explanation:
Solve for q:
2 q - 4 = 20
Add 4 to both sides:
2 q + (4 - 4) = 4 + 20
4 - 4 = 0:
2 q = 20 + 4
20 + 4 = 24:
2 q = 24
Divide both sides of 2 q = 24 by 2:
(2 q)/2 = 24/2
2/2 = 1:
q = 24/2
The gcd of 24 and 2 is 2, so 24/2 = (2×12)/(2×1) = 2/2×12 = 12:
Answer: q = 12
The distance might be the slope of the line
Answer:
B
Step-by-step explanation:
Just make sure that the x value is not being repeated twice. because one of the given points is (4,9) you cannot have the point (4,0). X value cannot repeat in a valid function.
Answer:
I used the function normCdf(lower bound, upper bound, mean, standard deviation) on the graphing calculator to solve this.
- Lower bound = 1914.8
- Upper bound = 999999
- Mean = 1986.1
- Standard deviation = 27.2
Input in these values and it will result in:
normCdf(1914.8,9999999,1986.1,27.2) = 0.995621
So the probability that the value is greater than 1914.8 is about 99.5621%
<u><em>I'm not sure if this is correct </em></u><em>0_o</em>
