K’ (-3,-3)
L’(1,-1)
M’(-1,-5)
N’(-5,-7)
D’(0,4)
E’(4,3)
F’(2,-5)
G’(-2,-4)
Answer:
203.5 N
Step-by-step explanation:
Force = mass × acceleration
F = ma
F = (55 kg) (3.7 N/kg)
F = 203.5 N
The intersection point is only one. Then the equation of the line is tangent to the circle at point (1, -2).
<h3>What is a circle?</h3>
It is a locus of a point drawn an equidistant from the center. The distance from the center to the circumference is called the radius of the circle.
Prove algebraically that the straight line with equation x = 2y + 5 is a tangent to the circle with equation x² + y² = 5.
x = 2y + 5 ...1
x² + y² = 5 ...2
If the intersection of the point of the circle and line is one. Then the line is tangent to the circle.
Then from equations 1 and 2, we have
(2y + 5)² + y² = 5
4y² + 25 + 20y + y² - 5 = 0
5y² + 20y + 20 = 0
5y² + 10y + 10y + 20 = 0
5y (y + 2) + 10(y + 2) = 0
(5y + 10)(y + 2) = 0
y = -2, -2
Then the value of y is unique then the value of x will be unique.
The value of x will be
x = 2(-2) + 5
x = -4 + 5
x = 1
The intersection point is only one. Then the equation of the line is tangent to the circle at point (1, -2).
More about the circle link is given below.
brainly.com/question/11833983
#SPJ1
Answer:
5x - 13y + 2
Step-by-step explanation:
Given: 2x – 4y + 6 + 3x – 9y – 4
Combine all like terms:
5x - 13y + 2
So 15 minus the flat fee is 11.25 divided by .85 is 13.23 km rounded 13km
