Given that the number of years should be represented with x, the number of fish in the pond after x years should best be represented with f(x). The equation that would best show the given scenario in the problem above is,
f(x) = 500(2^x)
From the given, 500 is used as the initial population of the fish.
Sec(theta) = 1 / cos (theta) = hypotenuse / x -coordinate
hypotenuse = 1 (because it is the radius of the unit circle)
sec (theta) = 1 / (-3/5) = - 5/3
cot (theta) = 1 / tan(theta) = x-coordinate / y - coordinate
cot (theta) = -3/5 / y
y^2 + (-3/5)^2 = 1 => y^2 = 1 - 9/25 = 16/25 = y = +/- 4/5
Third quadrant => y = -4/5
=> cot (theta) = (-3/5) / (-4/5) = 3/4
Answer:
10
Step-by-step explanation:
itsss 10 :))))))))))>))))))))
Answer:
a. 96 square units
Step-by-step explanation:
The figure is a rectangle with width AB = (20-12) = 8 units and height BC = (20-8) = 12 units.
The area of the rectangle is (8 units)×(12 units) = 96 square units.
Answer:
(3, 1)
Step-by-step explanation:
(a) Algebraic solution
(1) y = -⅔x + 3
(2) y = 2x - 5
Set Equation (1) equal to Equation (2)
-⅔x + 3 = 2x - 5
Multiply each side by 3
-2x + 9 = 6x - 15
Add 15 to each side
-2x + 24 = 6x
Add 2x to each side
24 = 8x
Divide each side by 3
(3) x = 3
Substitute (3) into (2)
y = 2×3 - 5 = 6 - 5 = 1
The ordered pair that makes both equations true is (3, 1).
(b) Graphical solution
In the diagram below, the red line is the graph of Equation (1). The blue line is the graph of Equation (2). The point of intersection is at (3, 1).