9514 1404 393
Answer:
nπ -π/6 . . . for any integer n
Step-by-step explanation:
tan(x) +√3 = -2tan(x) . . . . . given
3tan(x) = -√3 . . . . . . . . . . . add 2tan(x)-√3
tan(x) = -√3/3 . . . . . . . . . . divide by 3
x = arctan(-√3/3) = -π/6 . . . . use the inverse tangent function to find x
This is the value in the range (-π/2, π/2). The tangent function repeats with period π, so the set of values of x that will satisfy this equation is ...
x = n·π -π/6 . . . . for any integer n
The equation x=2t-1 represent the x-coordinate. To get the value of x, substitute t with -1.
That is: x = 2(-1) - 1
= -2-1
= -3
Likewise, the equation y=t∧-3 represent the y-coordinate. Substituting t with -1,
y = (-1)∧-3
= -1
The point that correspond to t=-1 is (-3,-1).
Answer:
16.5 ft by 25.5 ft
Step-by-step explanation:
Let w represent the width of the garden in feet. Then w+9 is the garden's length, and w(w+9) represents its area.
The surrounding walkway adds 8 feet to each dimension, so the total area of the garden with the walkway is ...
(w+8)(w+9+8) = w^2 +25w +136
If we subtract the area of the garden itself, then the remaining area is that of the walkway:
(w^2 +25w +136) - (w(w+9)) = 400
16w + 136 = 400 . . .simplify
16w = 264 . . . . . . . . subtract 136
264/16 = w = 16.5 . . . . . width of the garden in feet
w+9 = 25.5 . . . . . . . . . . .length of the garden in feet
The area of the figure is 28 yd^2
Answer:
1). 875 seats
2). 25 rows in each section
3). $8400
4). Saving of $360
5). 2105 tickets remained unsold
6). x = 16
