The base is a right triangle....its area = product of the legs / 2 = 24 * 7 / 2 = 84 m^2
The volume = base area * height / 3 = 84 * 14 / 3 = 392 m^3
Note that we can find the hyppotenuse^2 of the triangle on the right as
8^2 + 4^2 = h^2
64 + 16 = h^2 = 80
So...on the left....x is the hypotenusr of that triangle so
h^2 + 7^2 =x^2
80 + 49 = x^2
129 = x^2
sqrt (129) = x
Note k * log y x can be written as log y x^k
Using the change-of-base we have that
log x^3 / log y^5 = log x^k / log y and we can write
3 * log x / 5* log y = k * log x / log y multiply both sides by log y /log x and we get that
3 / 5 = k
1.
Let the equal number of tickets = n
So
5n + 8n ≤ 75
13n ≤ 75
n ≤ 75 /13
n ≤ 5.76
She can buy 5 tickets of each type = 10 tickets total
Proof
5(5) + 8(5) = $65
If she bought 6 tickets each....she would spend
5(6) + 8(6) = $78 (too much)
The exterior angles of a polygon sum to 360°
58 + 39 + 50 + 48 + 59 + 2x = 360
254 + 2x = 360
2x = 360 - 254
2x = 106
x = 53°
Period = pi
y int = 0
What choices are given for the points ???
3sin (x - pi/4) = 0 divide both sides by 3
sin ( x - pi/4) = 0
sin A = 0 at A = 0 and A =pi
A = x -p/4 = 0 A = x - pi/4 = pi
x = pi/4 x = pi + pi/4 = 5pi/4
-2cos(2x) + 1 = 0
-2 cos (2x) = -1
cos (2x) = 1/2
Note
cos (A) = 1/2 at x= 60° and x = 300°
A = 2x = 60 and A = 2x = 300
x= 30° = pi/ 6 x = 150° = 5pi/6
.80P = 15 where P is the original price
P = 15 / .80 = $18.75
The diagonals bisect each other
One side length = sqrt [ ( 12/2 )^2 + (16/2)^2 ] = sqrt [ 6^2 + 8^2 ] = sqrt [ 100] = 10
Perimeter = 4 * 10 = 40
