Answer:
ΔtH = (d -s) / vH
Step-by-step explanation:
Using time = distance / velocity,
Let dH be the distance the hare ran and replace in the equation
ΔtH = dH / vH
Since the tortoise won by a shell (s) and the the length of the racetrack is d, the hare ran a distance equal to
dH = d -s
now dH is replace to obtained an equation for ΔtH
ΔtH = (d -s) / vH
If x is in quadrant I, then both sin(x) and cos(x) are positive. The angle x/2 also belongs to quadrant I, and hence each of sin(x/2), cos(x/2), and tan(x/2) are positive.
Recall that for all x,
cos²(x) + sin²(x) = 1
and multiplying through both sides by 1/cos²(x) = sec²(x) gives another flavor of this identity,
1 + tan²(x) = sec²(x)
It follows that
sec(x) = √(1 + tan²(x)) = √937/19
which immediately gives us
cos(x) = 19/√937
and from the identity above we find
sin(x) = √(1 - cos²(x)) = 24/√937
Recall the half angle identity for cos :
cos²(x) = (1 + cos(2x))/2
which means
cos(x/2) = + √[(1 + cos(x))/2] = √[1/2 + 19/(2√937)]
Then
sin(x/2) = + √(1 - cos²(x/2)) = √[1/2 - 19/(2 √937)]
and by definition of tan,
tan(x/2) = sin(x/2) / cos(x/2) = 1/12 √[649/2 - 19 √937/2]
Just to be clear, the solutions are



Answer:
Step-by-step explanation:
hope this helped<3
Answer:
only (-4,0) is a reasonable solution.
Step-by-step explanation:
In order to see which ones are reasonable we need to apply the points to the system and check if they make both equations valid at the same time.
y = -x - 4
y = x² + x - 12
For (-4, 0):
0 = -(-4) - 4 -> 0 = 4 -4 -> 0 = 0 (valid)
0 = (-4)² - 4 - 12 -> 0 = 16 - 16 -> 0 = 0 (valid)
For (-3, -1):
-1 = -(-3) -4 -> -1 = 3 -4 -> -1 = -1 (valid)
-1 = (-3)² -1 - 12 -> -1 = 9 -13 -> -1 = -4 (invalid)
For (2, -6):
2 = -(-6) - 4 -> 2 = 6 -4 -> 2 = 2 (valid)
2 = (-6)² -6 -12 -> 2 = 36 -18 -> 2 = 18 (invalid)
Therefore only (-4,0) is a reasonable solution.