anybody could help me with this math problem please?

Question

go here
http://www.nysedregents.org/testing/mathre/b-607.pdf
# 31 and #32
i know the ans but dont know how to do the work and i'm taking the math b test nxt week

thanx

Answer 1

Hi,

#31 10 inches is the minimum snowfall. It occurred in 1975, 1985, and 1995. <== answer

The normal period of the cosine graph is 2Pi. When the constant following cosine is multiplied by a number, you must divide the normal period of 2Pi by this number to get the new period of your graph.

Given S(t) = 20 + 10 cos(Pi/5t), the period of the graph is found by 2Pi/(Pi/5) which equals 10. 10 years is the period of the graph. Since the cosine graph starts at its highest point and hits its lowest point after half of the period, the lowest point is after 10 /2 or 5 years later in 1975. Since the period is 10, it will hit a low point again in 1985 and in 1995 as well.

The baseline at the center of the cosine graph is at y = 20.The graph goes 10 inches higher and 10 inches lower because the amplitude in the equation is 10. Therefore the deepest snow is 30 inches and the least amount of snow is 10 inches.


#32 The maximum height was 64 feet.
It hit the ground after 4 seconds.

The maximum height occurs at the parabola's axis of symmetry, which is found from the formula x = -b/(2a). In this case "a" = -16 and "b" = 64, so x = -64/(2*-16) = -64/-32 = 2.
The maximum height will occur when the time t equals 2 seconds. To find the height, substitute t = 2 into the formula. The maximum height is 64 feet.

To find when it hits the ground, set 64t - 16t² = 0 and solve.

64t - 16t² = 0
16t(4 - t) = 0

If 16t = 0
t = 0 The rocket is on the ground at 0 seconds when it is fired.

4 - t = 0
t = 4 The rocket will hit the ground after 4 seconds.

The maximum height of the rocket is 64 feet. <== answer
It will hit the ground after 4 seconds. <== answer

Answer 2

the min value of cos(x) is -1, so the min value of 20 + 10 cos( whatever) is 20 + 10(-1) = 20 - 10 = 10 inches. cos(x) takes on its min value when x = π, so this function is at its min when πt/5 = π, so t/5 = 1, t = 5. the period of cos is 2π, so we should really solve

πt/5 + 2kπ = π
t/5 + 2k = 1
t/5 = 1 - 2k
t = 5 - 10k, where k is an integer. 10k is the period of the function. so we have min snow when t = 5 (k = 0), which is 1975, when t = 15 (k = -1), which is 1985, and obviously when t = 25 (k = -2), which is 1995.

if you remember that the coefficient of t inside the parentheses, the π/5, divided into 2π, give you the period (2π / (π/5) = 10), then once you have t = 5 (1975) as your first min, you just repeatedly add 10 years.

the rocket prob is simple. 64 is initial upward velocity, -16 is half acceleration of gravity, s(t) is height above ground. solve

s(t) = 0 = 64t - 16t² since height on ground is 0.
0 = 16t(4 - t)
t = 0 (at start) or t = 4, which answers 2nd question, back on ground in 4 sec.

since the path is a parabola, vertex is on line of symmetry halfway between x intercepts, (0+4)/2 = 2. so plug in 2 for t:
s(2) = 64(2) - 16(2)²
max height = 128 - 64 = 64 ft.

if you've learned a little calculus, you'd solve
s'(t) = 64 - 32t = 0
t = 2 sec, and then plug in the 2.

Answer 3

31.
S(t) = 20 + 10cos((π/5)t), t = year - 1970
cosθ varies from + 1 to - 1 as θ varies from 0 to π.
The function will be a minimum when
cos((π/5)t) = - 1, or S(t) = 20 - 10 = 10

cos((π/5)t) = - 1 when (π/5)t = π
t = 5
year = 1970 + 5 = 1975

32.
s(t) = 64t - 16t^2
s(t) = - 16(t^2 - 4t)
s(t) = - 16(t^2 - 4t + 4 - 4)
s(t) = - 16(t - 2)^2 + 64
which has a maximum value of 64 when t = 2
When s=0
- 16(t^2 - 4t) = 0
t(t - 4) = 0
t = 0, 4

Answer 4

32)

calculus method
s(t) = 64t-16t^2

s'(t) = 64 -32 t

s"(t) = -32

turning values s'(t) = 0 implies t = 2

s"(2) <0 hence max.

max ht = s(2) = 64*2 - 16 *(2^2) = 64 feet

total time of travel = 2 + 2 =4 seconds

qn 31 is similar or use graphing method

Answer 5

Fp ? 0.15p the placement Fp is finished value & ? 0.15p is the decrease value: 15%=.15 (5-(-3)) / (0-6) = -4/3 slope + 5 y-intercept then plug in a fee of the inequality like (10,10) 10 = -4/3(10) + 5 10 = -40/3+5 10 = -13-(a million/3) +5 once you communicate approximately that 10 > -8.33333333 y > (-4/3) x + 5 is your equation

Answer 6

s(t) = 64t - 16t^2
= -16 ( t^2 - 4t )
= -16 [ t^2 - 4t + (-4/2)^2 - (-4/2)^2 ]
= -16 [ (t - 2)^2 - 4 ]
= -16(t - 2)^2 + 64
Therefore, max occurs at s=64

s(t) = 64t - 16t^2
When rocket hits the ground, s(t) = 0
0 = 64t - 16t^2
0 = 16t(4 - t)
Therefore
t=0(time when rocket is launched) or
t=4(time when rocket hits the ground)

Answer 7

It's cheating!!!
You have already obtained the questionnaires so do it yourself