Solve for the missing side. 4) 12 mi X 15 mi
1 answer:
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Problem 1
See the attached image (figure 1)
16pi seems like a typo. I'm going to assume that it's a fraction and it is 1/(6pi)
f = 1/(6pi) = frequency
T = 1/f = 1/(1/(6pi)) = 6pi
Amplitude = 2
a = 2
b = 2pi/T = 2pi/(6pi) = 1/3
Midline: y = 3
d = 3
The function is
y = a*sin(bx-c)+d
y = 2*sin(1/3*x-0)+3
y = 2*sin(x/3)+3
===============================================
Problem 2
See the attached image (figure 2)
T = 12 is the period
a = 4 is the amplitude
b = 2pi/T = 2pi/12 = pi/6
y = 1 is the midline so d = 1
The y intercept is (0,1) which is the midline, which indicates no phase shifts have occurred so c = 0
The function is
y = a*sin(bx-c)+d
y = 4*sin((pi/6)x-0)+1
y = 4*sin((pi/6)x)+1
===============================================
Problem 3
See the attached image (figure 3)
Period = 4pi
T = 4pi
b = 2pi/T = 2pi/(4pi) = 1/2 = 0.5
Amplitude = 2
a = 2
Midline: y = 3
d = 3
y-intercept: (0,3)
The function is a reflection of its parent function over the x-axis, so 'a' is negative meaning a = -2 instead of a = 2
The function is
y = a*sin(bx-c)+d
y = -2*sin(0.5x-0)+3
y = -2*sin(0.5x)+3
===============================================
Problem 4
See the attached image (figure 4)
a = 10 which is half of the distance between the highest and lowest points
T = 8 is the period
b = 2pi/T = 2pi/8 = pi/4
c = -pi/2 is the phase shift since its really a cosine graph
d = 0 is the midline
The function is
y = a*sin(bx-c)+d
y = 10*sin((pi/4)*x+(-pi/2))+0
y = 10*sin((pi/4)*x+pi/2)
===============================================
Problem 5
See the attached image (figure 5)
a = 2 is the amplitude since it bobs up and down this distance from the midline
T = 8 seconds is the period (double that of the time it takes for it to go from the highest to the lowest point)
b = 2pi/T = 2pi/8 = pi/4
c = 0 is the phase shift as the buoy starts at normal depth of 20 meters
d = 20 is the midline
The function is
y = a*sin(bx-c)+d
y = 2*sin((pi/4)x-0)+20
y = 2*sin((pi/4)x)+20
===============================================
See the attached image (figure 1)
16pi seems like a typo. I'm going to assume that it's a fraction and it is 1/(6pi)
f = 1/(6pi) = frequency
T = 1/f = 1/(1/(6pi)) = 6pi
Amplitude = 2
a = 2
b = 2pi/T = 2pi/(6pi) = 1/3
Midline: y = 3
d = 3
The function is
y = a*sin(bx-c)+d
y = 2*sin(1/3*x-0)+3
y = 2*sin(x/3)+3
===============================================
Problem 2
See the attached image (figure 2)
T = 12 is the period
a = 4 is the amplitude
b = 2pi/T = 2pi/12 = pi/6
y = 1 is the midline so d = 1
The y intercept is (0,1) which is the midline, which indicates no phase shifts have occurred so c = 0
The function is
y = a*sin(bx-c)+d
y = 4*sin((pi/6)x-0)+1
y = 4*sin((pi/6)x)+1
===============================================
Problem 3
See the attached image (figure 3)
Period = 4pi
T = 4pi
b = 2pi/T = 2pi/(4pi) = 1/2 = 0.5
Amplitude = 2
a = 2
Midline: y = 3
d = 3
y-intercept: (0,3)
The function is a reflection of its parent function over the x-axis, so 'a' is negative meaning a = -2 instead of a = 2
The function is
y = a*sin(bx-c)+d
y = -2*sin(0.5x-0)+3
y = -2*sin(0.5x)+3
===============================================
Problem 4
See the attached image (figure 4)
a = 10 which is half of the distance between the highest and lowest points
T = 8 is the period
b = 2pi/T = 2pi/8 = pi/4
c = -pi/2 is the phase shift since its really a cosine graph
d = 0 is the midline
The function is
y = a*sin(bx-c)+d
y = 10*sin((pi/4)*x+(-pi/2))+0
y = 10*sin((pi/4)*x+pi/2)
===============================================
Problem 5
See the attached image (figure 5)
a = 2 is the amplitude since it bobs up and down this distance from the midline
T = 8 seconds is the period (double that of the time it takes for it to go from the highest to the lowest point)
b = 2pi/T = 2pi/8 = pi/4
c = 0 is the phase shift as the buoy starts at normal depth of 20 meters
d = 20 is the midline
The function is
y = a*sin(bx-c)+d
y = 2*sin((pi/4)x-0)+20
y = 2*sin((pi/4)x)+20
===============================================