Which composition of transformations maps figure EFGH to figure E"F"G"H"?
1 answer:
The composition of transformations maps figure EFGH to figure E"F"G"H" is Option A.a reflection across line k followed by a translation down.
<h3>What is translation?</h3>A translation can be explained as the movement of a shape in a direction which could be up, down but the appearance of the figure remain unchanged in any other way.
And from the figure, we can see that a translation is used, because of the movement of EFGH to figure E"F"G"H" which occurred in the same direction.
Hence, composition of transformations maps figure EFGH to figure E"F"G"H" is Option A.a reflection across line k followed by a translation down.
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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
===============================================
Answer:
Step-by-step explanation:
Volume of cylinder:
<u>Bottom cylinder:</u>
1 inch = 2.54 cm
diameter = 14 in
r = 14/2 = 7 in
r = 7 *2.54 = 17.78 cm
h = 4 in = 4*2.54 = 10.16 cm
Volume = 3* 17.78 * 17.78 * 10.16
= 9635.59 in³
<u>Upper cylinder:</u>
diameter = 12 in
r = 12/2 = 6in
r = 6*2.54 = 15.24 cm
h = 12 in = 12*2.54 = 30.48 cm
Volume = 3 * 15.24 *15.24 * 30.48
= 21237.63 in³
Volume of the object = 9635.59 + 21237.63
= 30873 cm³