Amplitutde:
Normally for sinx amplitude is 2 (1 +1). In the case when it is half sine amplitude is halved, therefore 0.5 +0.5 = 1 Paramers alpha = 2x does not matter because it only stretches the graph horizontally.
4.A store pays $85 for a table. The markup is 90%. What is the selling price?
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Amplitutde:
Normally for sinx amplitude is 2 (1 +1). In the case when it is half sine amplitude is halved, therefore 0.5 +0.5 = 1 Paramers alpha = 2x does not matter because it only stretches the graph horizontally.
Have a look at the image attachment. I've added the points A, B, C, D such that
A = base of the diving board pole, or location of the pool deck
B = the location 1.6 meters directly above point A
C = location of the person's eyes
D = diving board location
The goal is to find the length of segment AD
We are given
AB = 1.6
BC = 3.67
what we want to find is
BD = x
Due to the fact we have similar right triangles ABC and CBD, we can form the proportion below and solve for x
AB/BC = BC/BD
1.6/3.67 = 3.67/x
1.6*x = 3.67*3.67
1.6*x = 13.4689
1.6*x/1.6 = 13.4689/1.6
x = 8.4180625
So BD is 8.4180625 meters
Use this to find the length of AD
AD = AB + BD
AD = 1.6 + 8.4180625
AD = 10.0180625
which rounds to 10.0 meters when rounding to the nearest tenth (one decimal place)
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Answer: choice D) 10.0 m
A = base of the diving board pole, or location of the pool deck
B = the location 1.6 meters directly above point A
C = location of the person's eyes
D = diving board location
The goal is to find the length of segment AD
We are given
AB = 1.6
BC = 3.67
what we want to find is
BD = x
Due to the fact we have similar right triangles ABC and CBD, we can form the proportion below and solve for x
AB/BC = BC/BD
1.6/3.67 = 3.67/x
1.6*x = 3.67*3.67
1.6*x = 13.4689
1.6*x/1.6 = 13.4689/1.6
x = 8.4180625
So BD is 8.4180625 meters
Use this to find the length of AD
AD = AB + BD
AD = 1.6 + 8.4180625
AD = 10.0180625
which rounds to 10.0 meters when rounding to the nearest tenth (one decimal place)
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Answer: choice D) 10.0 m
By the divergence theorem, the surface integral can be evaluated by computing the triple integral