There is $100 it goes up 10% how much is it
2 answers:
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Answer:
318 cm.
Step-by-step explanation:
Let x represent the distance between Bill and fulcrum.
We have been given that Laura has a mass of 60 kg and is sitting 265 cm from the fulcrum of a seesaw. Bill has a mass of 50 kg.
To balance the seesaw, the product of Laura's weight and her distance from fulcrum of seesaw should be equal to the product of Bill's weight and his distance from fulcrum of seesaw as:
Therefore, Billy should be 318 cm far from the fulcrum to balance the seesaw.
Answer:
first exercise
V = 16 π
second exercise
V= 68π/15
Step-by-step explanation:
Initially, we have to plot the graph x = (y − 5)2 rotating around y = 3 and the limitation x = 4
<em>vide</em> picture 1
The rotation of x = (y − 5)2 intersecting the plane xy results in two graphs, which are represented by the graphs red and blue. The blue is function x = (y − 5)2. The red is the rotated cross section around y=3 of the previous graph. Naturally, the distance of "y" values of the rotated equation is the diameter of the rotation around y=3 and, by consequence, this new red equation is defined by x = (y − 1)2.
Now, we have two equations.
x = (y − 5)2
xm = (y − 1)2 (Rotated graph in red on the figure)
The volume limited by the two functions in the 1 → 5 interval on y axis represents a volume which has to be excluded from the volume of the 5 → 7 on y axis interval integration.
Having said that, we have two volumes to calculate, the volume to be excluded (Ve) and the volume of the interval 5 → 7 called as V. The difference of V - Ve is equal to the total volume Vt.
(1) Vt = V - Ve
Before start the calculation, we have to take in consideration that the volume of a cylindrical shell is defined by:
f(y) represents the radius of the infinitesimal cylinder.
Replacing (2) in (1), we have
V = 16 π
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Second part
Initially, we have to plot the graph y = x2 and x = y2, the area intersected by both is rotated around y = −7. On the second image you can find the representation.
<em>vide</em> picture 2
As the previous exercise, the exclusion zone volume and the volume to be considered will be defined by the interval from x=0 and y=0, to the intersection of this two equations, when x=1 and y = 1.
The interval integration of equation y = x2 will define the exclusion zone. By the other hand, the same interval on the equation x=y2 will be considered.
Before start the calculation, we have to take in consideration that the volume of a cylindrical shell is defined by:
Notice that in the equation above, x and y are switched to facilitate the calculation. f(x) is the radius of the infinitesimal cylinder
Having this in mind, the infinitesimal radius of equation (3) is defined by f(x) + radius of the revolution, which is 7. The volume seeked is the volume defined by the y = x2 minus the volume defined by x=y2. As follows:
V= 68π/15
