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Ronaldo95163 wrote:The Sharp EL-W516X is by far the best I have ever used thus far!
stev wrote:anyone has nice shortcuts to share? like easy ways of calculated long questions?
see this guy:
Ms. Arroyo asked the class to see if they could find the sum of the first 50 odd numbers. As everyone settled down to their addition, Terry ran to her and said, "The sum is 2,500." Ms. Arroyo thought, "Lucky guess," and gave him the task of finding the sum of the first 75 odd numbers. Within 20 seconds, Terry was back with the correct answer of 5,625.
How does Terry find the sum so quickly?
stev wrote:ok this is mess. lol
An area in the first Cartesian coordinate is bounded by one function, the line x = 1, and the x axis (y = 0).
The boundary function is y = (1/x) starting at x =1 and it goes to x = infinity.
Rotate this area about the x axis so it creates a volume.
Calculate the surface area and the volume of this object.
redmanjp wrote:How bout this
A carpenter makes tables and chairs. Each table can be sold for a profit of £30 and each chair for a profit of £10. The carpenter can afford to spend up to 40 hours per week working and takes six hours to make a table and three hours to make a chair. Customer demand requires that he makes at least three times as many chairs as tables. Tables take up four times as much storage space as chairs and there is room for at most four tables each week.
To get maximum profit, how much tables & chairs should he make & how much profit would he get?
stev wrote:^^^correcto!!!
that was the answer!!! well done corn bird
Trini Hookah wrote:888 + 88 + 8+ 8 + 8 = 1000
Ronaldo95163 wrote:Corn Bird what version of Maple is that?
Ronaldo95163 wrote:Nice^
re-re-re-re-related rates
A sphere is increasing in volume at the rate of 20∏cm^3/s. Given that the volume of a sphere of radius r is (4/3)∏r^3, calculate the radius of the sphere at the instant when the radius is increasing at the rate of 0.2cm/s
The surface area of a sphere is increasing at a constant rate of 6cm^2/s. Given that the surface area of a sphere of radius r is 4∏r^2 and the volume is (4/3)∏r^3, find the rate of increase of the radius and the volume at the instant when the radius is 5cm
DVSTT wrote:Ronaldo95163 wrote:Nice^
re-re-re-re-related rates
A sphere is increasing in volume at the rate of 20∏cm^3/s. Given that the volume of a sphere of radius r is (4/3)∏r^3, calculate the radius of the sphere at the instant when the radius is increasing at the rate of 0.2cm/s
The surface area of a sphere is increasing at a constant rate of 6cm^2/s. Given that the surface area of a sphere of radius r is 4∏r^2 and the volume is (4/3)∏r^3, find the rate of increase of the radius and the volume at the instant when the radius is 5cm
These kind of questions are so damn annoying! Always get trouble with it!
Ronaldo95163 wrote:
i) If a cyclist of mass 70kg uses a bicycle of mass 7kg, how much work must the cyclist do against gravity in order to ascend to 2100m from sea level(0m)?
ii)One particular descent goes from 2100m to 1600m. Assuming the work done against friction is 90% of the potential energy change of the cyclist and the cycle, what increase in speed in km/h can a rider attain by the end of the descent?
iii)What is the average rate of energy conversion of the cyclist and cycle if the descent in part(ii) takes 1 minute at constant speed?
Use acceleration due to gravity as 10ms^-2
Corn Bird wrote:Ronaldo95163 wrote:
i) If a cyclist of mass 70kg uses a bicycle of mass 7kg, how much work must the cyclist do against gravity in order to ascend to 2100m from sea level(0m)?
ii)One particular descent goes from 2100m to 1600m. Assuming the work done against friction is 90% of the potential energy change of the cyclist and the cycle, what increase in speed in km/h can a rider attain by the end of the descent?
iii)What is the average rate of energy conversion of the cyclist and cycle if the descent in part(ii) takes 1 minute at constant speed?
Use acceleration due to gravity as 10ms^-2
(i) mgh=(77)(10)(2100)
(ii) increase in kinetic energy=(.1)(77)(10)(500); assuming that cyclist begins descent at
zero velocity then increase in speed dv is the square root of (2)(.1)(10)(500). it's probably
a little harder to show that if the cyclist starts at positive speed then the increase in
speed will be smaller than this
(iii) no energy is converted to k.e. as cyclist travels at constant speed.
cyclist converts/loses his p.e. of (77)(10)(500) to friction in 60 s, so rate of conversion is
(77)(10)(500)/60 W
by the way gracen, this site for math students at any level may be interesting
http://math.stackexchange.com/
while this site
http://mathoverflow.net/
is apparently for research mathematicians
Ronaldo95163 wrote:More Calculus
A box with square base has no top. If 64cm^2 of the material is used, what is the maximum possible volume for the box?
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