# Tutor profile: Frank H.

## Questions

### Subject: Spanish

Can anyone explain to me the difference between Ser and Estar?

Both ser and estar are translated as to be (you probably knew that) I'll make some questions: Who ARE you? i AM a 21 years old guy What ARE you? i AM a doctor Where ARE you? i AM at school How ARE you? i AM sad. In Spanish: Quien ERES tu? yo SOY un chico de 21 años Que ERES tu? yo SOY un doctor Donde ESTAS tu? yo ESTOY en la escuela Como ESTAS tu? yo ESTOY triste As you can see in English it's the same.... In Spanish it's quite diferent for WHO/WHAT questions/answers we use SER for WHERE/HOW questions/answers we use Estar

### Subject: Calculus

Worked solution to differentiation of: (e^x - e^-x)/(e^x + e^-x)

You need to use the quotient rule: {f/g}' = {g*f' - f*g'}/g^2 In your case, f =(e^x - e^-x), g = (e^x + e^-x) {f/g}' =[(e^x + e^-x)*(e^x - e^-x)' - (e^x - e^-x)*(e^x + e^-x)']/(e^x + e^-x)^2 {f/g}' =[(e^x + e^-x)*(e^x - e ^-x(−1)) - (e^x - e^-x)*(e^x + e ^-x(−1))]/(e^x + e^-x)^2 {f/g}' =[(e^x + e^-x)*(e^x + e^-x) - (e^x - e^-x)*(e^x - e^-x)]/(e^x + e^-x)^2 {f/g}' = [(e^x + e^-x)^2 - (e^x - e^-x)^2]/(e^x + e^-x)^2 Multiply out [(e^2x + 2e^x*e^-x + e^-2x) - (e^2x -2e^x*e^-x +e^-2x)]/(e^x + e^-x)^2 (e^2x + 2e^x*e^-x + e^-2x - e^2x + 2e^x*e^-x - e^-2x)/(e^x + e^-x)^2 {f/g}' = 4e^x*e^-x /(e^x + e^-x)^2 Solution is {f/g}' = 4/(e^x + e^-x)^2

### Subject: Physics

Pat accidentally drops his cell phone (0.25kg) out of a window 10m above the ground.?

Strangely you need a spherical phone for this to work out. If a normal phone hits the trampoline one edge hits first and a fair bit of energy goes into spinning the phone. But don't take my word for it. Throw your phone onto a trampoline from 10 m and see how many times it comes straight back to you and how many times it fires off at some weird angle. If all the energy goes into the trampoline then kinetic energy -> potential ( spring) energy m g h = 1/2 k x^2 2mgh / k = x^2 x = sqrt( 2mgh / k ) = sqrt(2* 0.25 * 9.8 * 10 / 500) = 0.31 m approx.

## Contact tutor

needs and Frank will reply soon.