#1

okay i have a problem.

if a defect thermometer shows 1 degree C in a melting ice cube and 105 degrees in the steam of boiling water, how much will it show in an environment of 17 degrees?

possible answers:

A 15,43 °C

B 15,41 °C

C 15,39 °C

D 15,28 °C

E 15,17 °C

please hurry!

EDIT:

if anyone could integrate this for me would be awesome as well.

if a defect thermometer shows 1 degree C in a melting ice cube and 105 degrees in the steam of boiling water, how much will it show in an environment of 17 degrees?

possible answers:

A 15,43 °C

B 15,41 °C

C 15,39 °C

D 15,28 °C

E 15,17 °C

please hurry!

EDIT:

if anyone could integrate this for me would be awesome as well.

*Last edited by CoreysMonster at May 4, 2008,*

#2

A, definitely A

#3

....wouldn't this be a chemistry question?

#4

....wouldn't this be a chemistry question?

not really

A, definitely A

C, the answer is always C

actually in seriousness, i have no idea

though i'm probably doing something wrong, i can't really think at midnight on a sunday

1 degree C in a melting ice cube

0 degrees IRL = 1 degrees on the thermometer

105 degrees in the steam of boiling water

100 degrees IRL = 105 degrees on the thermometer

thus

thermometer temp= 1+actual temeperature*104/100

1+17*104/100 = 18.68 = wtf

[edit: if the thermometer was showing 17° that'd mean its actually 15.3846154, maybe you got the question wrong?]

edit: come on, that integral is easy

since its a sum it just splits up into 3 integrals

and

int cosx dx = sin x

int sinx dx = -cos x

the constants come out before each of them

and then you just insert the boundries

edit: 2

*Last edited by seljer at May 4, 2008,*

#5

edit: come on, that integral is easy

since its a sum it just splits up into 3 integrals

and

int cosx dx = sin x

int sinx dx = -cos x

the constants come out before each of them

and then you just insert the boundries

So i guess it's 5sin(x)-cos(x)?

which would be -1, but since Integrals are always positive the answer is 1, right?

EDIT: and i had the same approach as you did, to the physics one, didn't make any sense either.

#6

I would tell you the answer because I just worked it out, but you won't learn by cheating.

Sorry. Read your science/physics text book.

EDIT: Meh, good mood. It's E.

According to my calculations it should be less though... E is the closest/most possibly answer.

Sorry. Read your science/physics text book.

EDIT: Meh, good mood. It's E.

According to my calculations it should be less though... E is the closest/most possibly answer.

*Last edited by Seth Shadows at May 4, 2008,*

#7

I would tell you the answer because I just worked it out, but you won't learn by cheating.

Sorry. Read your science/physics text book.

dude, this is for an online test for getting into a university. I can only do it once, and i didn't notice that until i started the test. i have 20 minutes left and have worked out 20 problems right now at 12.22 at night. this is the only one i'm stuck on, so bear with me and please give me the answer.

EDIT: alright, thanks guy!

#8

So i guess it's 5sin(x)-cos(x)?

which would be -1, but since Integrals are always positive the answer is 1, right?

EDIT: and i had the same approach as you did, to the physics one, didn't make any sense either.

nope, integrals can be negative too (if its under the graph that part of the area comes out negative)

yep, 5sin(x)-cos(x) is right

and when you insert the numbers

5sin(2pi)-cos(2pi)-(5sin(0)-cos(0))=

5*0-1-(5*0-1) = 0

yeah, the answer is 0

that 2 i got up there was only for 0 to pi, not 2*pi

#9

nope, integrals can be negative too (if its under the graph that part of the area comes out negative)

yep, 5sin(x)-cos(x) is right

and when you insert the numbers

5sin(2pi)-cos(2pi)-(5sin(0)-cos(0))=

5*0-1-(5*0-1) = 0

yeah, the answer is 0

that 2 i got up there was only for 0 to pi, not 2*pi

ah ****!! I forgot that sin(0) isn't 0! I'm such an idiot... well, as it turns out i only got 58% of the test right... not cool. oh well, thanks guys.

#10

Seriously?

Ouch. Sorry man if part was my fault in calculations.

Try again?

Ouch. Sorry man if part was my fault in calculations.

Try again?

#11

Seriously?

Ouch. Sorry man if part was my fault in calculations.

Try again?

it doesn't show the right answers. oh well, i don't think they expected me to know everything, one of the questions was "estimate how many new automobiles are produced each year". I'm sure it's just to give an impression of what kind of things you'll be learning.. thanks a bunch though

#12

it doesn't show the right answers. oh well, i don't think they expected me to know everything, one of the questions was "estimate how many new automobiles are produced each year". I'm sure it's just to give an impression of what kind of things you'll be learning.. thanks a bunch though

That reminds me of a scholarship test I had to take that had a section for "current events." This was in 2006, and one of the questions was "Who won the 1998 World Series?" I lol'd hard.