Question

NO ONE CAN ANSWER THIS :((

WILL GIVE MEDALS TO ALL WHO TRY

Answer 1

Wow, that is hard.
I wish I can help you though.
What subject of Math is it?

Answer 2

just calculus. i have no idea where to begin i tried plugging in the x variable into the original equation

Answer 3

Oh okay, sorry haven't been to Calculus I am just learning Algebra

Answer 4

Are you good at Algebra 1?

Answer 5

@kahsdf ?

Answer 6

Sorry I have a couple of exams to study for. I can't help any of your problems atm.

Answer 7

Oh

Answer 8

Is that the question?

Answer 9

the question that needs to be answered has ????? in the answer box

Answer 10

you have to plug in 4 into the first derivative i.e. y'

Answer 11

the problem is that i am retarded and can't put the 4=x into y=mx+b form. I was trying for the longest time with point slope but just can't do the calculation

Answer 12

\[\sqrt{x}+\sqrt{y}=11\]
\(\frac{1}{2}x^{-\frac{1}{2}}+\frac{1}{2}y^{-\frac{1}{2}}\frac{dy}{dx}=0\)
so
\(\frac{1}{2}y^{-\frac{1}{2}}y'=-\frac{1}{2}x^{-\frac{1}{2}}\)

Answer 13

when \(x=4\)
\(y=81\)

Answer 14

so y'(4) = -9/2

Answer 15

everything else is correct in this problem...so y'(x)=-sqrt(y)/sqrt(x)

its not 81, have a medal tho for trying

Answer 16

ok -9/2 is good thanks

Answer 17

yep

Answer 18

could you help me with one last problem?

Answer 19

sure

Answer 20

again another problem involving point slope

Answer 21

hmm i didnt use point slope at all on that...

Answer 22

its not loading

Answer 23

i meant this one involves point slope i believe. my brains is kind of fried from doing calc all day, was working on related rates ughh..interesting though

Answer 24

hmm I dont know if it is me, but its not loading. Just keeps trying

Answer 25

http://imgur.com/w3asHvS

Answer 26

i don't know sorry

Answer 27

plug in 1 for x, and what do you get?

Answer 28

in the f'....

Answer 29

1/2(13)^(-1/2)*(13)

Answer 30

\(\frac{\sqrt{13}}{2}\) i do believe?

Answer 31

those are equivalent

Answer 32

but i know how to proceed from here thanks a million!!

Answer 33

@kahsdf Those problems are long not difficult. :P

Answer 34

that's why they are tricky.. lots of simple mistakes can be made

Answer 35

so this is your slope.
so
\(y = mx+b \implies y=\frac{\sqrt{13}}{2}x+b\)use the point given

\(\sqrt{13}=\frac{\sqrt{13}}{2}*1+b\) so
\(b=\frac{\sqrt{13}}{2}\)
so

\(\large y = \frac{\sqrt{3}}{2}x+\frac{\sqrt{3}}{2}\)

Answer 36

oh, sorry. I didn't see that you had it from here.

Answer 37

its all good i appreciate the help

Answer 38

np