Recent content by fashion_fever

what is the pH of a 0.140M NaOCN solution? ka HOCN=3.5x10^-4 (rule of 100 can simplify 0.140-x to 0.140) i know that ka=HOCN*OH^-/OCN^-=x^2/0.140 but i don't know the left hand side...i don't know what to do with HOCN=3.5x10^-4..

um...i don't get it..i didn't find i wrote 9(36+3x)^1/2..so what should it be? and for the x=144,x=36, i do substitute those in right?...

so it would just be: 2(468)^3/2 - 3456 ?

Homework Statement evaluate: higher limit of 36 lower limit of 0 (36+3x)^1/2 dx Homework Equations i thought of using subsititution? The Attempt at a Solution g(x)=36+3x g'(x)=3 when x=0, u=36+3(0)=36 when x=36, u=36+3(36)=144 from lower limit of 36 to higher...

oh! omg..okay i got it it was suppose to be +40.2cos58.6.. tytytytyty!

Homework Statement Three astronauts equipped with jetpacks are pushing a satellite, having a mass of 127 kg, through space. The forces being exerted on the satellite are F1=31.7 N at an angle of 30.8 degrees above the horizontal, F2=40.2 N at 58.6 degrees below the horizontal, and F3=53.9 N...

oh, okay good.then i try to simply it i got [e^x^2+x^2e^x+6e^x+6]/(x+e^x)^2...which is not right...

[(-e^x-xe^x)(x+e^x)-(1+e^x)(6-xe^x)]/(x+e^x)^2 is this step right?

sorry, i changed it to du. For the differentiate, I did use the quotient and product rule, but somehow the answer is wrong.

Differentiate: (6-xe^x)/(x+e^x)=... Evaluate Intergral: (2u^5-8u^3+5u^2)du, from lower limit -1 to higher limit 0. (i got 10/3 for some reason...) (u^7+6u^13)/u^9 du, from lower limit 1 to higher limit 2. (n i got 47 for this..) I tried these a few times but my answer is still...

got it thanks.

The velocity-versus-time graph is shown for a particle moving along the x-axis. Its initial position is xo=4.80 m at to=0.00 s. What is the particle's position at t=1.25 s if the total time the particle moves is 5.00 s and the maximum velocity is 2.00 m/s? 1)At t=3.75 s, what is the...