yes it looks right to me Because all you needed to do is to plug in the value 39cm for x and calculate the answer, so u are correct.
Math - Implict differentiation
How do I find the value of d^2y/dx^2 for the function defined implicitly by xy^2 + y = 2 at the point (1,-2)?
What are two dimensional arrays and can they be seen anytime as a one dimensional array?
Differentiate in terms of x: 1] y = cos^3 5x 2] y = sec^2 3x
Hey I am doing a Process Analysis assignment. I chose "How a magician saws a woman in half" I need some help in finding information for this topic as it relates to the stages and steps taken in doing this magic. Also i will like to find some visual of how the magic i...
Somtething seems to be wrong with this question to me. This is what i got:[substitution method] y+2x=3 (1) y+2x=-4 (2) from equation (2)y+2x+-4 => y=-4+2x call this equation (3) substitute equation (3) in equation (1) -4+2x+2x=3 4x=7 x=7/4 for x =7/4, substitute x in equati...
ok here is what i got: 8x+5y=184 (1) x-y=3 (2) from equation (1) 8x+5y=184 => 8x=184-5y x=184-5y/8 call this equation (3) substitute equation (3) in equation (2) 184-5y/8-y =3 (184/8)-(5y/8)-y=3 (184/8)-(13y/8)=3 (184/8)-3=13y/8 160/8=13y/8 20=13y/8 [cross multiply] 20(8)=1...
This is the substitution method y-2x=8 (1) 2y+x=1 (2) From equation (1) y-2x=8 would now be y=8+2x you can call this equation (3) so now, we substitute equation (3) in equation (2) 2(8+2x)+x=1 16+4x+x=1 16+5x=1 5x=1-16 5x=-15 x=-15/5 x=-3 for x=-3, substitute it in equation (3...
Well these would be two equation so: y=6-x (1) y=x+2 (2) so, substitute equation (1) y=6-x into equation (2) therefore, 6-x=x+2 6-2=x+x 4=2x 4/2=x 2=x for x=2, substitute x in equation (1) y=6-2 y=4 x=2 , y=4
how do i evaluated these limits: lim x --> 3 ((square root of x^2 +16)- 5)/(x^2 - 3x) lim x--> to infinity (2+ 100/x)
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