Troubling math!
- Thread starter RD4life
- Start date
That may be true, but you can still be terrible at Math and excel at Science.
I'm awesome (with what I know) at Math, but I don't particularly like anything else. Probably because you don't need to remember as much with Math, in terms of definitions, and the like...
What you do here is incorrect, you have to do the same on both sides, you can't substract with 6 on one side and multiply the other side with -6. And then you express Y in X, not X in Y.
Y in X would be y = 3/(x-5) - 6.
I have some spare time, so I can do the first problem you have up there.
1/5p+3=3/10p-8
I'm assuming you need to have p by itself in this equation, so that's what I'll be aiming for.
2/10p+3=3/10p-8
Subtract 3/10p to both sides.
-1/10p+3=3-8
Subtract 3 to both sides.
-1/1op=-8
divide -1/10 to both sides.
p=80
OR
-1/10p=-8
multiply 10 to both sides to cancel the denominator.
-1p=-80p
divide -1 to both sides.
p=80
Pretty certain that this is the right answer. I'm not that much confident in my algebra solving, but if you're confused with any parts of the work that I did, feel free to ask.
1/5p+3=3/10p-8
I'm assuming you need to have p by itself in this equation, so that's what I'll be aiming for.
2/10p+3=3/10p-8
Subtract 3/10p to both sides.
-1/10p+3=3-8
Subtract 3 to both sides.
-1/1op=-8
divide -1/10 to both sides.
p=80
OR
-1/10p=-8
multiply 10 to both sides to cancel the denominator.
-1p=-80p
divide -1 to both sides.
p=80
Pretty certain that this is the right answer. I'm not that much confident in my algebra solving, but if you're confused with any parts of the work that I did, feel free to ask.
I think you're gonna get a lot of mixed answers here. The way you're writing them is a bit unclear, for example;
15/2x-1=3 could mean (15/2x) -1 = 3 or (15/2x-1) = 3
If you could clarify them I'd be happy to attempt them :)
15/2x-1=3 could mean (15/2x) -1 = 3 or (15/2x-1) = 3
If you could clarify them I'd be happy to attempt them :)
Well, you'd have to use the product rule by taking the derivative of x^3 and multiplying that by sin(x), then add that whole mess by the derivative of sin(x) multiplied by x^3. Therefore your answer would be f`(x)= 3x^2(sin(x)) + x^3(cos(x)). You could simplify this further by factoring out x^2 leaving you with f`(x)= x^2(3sin(x) + x(cos(x)))
Calculus!
as for the problems in the orinigal post, WITH STEPS!
6+y=3/t with T=x-5
Express X in Y.
6+y=3/(x-5)
(x-5)(6+y)=3 Multiply both sides by (x-5)
x-5=3/(y+6) divide both side by (y+6)
x=(3/(y+6))+5 add 5 to both sides
1/5p+3=3/10p-8
need clarification on what is in each of the denominators
15/2x-1=3
15/2x = 4 add 1 to both sides
15=8x multiply both sides by 2x
15/8=x divide both sides by 8
2(q=1)=4p-5 (Express P in Q)
2(q-1)+5=4p add 5 to both sides
2q-2+5=4p multiply the 2 into the parentheses
(2q+3)/4 = p divide both sides by 4
q+8-1/p+6=0 (Express P in Q)
need some clarification on this one too
is it supposed to be (q+8)-(1/p)+6, because that's how I'm reading it.
I don't normally do other people's homework, but if you're truly having trouble figuring these out and not being lazy, I'll be happy to help.
Wasn't sure if you meant f(x)=x^(3sin(x)) or f(x)=(x^3)(sin(x)). If you meant the second then do what soapless said. Otherwise:
y=x^(3sin(x))
lny=ln(x^(3sin(x)))
lny=(3sin(x))(ln(x))
(1/y)(dy/dx)=(3sin(x))(1/x)+(cos(x))(ln(x))
dy/dx=y(3sin(x))(1/x)+(cos(x))(ln(x))
dy/dx=(x^(3sin(x))((3sin(x)/x)+cos(x)ln(x))