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AN630078's latest activity
AN630078
posted the thread
Iterative root finding for the cube root of 17
in
Precalculus Mathematics Homework Help
.
Firstly, the cube root of 17 is 2.571281591 which is 2.57 to 3.s.f. Initially, I thought about just approaching this problem using the...
Nov 12, 2020
AN630078
replied to the thread
Modulus Inequalities -- Help Needed Please
.
Thank you for your reply. Yes, I found found that with nonlinear inequalities. Graphs certainly have been helpful to my understanding...
Oct 16, 2020
AN630078
reacted to
haruspex's post
in the thread
Modulus Inequalities -- Help Needed Please
with
Like
.
Yes, that's the right graph now. Solving nonlinear inequalities does tend to be messy, often needing to be broken down into cases...
Oct 16, 2020
AN630078
reacted to
haruspex's post
in the thread
Modulus Inequalities -- Help Needed Please
with
Like
.
For 1 I would start with |3x-2| is nonnegative so |3x-2|<1/x implies x>0. I didn't follow your reasoning here: When 0<x<2/3 rewrite...
Oct 15, 2020
AN630078
reacted to
epenguin's post
in the thread
Solving Reciprocal Trigonometric Equation cot^2θ+5cosecθ=4
with
Like
.
OK, equations involving the reciprocal trig functions - reciprocal.equations are someting else (equations where the reciprocals of roots...
Oct 15, 2020
AN630078
replied to the thread
Modulus Inequalities -- Help Needed Please
.
Thank you for your reply. For 1, yes sorry my workings should simplify -3x+2-1/x<0 by converting to a fractional form; -3xx/x+2x/x-1/x<0...
Oct 15, 2020
AN630078
replied to the thread
Solving Reciprocal Trigonometric Equation cot^2θ+5cosecθ=4
.
No do not worry, I am pleased you have used it, now I know its meaning 😁
Oct 8, 2020
AN630078
reacted to
fresh_42's post
in the thread
Solving Reciprocal Trigonometric Equation cot^2θ+5cosecθ=4
with
Like
.
My fault, and worst of all: such an internet speak is against the rules ...:mad: Not sure whether I can give myself a warning?!
Oct 8, 2020
AN630078
reacted to
fresh_42's post
in the thread
Solving Reciprocal Trigonometric Equation cot^2θ+5cosecθ=4
with
Like
.
Nine point eight degrees something, i.e. ##9.8\ldots °##
Oct 8, 2020
AN630078
replied to the thread
Solving Reciprocal Trigonometric Equation cot^2θ+5cosecθ=4
.
Oh, ok I was confused by the "something". Thank you for explaining 👍
Oct 8, 2020
AN630078
replied to the thread
Solving Reciprocal Trigonometric Equation cot^2θ+5cosecθ=4
.
Thank you for your reply. Oh splendid, just to ask what do you mean be 9.8°sth?
Oct 8, 2020
AN630078
reacted to
fresh_42's post
in the thread
Solving Reciprocal Trigonometric Equation cot^2θ+5cosecθ=4
with
Like
.
Yes. I haven't checked the digits behind the point though, but it was 9.8°sth. I checked whether it is a better number in radians, but...
Oct 8, 2020
AN630078
replied to the thread
Solving Reciprocal Trigonometric Equation cot^2θ+5cosecθ=4
.
Yes, that is what I meant 😁
Oct 8, 2020
AN630078
reacted to
fresh_42's post
in the thread
Solving Reciprocal Trigonometric Equation cot^2θ+5cosecθ=4
with
Like
.
Maybe the co's: cosec, cotan; all inverses of the 'normal'.
Oct 8, 2020
AN630078
replied to the thread
Solving Reciprocal Trigonometric Equation cot^2θ+5cosecθ=4
.
The reciprocal of sinθ is cosecθ.
Oct 8, 2020
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