Register
Login
Username
Password
Login
forgot your password?
Home
Forum
+0
Formulary
Maths
Help
Complex Numbers
Integral Calculus
Differential Calculus
Equations
Graphs
Linear Algebra
Number theory
Percent
Standard Functions
Statistics
Trigonometry
Unit Conversion
Units
About
Imprint
Privacy Policy
Terms of Service
Credits
Google+
Facebook
Contact Email
scrutinizer
Username
scrutinizer
Score
259
Membership
Stats
Questions
1
Answers
125
1 Questions
125 Answers
0
1038
1
+259
How to convert Mbps in MBps
Hi, I wanted to use the sci calc, but don't know how the command line should look like for converting data speed of 9.98 Mbps in MBps. Could you give me an advise?
Thanks
Melody
●
scrutinizer
Nov 30, 2014
#1
+259
0
9 = 3*3, 15 = 3*5. 9 consists of 2 numbers 3, 15 contains one number 3. Hence the common lower multiple is 3.
scrutinizer
Sep 20, 2013
#2
+259
0
d
means "differentiation operator". If you tell this to you teacher he will be agreeably surprised, assure you.
scrutinizer
Sep 18, 2013
#2
+259
0
x-o.2x+o.o7(x-0.2x)=73.53 => (x - 0.2x)(1 + 0.07) = 73.53 => (x - 0.2x)*1.07 = 73.53 => 0.8x*1.07 = 73.53 => 0.856x = 73.53 => 856x = 73530 => x = 36765/428
solve(x-(0.2x)+0.07(x-(0.2x))=73.53,x)
scrutinizer
Sep 13, 2013
#1
+259
0
log ab^2 = log(b^2, a) = 2log(b,a) = 2log(7,3)
log(7^2, 3)
scrutinizer
Sep 12, 2013
#1
+259
0
5/3
scrutinizer
Sep 12, 2013
#2
+259
0
However, there's another way to solve this equation using 2 Bezout theorems about roots of polynomial. According to the 1st one, the polynomial p(x) can be exactly divided by the binomial (x - a) if x = a and the quotient is p(a), and according to the 2nd one if the polynomial with the whole coefficients have roots that are the whole numbers then these roots are divisors of the polynom's constant term.
In this case we should apply the theorem 2 first and look for divisors of 6. They are -1,1;-2,2;-3,3;-6;6. Then we insert into the equation them to make sure whether one of them fits it. The suitable integer is 3. As the coeffs of the expression are the whole numbers the polynomial is being divided by the binomial x - 3 exactly, so its 1rst root is x = 3. To find the other roots we should divide the polynomial x^3-x^2-4x-6 by x - 3. Performing long division we get the quotient x^2 + 2x + 2, so the full arrangement will look as (x - 3)(x^2 + 2x + 2) = 0 The first factor has the root already had been calculated (x = 3). The second equation has no roots among the real numbers but has two roots among the complex numbers: x2 = -1+ i, x3 = -1 - i
solve(x^3-x^2-(4x)-6 = 0,x)
scrutinizer
Sep 11, 2013
#1
+259
0
Strongly recommend to read this on how to solve cubic equations:
http://mathforcollege.com/nm/mws/gen/03nle/mws_gen_nle_bck_exactcubic.pdf
scrutinizer
Sep 9, 2013
#1
+259
0
(2 sqrt50 - 3 sqrt8 )^2. First you could simplify the expression: (2 sqrt{25*2} -3 sqrt{4*2} )^2 = (2 *5 *sqrt{2} -3*2* sqrt{2} )^2 = (10sqrt{2} - 6* sqrt{2} )^2 = (4sqrt(2))^2 = 16*2 = 32
scrutinizer
Sep 9, 2013
#2
+259
0
More detailed version: 2/3h-9=6-2/3h => 2 - 9*(3h) = 6*(3h) - 2 =>15*(3h) = 4 =>3h = 4/15 => h = 4/45
scrutinizer
Sep 9, 2013
#1
+259
0
When something "lies between" it's always two objects (numbers) it lies in between. So, "between sqrt(26) and..."?
scrutinizer
Sep 9, 2013
«
latest
11
10
..
9
8
7
6
5
..
2
1
»