Number of results: 209,662
math
Are you doing pre-calculus or calculus? In calculus, the maximum is obtained by setting f'(t)=derivative with respect to t = 0 and solving for t. In pre-calculus, you can transform the expression by completing the squares. Ax² + Bx + C =A(x²+ (B/A)x + C/A...
Tuesday, September 8, 2009 at 10:22pm by MathMate
Calculus
Simplify. State the nonpermissible values. 4a^2-1/4a^2-16*2-a/2a-1 This is what I did to try and solve it out: (2a+1)(2a-1)/(2a+4)(2a-4)*2-a/(2a-1) I cancelled out the (2a-1)'s and don't get how to solve. I'm supposed to get the answer : -2a-1/4(a+2) But I don'...
Friday, December 23, 2011 at 5:52pm by -Untamed-
math exam prep question
1/(4a+2) = 1/2(2a+1) 4/(6a+3) = 4/3(2a+1) Now you have 1/2(2a+1) + 4/3(2a+1) Change it all to be over 6(2a+1): 3/6(2a+1) + 8/6(2a+1) = 1/6(2a+1) How did yo get 14? forget the (2a+1) for now 1/2 + 4/3 = 3/6 + 8/6 = 11/6
Tuesday, January 22, 2013 at 3:55pm by Steve
Calculus--Reiny
Calculus - Reiny, Friday, December 23, 2011 at 6:59pm I read that as (4a^2-1)/(4a^2-16) * (2-a)/(2a-1) , my brackets are necessary = (2a+1)(2a-1)/( 4(a+2)(a-2) * (2-a)/(2a-1) the (2-a)/(a-2) are opposite, they will give you the -1 so.. = -(2a+1)/(4(a+2)) , again you have to ...
Sunday, December 25, 2011 at 8:43pm by -Untamed-
Calculus--Reiny
(2a+1)(2a-1)/( 4(a+2)(a-2) * (2-a)/(2a-1) (2a+1)(2a-1) (2-a) ==================== 4(a+2)(a-2)(2a-1) (2a+1)(2a-1) (2-a) ==================== 4(a+2)(-1)(2-a)(2a-1) -(2a+1)(1) (2-a) ==================== 4(a+2)(2-a)(1) -(2a+1)(1) (1) ==================== 4(a+2)(1)(1) -(2a+1...
Sunday, December 25, 2011 at 8:43pm by Damon
pre-calculus
using the quadratic formula, the discriminant is (a^2+2a)^2 - 8a^3 = a^2((a+2)^2 - 8a) = a^2(a-2)^2 So, we see that x = ((a^2+2a) ± a(a-2))/2 = a^2 or 2a so, if the roots are a^2 and 2a, the separation is 2a-a^2. When is that a maximum? When a=1.
Monday, January 28, 2013 at 3:49am by Steve
AP Calculus
2 f ( a ) = 2 ( 3 a ^ 2 - a + 2 ) = 6 a ^ 2 - 2a + 4 f ( 2a ) = 3 [ ( 2a ) ^ 2 ] - 2a+ 2 = 3 ( 4a ^ 2 ) - 2a + 2 = 12 a ^ 2 - 6 a + 6 f ( a ^ 2 ) = 3 [ ( a ^ 2 ) ^ 2 ] - a ^ 2 + 2 = 3 a ^ 4 - a ^ 2 + 2 [f ( a ) ] ^ 2 = ( 3 a ^ 2 - a + 2 ) ^ 2 = 9 a ^ 4 - 6 a ^ 3 +1 3 a ^ 2 - 4...
Friday, September 2, 2011 at 12:31am by Anonymous
Calculus
I read that as (4a^2-1)/(4a^2-16) * (2-a)/(2a-1) , my brackets are necessary = (2a+1)(2a-1)/( 4(a+2)(a-2) * (2-a)/(2a-1) the (2-a)/(a-2) are opposite, they will give you the -1 so.. = -(2a+1)/(4(a+2)) , again you have to use brackets to write it on here.
Friday, December 23, 2011 at 5:52pm by Reiny
calculus
That is correct. The minimum ofr maximum (vertex) is at x = -b/2a You already know that from the quadratic equation. x = -b/2a +/- (b/2a)sqrt (b^2-4ac) Now if x = -b/2a what is y? y = a (b^2/4a^2) + b(-b/2a) + c = b^2/4a -b^2/2a + c = -b^2/4a+c for max or min f" = 2 a if ...
Sunday, May 8, 2011 at 11:34pm by Damon
Calculus
f'(x) = ( a(x^2 - 5x + 4) - (2x-5)(ax+b) )/(x^2 - 5x +4)^2 = 0 for a horizontal line when x = 2 a(-2) + (2a+b) = 0 -2a + 2a + b = 0 b = 0 also f(2) = -1 2a/(4 - 10 + 4) = -1 2a/-2 = 1 a = 1 a=1 , b=0 so f(x) = x/(x^2 - 5x + 4)
Monday, May 28, 2012 at 9:07pm by Reiny
Intermedia Algebra
2a^2 - 3a = 5. 2a^2 - 3a - 5 = 0, Use the AC Method: A*C = 2*(-5) = 10, 2a^2 + (2a - 5a) -5 = 0, Group the 4 terms into 2 factorable pairs: (2a^2 + 2a) + (-5a - 5) = 0, Factor each binomial: 2a(a + 1) -5(a + 1) = 0, Factor out a+1: (a + 1)(2a - 5) = 0, a + 1 = 0, a = -1. 2a - ...
Sunday, June 19, 2011 at 11:21am by Henry
Calculus
from (A) we would know that 1 = a + b + c from (b) we know 1 = -b((2a) or b = -2a from (c) we know that at the y-intercept x = 0 so 6 = 0 + 0 + c, so c=6 so back in the first equation a+b+c = 1 a + (-2a) + 6 = 1 -a = -5 a + 5, then b = -10 and c=6 so f(x) = 5x^2 - 10x + 6
Tuesday, September 2, 2008 at 7:50pm by Reiny
Math
start this way: x^2 + b/a x = -c complete the square x^2 + b/a x + (b/2a)^2=(b/2a)^2-c (x+b/2a)^2=(b/2a)^2 -c take the sqrt of each side... x= -b/2a +- sqrt (b^2-4ac) /2a
Saturday, February 25, 2012 at 8:11pm by bobpursley
math
By the way, the quadratic equation really is completing the square a x^2 + b x + c = 0 x^2 + (b/a) x = -(c/a) x^2 + (b/a) x + (b/2a)^2 = -(c/a) +(b/2a)^2 (x + b/2a)^2 = [-c/a + (b/2a)^2 ] so x+ b/2a = +/- sqrt (b^2/4a^2 - 4 a c/4a^2) or x = -b/2a +/- (1/2a)sqrt(b^2-4 a c)
Monday, January 28, 2008 at 7:24pm by Damon
calculus
So you want the Integral[1/x^3] from 1 to a = integral[1/x^3] from a to 2 (the integral of 1/x^3 is -1/(2x^2) ) then -1/(2a^2) - (-1/2) = -1/(2(4)) - (-1/2a^2) -1/(2a^2) + 1/2 = -1/8 + 1/(2a^2) 1/2 + 1/8 = 2/(2a^2) 5/8 = 1/a^2 5a^2 = 8 a^2 = 8/5 = 1.6 a = √1.6
Thursday, January 21, 2010 at 3:21am by Reiny
Simple Calculus
dy/dx = 3x^2 + 2ax + b when x = -1, dy/dx = 3 3 - 2a + b = 3 b = 2a also the point (2,64) lies on the curve, so 64 = 8 + 4a + 2b 4a + 2b = 56 2a + b = 28 but b = 2a 2a + 2a = 28 a = 7 b = 14 thus a+b = 21
Thursday, March 28, 2013 at 1:14am by Reiny
calculus!URGENT
dy/dx = 3x^2 at x=a , y = x^3 and dy/dx = 3a^2 equation of tangent: y = 3a^2x + b but (a,a^3) is on it, so a^3 = 3a^2(a) + b b = -2a^3 tangent equation: y = (3a^2)x - 2a^3 intersect that line with y = x^3 x^3 = 3a^2x - 2a^3 x^3 - 3a^2x + 2a^3 = 0 We already know that x-a is a ...
Wednesday, August 17, 2011 at 9:05pm by Reiny
Math
In this problem: -2/a+2 + 3a/2a-1 First I made the denominators alike: -2(2a-1)/(a+2)(2a-1) + 3a(a+2)/(a+2)(2a-1) = -4a+2/(a+2)(2a+1) + 3a^2+6a/(a+2)(2a-1) What I want to know is if I can eliminate the a+2 for the first fraction, on both the numerator and the denominator?
Wednesday, May 20, 2009 at 5:22pm by Angie
calculus
look at the transition points, they should be the same. when x = 1 h(x) = -x - 3 = -1-3 = -4 h(1) = a + b so a+b = -4 h(-2) for ax+b = -2a+b h(-2) for x^2 = 4 so -2a + b = 4 a+b = -4 -2a +b = 4 subtract: 3a = -8 a = -8/3, back in a+b= -4 -8/3+b = 4 b = 20/3 a = -8/3 , b = 20/3
Saturday, September 24, 2011 at 2:54pm by Reiny
alg
I get 17/2a - 3 are you sure you do not mean: 7/(a-1) + 3/(2a-2) ??? that is 14/(2a-2) + 3/(2a-2) which is 17/(2a-2) which is 17/[2(a-1)] which is your (b) BE CAREFUL ABOUT PARENTHESES WHEN USING A KEYBOARD FOR MATH EXPRESSIONS!!!
Wednesday, June 30, 2010 at 4:50pm by Damon
calculus
let the point of contact be P(a,1-a^2) dy/dx = -2x, so at our point P the slope of the tangent is -2a equation of tangent: y - (1-a^2) = -2a(x-a) y - 1 + a^2 = -2ax +2a^2 2ax + y = a^2 + 1 the base of the triangle is the x-intercept of this line, the height of the triangle is ...
Tuesday, November 24, 2009 at 5:55pm by Reiny
math
2a^2+9a+4=0 2a^2+a+8a+4=0 a(2a+1)+4(2a+1)=0 (2a+1)(a+4)=0 a=-1/2 and a=-4
Tuesday, October 28, 2008 at 9:28pm by Adithya
mathematics
(2a^2-3a-2)/(4a^2-1) Divided by (a^2-4)/(2a^2-5a+2) Invert divisor(2nd fraction) and multiply: (2a^2-3a-2)/(4a^2-1)*(2a^2-5a+2)/a^2-4. (a-2)(2a+1)/(2a+1)(2a-1)*(a-2)(2a-1)/ (a+2)(a-2) = (a-2)/(a+2).
Sunday, July 3, 2011 at 6:46pm by Henry
trig
tan A = 1/cotA 1/cotA^2 = sin^2A/cos^2A = (1 - cos^2A)/cos^2A = [1/cos^2A] - 1 1 + (1/cot^2A) = 1/cos^2A sec^2A = 1 + (1/cot^2A) The sign of secA will depend upon the quadrant A is in
Saturday, May 7, 2011 at 5:49am by drwls
Math/Algabra
B = 2A C = B - 3 = 2A-3 A + 2A + 2A - 3 =102 Solve for A, then B and C.
Sunday, February 13, 2011 at 5:45pm by PsyDAG
c) --- Calculus
inherently the same as Mgraph's, but presented in slightly different way dy/dx = f'(x) = 3x^2 - 2x - 4 at (a,b) dy/dx = 3a^2 - 2a-4 , which is the slope of the tangent at (a,b) but the slope of the tangent passing through (a,b) and (0,-8) is also = (b+8)(a-0) = (b+8)/a...
Monday, January 16, 2012 at 12:52am by Reiny
Calculus
answer to C question at the end: If 1st coordinate (x-value) is C, the 2nd coordinate (y-value) is C^2+7. Now, for the other question: At (x,y), the slope of the tangent line is dy/dx = 2x So, now we have a point (a,a^2+7) and a slope (2a). The equation of the tangent line ...
Sunday, April 8, 2012 at 6:08pm by Steve
Math 12 (Logs)
3^(2a+3) = 7^(a-4). (2a+3)*Log3 = (a-4)*Log7. Divide both sides by Log3: 2a+3 = 1.771(a-4). 2a+3 = 1.771a-7.085. 2a-1.77a = -7.085-3. 0.23a = -10.085. a = -43.85.
Sunday, April 15, 2012 at 7:34pm by Henry
Calculus - Partial Fractions
I've set a problem up, something like this. s^4: A+D=0 s^3: -2A+B-3D+E=0 s^2: 2A-B+C+3D-3E=0 s^1: -2A+B-D+3E=4 s^0: A+C-E=4
Monday, March 24, 2008 at 10:02pm by David
calculus
mA is the integral divided by the length of the interval, namely 2-0=2. I=∫sin(Ax)^3dx Substitute u=cos(Ax), du =-Asin(Ax)dx so I=-(1/A)∫(1-u²)du =-(1/A)[u-u³/3] from 0 to 2 =(cos^3(2A)-cos^3(0)-3cos(2A)+3cos(0))/3A =(cos^3(2A)-3cos(2A)+2)/3A ...
Wednesday, February 23, 2011 at 3:00pm by MathMate
Math Trig
Find the Cartesian form of the parametric equation. x = (2a)(cot T) y = (2a)(sin^2 T) how? lol here's what i got y = (sin^2 T) y = (2a)y^2 y = 2a then? do the same for X? i'm stuck there
Monday, May 31, 2010 at 5:44pm by kyu
Calculus
Thanks Reiny =) But I to get everything clear, the 2-a and the a-2 cancel right? After I cancel those I am left with (2a+1) as the numerator:S Sorry I am just confused. Also did you factor out the 2a+4 and 2a-4 to get 4(a+2)(a-2)? Could you please explain step by step?
Friday, December 23, 2011 at 5:52pm by -Untamed-
Calculus
an (2a-4)=2(a-2) or -2(2-a) which will cancel out your 2-a, then do the same for the 2a+4, and this should give you your answer
Friday, December 23, 2011 at 5:52pm by Lindsay
Calculus
Find constants a,b and c such that the graph of f(x)=x^3+ax^2+bx+c will increase to the point (-3,18), decrease to the point (1,-14) and then continue increasing. ============================== The derivative is zero at x = -3 and x = 1 f' = 3 x^2 + 2ax + b 0 = 3(9) +2a(-3...
Wednesday, March 10, 2010 at 7:20pm by Damon
Math
hints: 2a-3=-(3-2a) 9-4a^2=(3-2a)(3+2a)
Wednesday, February 29, 2012 at 9:24am by MathMate
calculus
the area of the parallelogram is the magnitude of the cross product so [a,1,-1] x [1,1,2 ] = [ 3 , -2a-1 , a-1] then | [ 3 , -2a-1 , a-1]| = √35 √(9 + 4a^2 + 4a + 1 + a^2 - 2a + 1) = √35 5a^2 +2a + 11 = 35 5a^2 + 2a - 24 = 0 (a - 2)(5a + 12) = 0 a...
Monday, May 24, 2010 at 1:05pm by Reiny
Algebra
I think that's safe. The symbols used don't really matter. It's easier to type Roman letters, so I'll use those ax^2 + bx + c = 0 a(x^2 + b/a x) = -c a(x^2 + b/a x + (b/2a)^2) = (b/2a)^2 - c a(x + (b/2a))^2 = (b^2 - 4ac)/4a (x + b/2a)^2 = (b^2 - 4ac)/4a^2 x + b...
Monday, February 27, 2012 at 1:57am by Steve
calculus
by definition, the sum of the two focal length is 2a check, distance from centre to the x vertex is a + a = 2a
Thursday, April 8, 2010 at 7:35pm by Reiny
Calculus
a)f(x)=(x+2)(x-1)(x-2) c)From the graph a>=2, b>=0. The equation of the tangent y=b+(3a^2-2a-4)(x-a) and if x=0 y=b-(3a^3-2a^2-4a) or y=a^3-a^2-4a+4-(3a^3-2a^2-4a) y=-2a^3+a^2+4=(2-a)(6+3a+2a^2)-8 If a>2 then y<-8 (a,b)=(2,0)
Monday, January 16, 2012 at 12:52am by Mgraph
4th grade math
XXXX axx(2a) a(b-a)b(2a) and 2a=b-1 so a cannot be 1 as that would make the hundreds digit the same as the ones if a is 2, then 2a=4, and b=5, and the thousands digit is also4. if a is 3, then 2a=6, and b=7, and thousands is 4 (7-3) number is 3476 lets see if there are others...
Tuesday, August 30, 2011 at 9:50pm by bobpursley
Mechanics (A level), Please help
Call the length of string below point B as h, then diagonal of triangle = a - h. angle to vertical of string = theta T + Tcostheta = W use pythagoras to find h (a - h)^2 = b^2 + h^2 so h = (a^2 - b^2)/2a and costheta = h/(a-h) = (a^2 - b^2) /(2a(a - [(a^2 - b^2)/2a] ) which ...
Sunday, April 1, 2012 at 3:37am by Bob
Calculus - Functions?
I will do #1 for you. Show some work or steps that you have done for #2 and#4 and I will evaluate your work. #1: f '(x) = 12x^2 + 2ax + b f ''(x) = 24x + 2a f '(-1) = 0 = 12 - 2a +b ---> 2a - b = 12 f ''(-2) = 0 = -48 + 2a ----> a= 24 then...
Monday, February 21, 2011 at 6:14pm by Reiny
Calculus questions..really need help, please!
4) Let f(x) = ax^2+bx, where a and b are constants. If the tangent line to the curve y=f(x) at the point (1,1) has equation y = 3x-2, then what's f(3)? What I did was first find f(1) and I got a+b Then I took the derivative of f'(x) and got 2a+b..so I then took the ...
Tuesday, November 4, 2008 at 7:55pm by Damon
Pre-Calculus
How can the following two identities be verified? 1. (cos^2y)/(1-siny)=1+siny 2. sin^2a-sin^4a=cos^2a-cos^4a Thanks! change cos^2a to 1-sin^2a=(1-sina)(1+sina) Thank you. I figured out the first one with your help, but I'm still stuck on the second. Any tips?
Sunday, May 20, 2007 at 9:05pm by Corin
Math
multiply everything out, then add like terms (2a+3b)(2a-3b) = 2a^2 +6ba -6ab - 9b^2 = 2a^2 - 9b^2
Tuesday, October 30, 2012 at 8:33pm by Jennifer
Math
y = 2 - x/2 y = 1 + ax 0 = 1 - x/2 - ax x(a + 1/2) = x(2a+1)/2 = 1 x = 2/(2a+1) y = 1 + 2a/(2a+1) You started out OK, but should have simplified the equation for x and used it to derive an equation for y.
Thursday, October 8, 2009 at 7:05pm by drwls
trig
determine wheather each of the following is an identity or not prove it 1. cos^2a+sec^2a=2+sina 2. cot^2a+cosa=sin^2a
Sunday, March 20, 2011 at 11:46am by Anonymous
Math
2(a-5)-5=3 2a - 10 - 5 = 3 2a = 3 + 15 2a = 18 a = 9
Thursday, October 7, 2010 at 9:23pm by Ms. Sue
Math help please
1. 4(2-a)^2 - 81 , difference of squares = (2(2-a) + 9)(2(2-a) - 9) =(4 - 2a +9)(4 -2a - 9) = (-2a + 13)(-2a - 5) or (2a-13)(2a+5) , I wanted to start the brackets with a positive so I multiplies by (-1)(-1) Do #4 the same way. 3. 3x^2 - 27(2-x)^2 = 3[x^2 - 9(2-x)} , again a ...
Monday, October 3, 2011 at 7:38am by Reiny
Math
the proof is simply to complete the square for a generalised quadratic equation. Like this: ax2(xsquared) + bx + c = 0 Take 'a' outside: a[x2 + bx/a + c/a] = 0 Divide through by 'a': x2 + bx/a + c/a = 0 Complete the square: (x + b/2a)2 - b2/4a2 + c/a = 0 ...
Friday, July 13, 2012 at 7:10pm by Mike
algebra 2
I am not certain what you have... IF it is this, 3/(2a+3) + 2a/(2a+3) then (3+2a)/(2a+3) = 1
Tuesday, September 16, 2008 at 9:00pm by bobpursley
Pre-Calc
It looks like you are finding the derivative by First Principles. f(a+h) = -1/(a+h)^2 f(a) = -1/a^2 then (f(a+h)-f(a))/h = [-1/(a+h)^2 - (-1/a^2)]/h = [-a^2 + (a+h)^2]/[(a^2(a+h)^2](1/h) = [-a^2 + a^2 + 2ah + h^2]/[(a^2(a+h)^2](1/h) = h(2a+h)/[(a^2(a+h)^2](1/h) = (2a+h)/[(a^2(...
Sunday, May 16, 2010 at 9:46pm by Reiny
Calculus 151
Let the point of contact of the tangent be P(a,b) then b = 4 - a^2 we know dy/dx = -2x, which is the slope of the tangent so at P the slope is -2a equation of the tangent is y = -2ax + k but (a,b) lies on it b = -2a(a) + k k = b + 2a^2 = 4-a^2 + 2a^2 k = 4 + a^2 tangent ...
Tuesday, November 16, 2010 at 10:02pm by Reiny
Calculus - Incomplete
f' = 3x^2 + 2ax + b f'' = 6x + 2a f''(0) = -2, so 6*0 + 2a = -2 a = -1 f = x^3 - x^2 + bx + c Hey! (0,-2) is a) not a closed interval b) not written as [low,hi] Watch this space for further correct info.
Monday, February 27, 2012 at 2:46pm by Steve
algebra
a^3-2a^2/2a^2-4a a^2(a-2)/2a(a-2) Reduces to a/2
Thursday, May 19, 2011 at 3:59pm by Math Nerd
math
b = a + 3 c = 2a -3 c^2 = a^2 + b^2 (Pythagorean theorem) Solve those three equations simultaneously. c^2 = 4a^2 -12a +9 c^2 = a^2 + (a+3)^2 = 2a^2 +6a +9 2a^2 -18a =0 a = 9 b = 12 c = 15
Saturday, March 5, 2011 at 3:30pm by drwls
algebra
Before you post more of these, just remember that the axis of symmetry is x = -b/2a This comes right from the quadratic formula: x = [-b +/- sqrt(b^2-4ac)]/2a = -b/2a +/- sqrt(b^2-4ac)/2a The roots are symmetrically located around the line x = -b/2a
Monday, October 24, 2011 at 10:37pm by Steve
Math
This probably should have been written as (2a^2+13a-7)/[(3a^3-27a)(4a^2-1)/9a^2] Try factoring 2a^2 +13a -7 into (2a -1)(a+7) and 3a^3-27a into 3a(a^2-9 = 3a(a+3)(a-3) and 4a^2 -1 into (2a +1) (2a -1) The 3a and (2a-1) terms will cancel.
Thursday, August 7, 2008 at 7:15pm by drwls
algebra
2a+11=a+5 _____ 3 help 2a+11/3=a+5 is that the question? if it is 2a+11/3=a+5/1, then you cross multiply 3(a+5)=2a+11, distribute the 3, which equals 3a+5=2a+11 so now solve for a , bring them to one side, a=-4
Friday, May 18, 2007 at 1:37am by debbie
Math
a bit easier to read .... divide by a x^2 + (b/a)x = -c/a complete the square , add b^2/(2a^2) to both sides x^2 + (b/a)x + b^2/(4a^2) = b^2/(4a^2) - c/a write the left side as a square and add the two terms on the right (x + b/(2a) )^2 = (b^2 - 4ac)/(4a^2) take √ of...
Friday, July 13, 2012 at 7:10pm by Reiny
pre-calculus
let f(x)=x^2 -(a^2 + 2a)x + 2a^3, where 0<a<2. for which value of a will the distance between the x-ints. of f be a maximum?
Monday, January 28, 2013 at 3:49am by cassie
math
I will assume you mean √(2a) + 5√(8a^3) = √(2a) + 5√( (2a)^3) = √(2a) + 5(2a)√(2a) = 11√(2a) or 11√2 √a
Saturday, November 12, 2011 at 11:41pm by Reiny
math
4C = 2A + 2B ----> C = (2A+2B)/4 2B = A+3C ------> C = (2B-A)/3 (2A + 2B)/4 = (2B-A)/3 6A + 6B = 8B - 4A 10A = 2B B = 5A into C = (2A+2B)/4 C = (2A + 10A)/4 C = 3A etc
Saturday, November 24, 2012 at 9:54am by Reiny
Algebra II
a ( x^2 + (b/a) x + (c/a)) a ( x^2 + 2(b/(2a)) x + (c/a)) a [ x^2 + 2(b/(2a)) x + (b/(2a))^2 + - ( (b/(2a))^2 - (c/a) )] a [( x + (b/(2a)))^2 - - ( (b/(2a))^2 - (c/a) )] a [( x + (b/(2a)))^2 - - ( (b^2 - 4ac/((2a)^2)] a [( x + (b/(2a)))^2 - - ( (b^2 - 4ac/((2a)^2)] a [( x + (b...
Tuesday, May 18, 2010 at 6:21pm by Anonymous
math
From 2a-4>3a, subtract 2a from each side of the inequality to get 2a-2a-4 > 3a-2a 0-4 > a 4>a exchange sides of inequality to get a<4.
Sunday, November 22, 2009 at 7:16pm by MathMate
Maths
Expand left side into sines and cosines: sqrt(sec^2A+csc^2A) =sqrt(1/cos^2A+1/sin^2A) =sqrt((sin^2A+cos^2A)/(cos^2A sin^2A)) =sqrt(1/(cos^2A sin^2A)) =1/cosA sinA Similarly expand right hand side: tanA+cotA =sinA/cosA + cosA/sinA =(sin^2A + cos^2A)/(cosA sinA) =1/(cosA sinA)
Tuesday, April 26, 2011 at 2:15pm by MathMate
Physics PLEASE
You muliplied both sides by 2a.. 2a(x-xo)=2a(V-vo)(V+vo)/2a 2a(x-x0)=(v-vo)(v+vo)
Friday, June 19, 2009 at 10:38pm by bobpursley
math
If the sides have length a and b, then, assuming b is the side parallel to the barn, a+a+b = 120, so b = 120-2a Area = a*b = a*(120-2a) = 120a - 2a² = 2a(60-a) Now, this is a parabola opening downward. The vertex (a maximum) is halfway between the roots, 0 and 60, at a=30...
Wednesday, September 28, 2011 at 3:42pm by Steve
verifying trigonometric identities
Learn your identites well in order to prove these. 1. (1+sina) (1-sina) = 1-sina+sina-sin^2a = 1-sin^2a = cos^2a (according to identity) 2. cos^2b-sin^2b you know that sin^2b = 1-cos^2b, so: cos^2b-(1-cos^2b) =cos^2b-1+cos^2b =2cos^2b-1 3. sin^2a - sin^4a= = sin^2a- (sin^2a)(...
Monday, March 3, 2008 at 4:49pm by ^quen^
trigonometry
are you sure it wasn't cscA expressed in terms of cotA ? If so, then form cos^2 A + cos^2A = 1 divide by sin^2A cot^2A + 1 = csc^2A cscA = √(cot^2A + 1) The way you have it, would make a very messy conversion.
Tuesday, May 17, 2011 at 5:33am by Reiny
Math
sum of 16 terms is -504 ---> 8(2a + 15d) = -504 2a + 15d = -63 sum of 9 terms is -126 --> (9/2)(2a + 8d) = -126 2a + 8d = -28 subtract: 7d = -35 d=-5 in 2a+8d=-28 2a - 40 = -28 2a = 12 a=6 sum(30) = 15(12 + 29(-5)) = -1995
Thursday, October 27, 2011 at 10:14am by Reiny
geometry
By symmetry, S must lie on the vertical line x=a. Let the coordinates of S be (a,y). Since the length of one side of the triangle is (2a-0)=2a, we can calculate the distance of (a,y) to (0,0) and equate to 2a accordingly. Using Pythagoras theorem, (2a)=sqrt(a²+y&...
Sunday, July 3, 2011 at 8:39pm by MathMate
Math
Which equation is not equivalent to A = 1/2bh 2A=bh h=2A/b 2b=A/h b=2A/h I think it is 2b=A/h. Is this correct?
Saturday, January 7, 2012 at 10:39am by Jane
math
2a/3 = 2 + 4a. 2a/3 - 4a = 2 Multiply both sides by 3, 2a - 12a = 6 -10a = 6 a = -0.6 Check: 2*(-0.6)/3 = 2 + 4*(-0.6) -0.4 = 2 - 2.4 -0.4 = -0.4
Monday, August 2, 2010 at 4:48pm by Henry
calc
"lim h->0 [(1+2a+2h)(1+a+h)]-[(1+2a)(1+a)]/h =4a+2 " Shouldn't it be {[(1+2a+2h)(1+a)-(1+2a)(1+a+h)]/[(1+a+h)(1+a)]}/h ={[h]/[(1+a+h)(1+a)]}/h =1/(1+a+h)(1+a) I'm sure you can do the rest.
Wednesday, October 14, 2009 at 11:05pm by MathMate
math
The standard notation for the length of medians: a=(1/2)*sqrt(2C^2+2B^2-A^2) b=(1/2)*sqrt(2C^2+2A^2-B^2) c=(1/2)*sqrt(2A^2+2B^2-C^2) square both sides: a^2=(1/4)*(2C^2+2B^2-A^2) b^2=(1/4)*(2C^2+2A^2-B^2) c^2=(1/4)*(2A^2+2B^2-C^2) multiply 4 on both sides: 4a^2=(2C^2+2B^2-A^2) ...
Sunday, March 23, 2008 at 11:53am by Qun
Math
Let a = the first number, b = the second number and c = the third number. Then, b = (a + 1) and c = (a + 2) Therefore, a + (a + 1) = 3(a + 2) + 3 yielding 2a + 1 = 3a + 6 + 3. .........2a + 1 = 3a + 9 .........2a + 1 - 9 = 3a .........2a - 8 = 3a ............-8 = 3a - 2a...
Monday, February 22, 2010 at 1:54pm by tchrwill
Intermediate Algebra
a=3 or a=1/2 2a^2+3=7a 2a^2-7a+3=0 2a^2-6a-a+3=0 2a(a-3)-(a-3)=0 (2a-1)(a-3)=0
Monday, April 2, 2012 at 6:31pm by Rezu
calcus/ math
Ok, dimension are AxB, perimeter is 2A+B area= AB but B= Perimeter-2A area= A(perimeter-2A) dArea/dA= perimeter-2A+ A(-2)=0 A= perimeter/4 solve for B
Tuesday, December 28, 2010 at 1:56pm by bobpursley
Math
What if I rewrote your first problem this way 8a^3+60a^2+150a+125 = (2a)^3 + 3(2a)^2 (5) + 3(2a)(5^2) + 5^3 ? can you see the pattern? do the same with the second problem, Post your answers please
Thursday, December 29, 2011 at 5:18pm by Reiny
Math
a-lenght of two sides of triangle c- lenght of third side of triangle c=2a-4 Perimeter: P=a+a+c= 2a+(2a-4)= 2a+2a-4= = 4a-4 =4(a-1) P=16 4(a-1)=16 Divided with 4 a-1= 16/4 a-1= 4 a=4+1= 5 a=5 meters c=2a-4 = 2*5-4 = 10-4= 6 c=6 meters Perimeter=a+a+c=5+5+6=10+6= 16 meters
Sunday, November 7, 2010 at 4:59pm by Bosnian
calculus
f(a+h)= (a+h)^3= (a^2+ah+h^2)(a+h)=a^3+ha^2+h^2a+ha^2+ah^2+h^3 f(a)= a^3 so f(a+h)-f(a)= a^3+ha^2+h^2a+ha^2+ah^2+h^3-a^3 = ha^2+h^2a+ha^2+ah^2+h^3 dividing that by h gives = a^2 + ah+ a^2 +ah^2 + h^2 = 2a^2 as the h approaches zero
Thursday, February 24, 2011 at 11:38am by bobpursley
gr 10 math
Solve using the quadratic formula. a) x^2 - 7x + 12 = 0 a = 1 b = -7 x = -b +/- the sr of b^2 - 4ac /2a 7 +/- the sr of -7^2 - 4(1)(12) /2a 7 +/- the sr of -49 - 48 /2a 7 +/- the sr of -97 /2a 7 +/- 97 /2a -90/2=-45 -104/2= 54 The roots are -45 and 54. did i solve this ...
Tuesday, May 31, 2011 at 11:37am by Rebecca
Trigonometry
after some of those others you posted, this one is pretty mechanical cos 120° = -1/2 sin 120° = √3/2 (-1/2)^2 + (cosa * (-1/2) - sina*(√3/2))^2 + (cosa * (-1/2) + sina*(√3/2))^2 1/4 + 1/4 (cos^2a + 2√3 sina*cosa + 3sin^2a) + 1/4...
Thursday, January 12, 2012 at 5:11am by Steve
Calculus
Let A(a,12-a^2) be the point of the triangle which lies on the parabola. Since it must be an isosceles triangle, c = 2a The area of the triangel = 1/2(c)(12-a^2) =1/2(2a)(12-a^2) =12a - a^3 Then d(Area)/da = 12 - 3a^2 = 0 for a max area 3a^2 = 12 a = +- 2, c = 2a so c = 4
Thursday, January 24, 2008 at 9:54pm by Reiny
integral Calculus
It's just a polynomial. Expand things out to get ʃ x^2(a^4 - 2a^2x^4 + x^8)^2 dx = ʃ a^4x^2 - 2a^2x^6 + x^10 dx = a^4x^3/3 - 2a^2x^7/7 - x^11/11 + C
Wednesday, December 7, 2011 at 11:56am by Steve
maths
ok cool... a) S/n=n/2 [2a+(n-1)d] 24=3/2[2a+2d] 48=3[2a+2d] 16=2a+2d 8=a+d B)i dont know from here hey... my name is marco
Sunday, February 21, 2010 at 8:10am by Anonymous
Algebra 1- Please Help !!!!!!!!:D
a+b=45 2a=5b+6, 2a-5b=6 Multiply the top equation by -2. -2a-2b=-90 2a-5b=6 Add the equations together. -7b=-84 b=12 a=33
Saturday, October 16, 2010 at 9:26pm by Jen
Trig
use the identity cos 2A = 2 cos^2 A - 1 on cos4A cos 4A = 2cos^2 2A - 1 so 2cos^2 2A - 1 + cos 2A = 0 (2cos 2A -1)(cos 2A + 1) = 0 cos 2A = 1/2 or cos 2A = -1 2A = 60° or 300° or 2A = 180° A = 30° or 150° or A = 90° I guess I should have kept x instead ...
Monday, November 14, 2011 at 11:15am by Reiny
math
look at it as an arithmetic series where a = ? d = 1 and n = 13 S(13) = (n/2)(2a + (n-1)d) 26 = (13/2)(2a + 12) 4 = 2a + 12 2a = -8 a = -4 term(13) = a + 2d = -4 + 12 = 8
Thursday, September 29, 2011 at 6:28pm by Reiny
maths
from y^2 = 4ax 2y dy/dx = 4a dy/dx = 2a/y ---- >slope of tangent to parabola from xy = c^2 xdy/dx + y = 0 dy/dx = -y/x -----> slope of tangent of hyperbola but we are told that they intersect at right angles, so the tangents must be perpendicular, making 2a/y = x...
Wednesday, March 23, 2011 at 7:04am by Reiny
Algebra
2 (3a - 4) - 2 (4a + 3) simplify 6a-8-8a-6, now combine like terms: -2a-14 2 (a + 4) = 4 (a - 3)solve the equation 2a+8=4a-12, 8=2a-12, 20=2a, a=10
Wednesday, March 31, 2010 at 7:33pm by Jen
Maths
factor out an a a(x^2 + b/a) + c = 0 compete the square and don't change the value a(x^2 + b/a + b^2/4a^2) - a(b^2/4a^2) + c = 0 a(x + b/2a)^2 - b^2/4a + c = 0 (x + b/2a)^2 - b^2/4a^2 + c/a = 0 (x+b/2a)^2 = (b^2 - 4ac)/4a^2 x + b/2a = ±√(b^2-4ac) / 2a x...
Monday, April 1, 2013 at 11:10pm by Anonymous
Calculus (Math 2A)
Whoops, pay attention to what bobpursley did.
Sunday, February 19, 2012 at 6:14pm by Damon
Calculus (Math 2A)
diameter = 19, radius = 9.5 I mean
Sunday, February 19, 2012 at 6:14pm by Damon
Math
noo...it's asking after u expand the entire thing, which term is a constant ....for example (2a-3)^2= (2a-3)(2a-3) 4a^2-12a+9 Then the third term would be "constant" cause it doesn't have any variables attached to it...
Saturday, March 5, 2011 at 7:50pm by Jowin
algebra
Let a be the larger number. a - b = 3 a*b = 2(a + b) a*(a-3) = 2a + 2(a-3) a^2 -3a = 2a + 2a -6 a^2 -7a +6 = 0 (a - 6)(a -1) = 0 a = 6 or 1 Forget the a = 1 solution, because b cannot be negative a = 6 and b = 3
Wednesday, April 27, 2011 at 6:37am by drwls
algebra
want to know if this right Area of a triangle with b base and height h A=1/2bh,for b 2A=b*h 2A/h=b b=2A/h
Wednesday, August 31, 2011 at 10:17pm by babygirl
math subject
2a + sqrt(2a^2) = 10, Isolate the radical: 2a - 10 = -sqrt(2a^2), Square both sides: 4a^2 - 40a + 100 = 2a^2, 4a^2 - 2a^2 - 40a + 100 = 0, 2a^2 - 40a + 100 = 0, Divide both sides by 2: a^2 - 20a + 50 = 0, Solve for a using the quadratic formula: a = (20 +- sqrt(400 - 200) / 2...
Tuesday, September 28, 2010 at 1:02pm by Henry
trig
a) LS = [sin^2a + 2sinacosa + cos^2a - 1] / [ cosa/sina - sinacosa] = [1 + 2sinacosa -1] / [ (cosa - sin^2acosa)/sina] = 2sinacosa/[ cosa(1 - sin^2a)/sina] = 2sinacosa(sina/(cosa(cos^2a)) = 2 sin^2a/cos^2a = 2tan^2a = RS b) The way you typed it, LS ≠ RS Did you mean...
Wednesday, November 30, 2011 at 8:09am by Reiny
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