Saturday

April 19, 2014

April 19, 2014

Number of results: 21,667

**correction - math**

1st one: if f(x) = x/(x+2) , then y = x/(x+2) step1: interchange the x and y variables to get the inverse as x = y/(y+2) step 2: solve this new equation for y x = y/(y+2) xy + 2x = y xy - y = -2x y(x-1) = -2x y = -2x/(x-1) or y = 2x/(1-x) f^-1 (x) = 2x/(1-x) now do the same ...
*Thursday, October 17, 2013 at 7:56pm by Reiny*

**Calculus **

the graph of |x|+|y| ≤ 1 is a square with vertices (1,0), (0,1), (-1,0) and (0,-1) y ≥ 2x^2 is the region above y = 2x^2 so we need their intersection: The line of the square in the first quadrant is y = -x + 1 then 2x^2 = -x + 1 2x^2 + x - 1 = 0 (2x-1)(x+1) = 0 x...
*Friday, August 12, 2011 at 4:56am by Reiny*

**math**

work the inner brackets first -[2x-(5x+2)]=2+(2x+7) -2[2x - 5x -2] = 2 + 2x + 7 -2[-3x - 2] = 9 + 2x 6x + 4 = 9 + 2x 6x - 2x = 9 - 4 4x = 5 x = 5/4
*Sunday, October 13, 2013 at 10:52pm by Reiny*

**Calculus**

my first line after using the product rule would be (2x-1)^4(-1)(4x+1)^-2(4) + (4x+1)^-1(5)(2x-1(^4(2) = -4(2x-1)^5(4x+1)^-2 + 10(4x+1)^-1(2x-1)^4 = (2x-1)^4(4x+1)^-2[-4(2x-1) + 10(4x+1)] = (2x-1)^4(4x+1)^-2(32x + 14) = 2(2x-1)^4(4x+1)^-2(16x+7) perhaps your teacher will not ...
*Tuesday, October 13, 2009 at 9:47pm by Reiny*

**Pre-Cal**

1)(1-cos^2x)(csc x) ? Is this equal to sec x? 2) Expand: log7 3 sqrt x / x^2y 3) Find and simplify: f(g) F(x) = x^2 + 2x g(x) = 2x x^2 + 2x(2x) = x^2 + 4x Is this correct?
*Monday, March 15, 2010 at 4:09pm by Hannah*

**Algebra 2**

is the equaton suppose to look like this (x/2x) + 1 + 1/4 = (2/2x) + 1 ? if it is then what you do is simplify ffirst (x/2x) = 1/2 cause x cancel (1/2) + 1 +(1/4) = (2/2x) + 1 and (2/2x) =(1/x) (1/2) + 1 + (1/4) = (1/x) + 1 1 + (3/4) = (1/x) + 1 subtract 1 from both side (3/4...
*Monday, April 30, 2012 at 8:19pm by visoth*

**integration by parts**

s- integral s ln (2x+1)dx ? = ln(2x+1)x - s x d( ln (2x+1)) = ln(2x+1)x- s x [(2x+1)'/ (2x+1)] dx = ln(2x+1)x- s x [(2)/ (2x+1)] ?... then i'm confused... "ln(2x+1)x- s x [(2)/ (2x+1)] ?... then i'm confused..." x [(2)/ (2x+1)] = 2x/(2x+1) = (2x+1-1)/(2x+1) = 1-1/(2x+1) B.t.w...
*Saturday, February 17, 2007 at 1:05pm by nicholas*

**Maths - Integration by Parts**

I'll do #1, you can show us where you get stuck on the others. Let I = ∫e^2x cos4x dx u=cos 4x du = -4sin4x dx dv = e^2x dx v = 1/2 e^2x I = uv - ∫v du = 1/2 e^2x cos4x - ∫-2e^2x sin4x dx Now do it again, letting u = sin4x I = 1/2 e^2x cos4x + (e^2x sin4x...
*Wednesday, November 7, 2012 at 10:15am by Steve*

**Pre-Cal(Please help)**

1)(1-cos^2x)(csc x) ? Is this equal to sec x? 2) Expand: log7 3 sqrt x / x^2y 3) Find and simplify: f(g) F(x) = x^2 + 2x g(x) = 2x x^2 + 2x(2x) = x^2 + 4x Is this correct?
*Monday, March 15, 2010 at 4:53pm by Hannah*

**Math**

1) Wrong. Cosines are negative in the second quadrant. 2) Wrong. (1-csc^2x)/cot^2x = [(sin^2x-1)/sin^2x]*sin^2x/cos^2x = -cos^2/cos^2 x = -1 3) Wrong. -5(cot^2x - csc^2 x) = -5(cos^2x -1)/sin^2x = -5(-sin^2x/sin^2x)= +5 4) Correct. Sec^2 is always 1 or greater. Adding tan^2 ...
*Thursday, February 14, 2008 at 1:00pm by drwls*

**algebra2**

[-2x + 16 ] - 9 = - 3. [-2x + 16] = 9 - 3, [-2x + 16] = 6, The number whose absolute value = 6 can be (+-): +-(-2x + 16) = 6, +(-2x + 16 = 6, -2x + 16 = 6, -2x = 6 - 16, -2x = - 10, x = - 10 / -2 = 5. -(-2x + 16) = 6, 2x - 16 = 6, 2x = 6 + 16, 2x = 22, x = 22 / 2 = 11. ...
*Sunday, November 21, 2010 at 3:06pm by Henry*

**Pre Cal**

the question is supposed to be: cos^2xtan^2xcos^2x=1 =cos^2x+ (sin^2x/cos^2x)(cos^2x/1) *the last 2 cos^2x will cancel out leaving you with: =cos^2x+sin^2x=1
*Sunday, March 25, 2012 at 7:56pm by olivia*

**11th grade**

factor out 5 u get 5(4x^2+8x +3) = 0 5(4x^2+2x+6x+3)=0 5(2x(2x+1)+3(2x+1))=0 5(2x+1)(2x+3)=0 x=-1/2 or x=-3/2
*Monday, December 13, 2010 at 8:22pm by chemistry*

**Algebra**

Here's (a): {[4x^2-49]/[8x^3+27]} *{ [4x^2+12x+9]/[2x^2-13x+21]} 4x^2-49 = (2x-7)(2x+7) 8x^3+27 = (2x+3)(4x^2-6x+9) 4x^2+12x+9 = (2x+3)(2x+3) 2x^2-13x+21 = (2x-7)(x-3) Putting it all together, you can cancel the (2x-7)(2x+3) to get (2x+7)(2x+3) / (x-3)(4x^2-6x+9) Do (b) ...
*Tuesday, February 19, 2013 at 5:56am by Steve*

**Algebra 1**

this is not an equation, it probably says to "simplify" it. (2x²+5x+3)-(x²+2x+1) how about pretending to see it as : 1(2x²+5x+3)-1(x²+2x+1) , then = 2x²+5x+3-x²-2x-1 = x²+3x+2
*Tuesday, September 8, 2009 at 8:18pm by Reiny*

**Math - Trig - Double Angles**

RS: = (cos2(2x)) + (4cos(2x)) + 3 = 2cos^2(2x) - 1 + 8cos^2(2x) - 1 + 3 = 2cos^2(2x) + 8cos^2(2x) - 2 + 3 = 2cos^2(2x) + 8cos^2(2x) + 1 I don't think I did it right, but I can't think of any other way of doing it...
*Saturday, November 17, 2007 at 6:17pm by Anonymous*

**Math (trigonometric identities)**

19). 4sin^2x/cos^2x - cos^2x/cos^2x 4(1-cos^2x) - cos^2x. All over cos^2x 4 - 4cos^2x - cos^2x. All over cos^2x 4 - 5cos^2x. All over cos^2x 4/cos^2x - 5
*Tuesday, December 4, 2012 at 10:46pm by Mac*

**turnford**

Identities (I'll use "x" for theta): sinx/cosx + cosx/sinx = 2cosec2x cos^2x - sin^2x = cos2x sin^2x + cos^2x = 1 Working with the left side: (sinx/cosx + cosx/sinx)(cos^2x - sin^2x) = (sin^2x/sinxcosx + cos^2x/sinxcosx)(cos^2x - sin^2x) = [(sin^2x + cos^2x)/sinxcosx](cos^2x...
*Wednesday, March 13, 2013 at 5:19pm by MathGuru*

**Math(Please help)**

a. (f + g)(x) = f(x) + g(x) =(2x - 3 + 2x^2. b. (f - g)(x) = f(x) - g(x) = (2x -3) - 2x^2. c. (f * g)(x) = f(x) * g(x) = (2x -3) * 2x^2 = 4x^3 - 6x^2. y^2 - x - 49 = 0. Solve for x: x = y^2 - 49 = 0, This x parabola is the difference of 2 squares: (y + 7) (y - 7) = 0, y + 7 = ...
*Wednesday, October 6, 2010 at 7:55pm by Henry*

**Calculus Help Please!!!**

f(x+h) = (x + h) + 1/(x+h) = [(x+h)^2 + 1] /(x+h) = [x^2 + 2 x h + h^2 + 1 ] / (x+h) f(x) = (x^2+1)/x f(x+h) - f(x) =[x^2+2xh+h^2+1]/(x+h) - (x^2+1)/x = x^3+2x^2h+h^2x+x -(x^2+1)(x+h) -------------------------------- x^2 + hx divide by h x^3+2x^2h+h^2x+x -(x^2+1)(x+h...
*Tuesday, February 25, 2014 at 7:18pm by Damon*

**trig**

sins^2xcscx^2-sin^2x = sin^2x(csc^2x - 1) = sin^2x(cot^2x) = sin^2x(cos^2x/sin^2x) = cos^2x sec^4x - tan^4x = (sec^2x - tan^2x)(sec^2x + tan^2x) = (tan^2x + 1 - tan^2x)(tan^2x + 1 + tan^2x) = 1(2tan^2x +1) = 2tan^2x + 1
*Friday, October 7, 2011 at 9:32pm by Reiny*

**calculus**

f = g(e^2x) f' = 2e^2x g' f'' = 4e^2xg' + 2e^2xg'' example: g(x) = x sinx f(x) = g(e^2x) = e^2x sin(e^2x) g' = sinx + xcosx g'' = cosx + cosx - xsinx = 2cosx - xsinx f' = 2e^2x sin(e^2x) + e^2xcos(e^2x)*2e^2x = 2e^2x(sin(e^2x) + e^2x cos(e^2x)) Letting g(e^2x) = g(u) f' = 2e^...
*Wednesday, December 7, 2011 at 12:08pm by Steve*

**Calculus**

When there is a negavtive exponent you need to flip the term to makee it positive so the problem should be written as: H(x)=(x^4-2x+7)[(1/x^3)+(2/x^4)] Then you simply multiply the two groups to get: H(x)=[(x^4-2x+7)/(x^3)]+[2(x^4-2x+7)/(x^4)] Then simplify the second fraction...
*Thursday, February 28, 2013 at 1:21pm by Trevor*

**oops---Chemistry**

Note that I dropped a - sign here on the change of -2x. The equil value of 0.1-2x is correct as written. b. ..........2H2S ==> 2H2 + S2 initial..0.1........0.....0 change....2x.......2x.....x equil...0.1-2x.....2x.....x
*Monday, April 9, 2012 at 8:58pm by DrBob222*

**College Algebra**

Which of the following is equivalent to (2x-3)^2=25 ? A) 2x - 3 = 5 B) 2x - 3 = -5 C) 2x - 3 = 5 and 2x - 3 = -5
*Tuesday, October 4, 2011 at 8:40pm by Nolan*

**Add the Polynomials**

(9X^3+2X^2-7)+(2X^3-2X+9) IS ANSWER: 11X^3+2X^2-2X+2
*Sunday, June 10, 2012 at 11:57pm by Adriana*

**Math-Derivatives. **

write it as y = (sec (2x))^3 now use the chain rule ... y' = 3(sec (2x)^2 (derivative of sec(2x)) = 3(sec (2x)^2(sec (2x))tan (2x))(2) = 6(sec (2x))^3 tan (2x)
*Wednesday, March 24, 2010 at 10:50pm by Reiny*

**Algebra**

No, you should use FOIL F - multiply the Firsts O - multiply the Outers I - multiply the Inners L - multiply the Lasts so (5 - 2x) (5 - 2x) = (5)(5) + (5)(-2x) + (-2x)(5) + (-2x)(-2x) = 25 - 20x + 4x^2
*Thursday, November 8, 2007 at 3:59pm by Reiny*

**trig**

RS = (3 - 4cos(2x) + cos^2(2x) - sin^2(2x))/8 = (3 - 4cos(2x) + cos^2(2x) - (1 - cos^2(2x)))/8 = (2cos^2(2x) - 4cos(2x) + 2)/8 = (2(cos^2(2x) - 2cos(2x) + 1))/8 = (2(cos(2x) - 1)^2)/8 = (2(1 - 2sin^2(x) - 1)^2)/8 = (2(-2sin^2(x))^2)/8 = (8sin^4(x))/8 = sin^4(x) = LS hope I ...
*Monday, February 23, 2009 at 8:15pm by Reiny*

**trigonometryy**

(sin^2x + cos^2x)(sin^2x - cos^2x) = 7/2(sinxcosx) 1(sin^2x - cos^2x) = 7/2(sinxcosx) cos^2x - sin^2x = -(7/2)(1/2)(2sinxcosx) cos 2x = -(7/4)sin 2x -4/7 = sin2x/cos2x tan2x = -4/7 so 2x is in the 2nd or 4th quadrants 2x = 150.26° or 2x = 330.255 x = 75.13° or x = 165.13° ...
*Saturday, May 8, 2010 at 8:50am by Reiny*

**calculus**

dy/dx = (1/2) (2x+1)^(-1/2) (2) = 1/(2x+1)^(1/2) or 1/√(2x+1) slope of line is -3 so slope of perpendicular is 1/3 so we have ... 1/√(2x+1) = 1/3 √(2x+1) = 3 square both sides ... 2x + 1 = 9 2x = 8 x = 4 f(4) = (9)^1/2) = 3 so the point of contact is (4,3) ...
*Thursday, February 28, 2013 at 9:19pm by Reiny*

**Algebra**

recall that area of triangle is given by: A = (1/2)*bh where A = area, b = base, and h = height. substituting, 2x^2 + 5x + 3 = (1/2)*(4x+6)*b we do factoring: 2x^2 + 5x + 3 = (1/2)*2*(2x+3)*b (2x+3)*(x+1) = (2x+3)*b then we divide both side by 2x+3. therefore, b = x+1 hope ...
*Thursday, October 13, 2011 at 9:28am by Jai*

**college trigonometry**

LS = (sinx/cosx - cosx/sinx)/(sinxcosx) = [(sin^2x - cos^2x)/(sinxcosx)]/sinxcosx = (sin^2x - cos^2x)/(sin^2x cos^2x) =sin^2x/(sin^2xcos^2x) - cos^2x/(sin^2xcos^2x) = 1/cos^2x - 1/sin^2x = sec^2x - csc^2x ≠ RS you have a typo, your right side should have been sec^2x - ...
*Tuesday, February 12, 2013 at 11:28pm by Reiny*

**math**

I asked this question before and I was told my work was correct by my friend insists she is correct HELP I did translate and simplify"six less than twice the sum of a number and three" choices are A)-2x b)2x c)-2x+9 d)2x-3 I set it up 2(x+3)-6 2x+6-6 answer=2x My friend did 2x...
*Tuesday, January 22, 2013 at 4:03pm by ES PUZZLED*

**Algebra**

I dont understand how to do the folliwing problems. Find and simplify the difference quotient f(x+h) -f(x)/h, h cannot equal 0 for the given formula. 1. f(x)=5x^2 options: a. 5, b.5(2x+h), c. 10/h +x+5h, d. 5(2x^2+2xh+h^2)/h 2. f(x)=x^2+3x+9 options: a. 2x+h+9, b.1, c.2x^2+2x+...
*Saturday, September 1, 2007 at 3:31pm by Soly*

**Algebra**

I dont understand how to do the folliwing problems. Find and simplify the difference quotient f(x+h) -f(x)/h, h cannot equal 0 for the given formula. 1. f(x)=5x^2 options: a. 5, b.5(2x+h), c. 10/h +x+5h, d. 5(2x^2+2xh+h^2)/h 2. f(x)=x^2+3x+9 options: a. 2x+h+9, b.1, c.2x^2+2x+...
*Saturday, September 1, 2007 at 3:40pm by Soly*

**Algebra 2**

Did you know that all complex roots come in conjugate pairs? So if -i is a zero, then so is +i and x^2+1 must be a factor of your expression (if x^2 + 1 = 0 , then x = ± i ) by long algebraic division, x^4 + 2x^3 + 2x - 1 = (x^2+1)(x^2 + 2x - 1) so solve x^2 + 2x-1 = 0 in this...
*Sunday, January 1, 2012 at 7:44am by Reiny*

**Calc.**

| x cos(x^2) dx | = integral symbol By half-angle cos^2 x = 1/2 (1 + cos 2x) | x 1/2 (1 + cos 2x) dx 1/2 | x + x cos 2x dx 1/2 | x dx + 1/2 | x cos 2x dx 1/4 x^2 + 1/2 | x cos 2x dx Integrate | x cos 2x dx by Integration by Parts u = x, dv = cos 2x dx du = dx, v = 1/2 sin 2x 1...
*Thursday, January 27, 2011 at 7:37pm by helper*

**Math - help really needed**

You actually get (1/tan^2x)+1, which is cot^2 + (csc^2x-cot^2x), which is csc^2x, which is 1/sin^2x. And that's your answer. (You could only cancel out the tan^2 if it were (Some #)(tan^2x)/tan^2x, rather than (1)+(tan^2x)/tan^2x. Dividing is the inverse of multiplying, so ...
*Tuesday, December 2, 2008 at 9:33pm by Bob*

**alg2 check?**

2x+2(x+3)=2(x+7) 2x + 2x + 6 = 2x + 14 4x + 6 = 2x + 14 2x = 8 x = 4 You're right.
*Thursday, July 18, 2013 at 1:09pm by Ms. Sue*

**Calculus**

The curve looks sort of bell-shaped. Let the point of contact in quadrant I be (x,y) then the base of the rectangle is 2x and its height is y A = 2xy = 2x(e^(-x^2)) d(A)/dx = (2x)(-2x)(e^(-x^2)) + 2(e^(-x^2)) = 0 for a max/min of A skipping some steps e^(-x^2)) ( -2x + 1) = 0 ...
*Monday, January 3, 2011 at 5:59pm by Reiny*

**Math**

Perimeter = S1 + S2 + S3 5x^4-2x^3+x-3= (2x^4 + 2x-1) + (2x^4+2x-1) + base 5x^4-2x^3+x-3 = 2(2x^4 + 2x-1)+ base 5x^4-2x^3+x-3 - 2(2x^4 + 2x-1) = BASE 5x^4-2x^3+x-3 -4x^4-4x+2= BASE 5x^4-4x^4-2x^3-3x-1 = BASE
*Tuesday, March 12, 2013 at 4:28pm by Lena*

**Alg**

2x/2x+1 is the same as (2x/2x) + 1 did you mean 2x/(2x + 1) ? If so, what do you want done with it?
*Saturday, February 14, 2009 at 9:47pm by Reiny*

**calculus**

ambiguous .... is it 4√(2x^2 + 2x + 1) or 4x√(2x^2 + 2x + 1)or 4√(2x^2) + 2x + 1 or ... can you see my point ? You will need brackets.
*Sunday, March 31, 2013 at 9:08pm by Reiny*

**Chemistry**

...............2SO2 + O2 ==> 2SO3 initial........1.7 M..1.7M...0 change..........-2x....-x......2x equil........1.7-2x....1.7-x...2x Substitute the equilibrium values from the ICE chart above into the Kc expression for the reaction and solve for x. 2x is what you want.
*Wednesday, June 15, 2011 at 8:41pm by DrBob222*

**Trig check my work please?**

Carry, where are you getting these, even though I enjoy doing them, they are getting to me, lol recall tan(A+B) = (tanA + tanB)/(1 - tanAtanB) so LS = (tanx + tan30)/1-tanxtan30)*(tan30-tanx)/(1+tanxtan30) = (tan^2 30 - tan^2 x)/(1 - (tan^x)(tan^2 30)) now tan^2 30º = 1/3 so ...
*Monday, February 23, 2009 at 8:45pm by Reiny*

**Math(Please, Please, help)**

b) the same way you did a) , except .. 2x - 3 - (2x^2) = (f-g)(x) = - 2x^2+ 2x - 3 c) in this case you multiply, (fxg)(x) = (2x-3)(2x^2 then fxg(2) = (4-3)(8) = .... (I just replaced the x with 2)
*Thursday, October 7, 2010 at 5:30pm by Reiny*

**math**

Assuming the usual carelessness with parentheses, we have (cot^2x-1)/csc^2x (cot^2x-1)*sin^2x cos^2x - sin^2x cos(2x)
*Wednesday, November 20, 2013 at 5:16pm by Steve*

**Math - Trig - Double Angles**

And I still get stuck... = cos2(2x) + 4(2cos^2(2x)-1) + 3 = cos2(2x) + 8cos^2(2x) - 4 + 3 = 2cos^2(2x) - 1 + 8cos^2(2x) - 4 + 3 = 2cos^2(2x) + 8cos^2(2x) - 2 What am I doing wrong now?
*Saturday, November 17, 2007 at 6:17pm by Anonymous*

**Precalculus**

Verify the identities. Cos^2x - sin^2x = 2cos^2x - 1 When verifying identities, can I work on both side? Ex. 1 - sin^2x - sin^2x = 1 - 2sin^2x 1 - 2sin^2x = 1 - 2sin^2x
*Wednesday, April 3, 2013 at 10:50pm by Shadow*

**math**

4(1/2x+1/2)=2x+2 4/2x+4/2+2x+2 2x+2=2x+2 2x+2x=2+2 4x=4 x=1
*Wednesday, November 14, 2012 at 7:29pm by brittney*

**Maths**

(1+2x)^n = 1 + n(2x) + n(n-1)/2! (2x) + n(n-1)(n-2)/3! (2x)^3 + .... so (1-x)(1+2x)^2 = (1-x)(1 + n(2x) + n(n-1)/2! (2x)^2 + n(n-1)(n-2)/3! (2x)^3 + ....) = 1 + 2nx + 4n(n-1)/2 x^2 + 8n(n-1)(n-2)/6 x^3 + ... - x - 2nx^2 - 4n(n-1)/2 x^3 - .. so the only two terms with a first ...
*Sunday, May 12, 2013 at 4:25am by Reiny*

**Math**

Are the tiles to be used for the border? let the width of the border be x so the whole frame = (10+2x)(10+2x) so area of border = (10+2x)(10+2x) - 100 but (10+2x)(10+2x) - 100 = 300 (10+2x)^2 = 400 10+2x = 20 2x = 10 x = 5
*Wednesday, November 17, 2010 at 7:49pm by Reiny*

**algebra**

if x is second side, first is 2x-5 third is 2x-5+2 = 2x-3 so, 2x-5 + x + 2x-3 = 47 5x = 55 x = 11 So, the 3 sides are 17,11,19
*Thursday, March 28, 2013 at 12:11pm by Steve*

**Math - Trig - Double Angles**

Ah okay, I get it now, so "2x" is like "x", they are always together I'm working on the second question now and I'm expanding the identity, but I'm kind of stuck... 8cos^4x = cos4x + 4cos2x + 3 LS: = 8cos^4x = 8(cos^2x)(cos^2x) = 4(cos^4(2x)) = 4(cos^2(2x))*(cos^2(2x)) = 4(...
*Saturday, November 17, 2007 at 6:17pm by Anonymous*

**math**

Integrate it in two pieces. For x<1.5, |2x -3| = -2x + 3 For x>or=1.5, |2x -3| = 2x + 3 I have no idea why you have the cube of zero in front of the |2x -3| What is the _0^3 supposed to be?
*Thursday, February 12, 2009 at 1:23am by drwls*

**Math**

Write the equation in standard form using integers. Can someone please check my work? Ax+By= C Equation: y- 4 = -2(x - 3) y-4 = -2x + 6 +2x +2x 2x+y-4=6 +4 +4 2x+ y=10
*Monday, January 7, 2013 at 2:58pm by Cassidy*

**Calculus**

Rationalize the denominator, that is, multiply top and bottom by √(2x+2) + 2 to give you lim [(x^2 - 1)/(√(2x+2) - 2) ] * (√(2x+2) + 2)/(√(2x+2) + 2) = (x+1)(x-1)(√(2x+2) + 2)/(2x+2 - 4) = (x+1)(x-1)(√(2x+2) + 2)/( 2(x-1)) = lim (x+1)(√...
*Monday, October 3, 2011 at 11:49pm by Reiny*

**math**

two digit number --- x multiplied it by 2 --- 2x added 18 ----- 2x + 18 divided by 8 --- (2x+18)/8 (2x+18)/8 = 9 2x + 18 = 72 2x = 54 x = 27
*Tuesday, March 1, 2011 at 7:57am by Reiny*

**Math**

do the reverse of PEMDAS 3√1/(2x+1) + 2 = 5 3/√(2x+1) = -3 1/√(2x+1) = -1 1 = -√(2x+1) squaring yields 1 = 2x+1 0 = 2x x = 0 check: 3/√1 + 2 = 5 yes Rather a poor exercise, imho
*Saturday, September 14, 2013 at 6:24pm by Steve*

**alg2 check?**

first off, did you check to see whether x=4 fits the equation? 2(4)+2(4+3) =?= 2(4+7) 8 + 14 =?= 22 yes, so your answer is correct 2x+2(x+3)=2(x+7) 2x+2x+6 = 2x+14 4x+6 = 2x+14 2x = 8 x = 4 Looks like you nailed it. Good job.
*Thursday, July 18, 2013 at 1:09pm by Steve*

**algebra**

x(-2x - 8) = x*(-2x) - x*8 = -2x^2 - 8x You put -2x instead of -2x^2
*Thursday, January 7, 2010 at 4:35pm by Marth*

**calculus**

Using the product rule (1)(e^-2x)+(x)(-2x)(e^-2x) (e^-2x)(1+2x^2)
*Friday, November 5, 2010 at 2:42pm by Goizueta*

**Math**

Which of the following is equivalent to (2x-3)^2=25 ? A) 2x - 3 = 5 B) 2x - 3 = -5 C) 2x - 3 = 5 and 2x - 3 = -5
*Tuesday, October 4, 2011 at 8:55pm by Christian*

**MathLog#4**

a^2-b^2 = (a+b)(a-b) (2x^n)^2 - 1 = (2x^n - 1)(2x^n + 1) so, dividing by (2x^n-1) leaves you with just 2x^n + 1
*Monday, April 2, 2012 at 12:02pm by Steve*

**precal**

if you mean solve (x^2-2x)^2 -11(x^2-2x)+24 = 0, then let u = x^2 - 2x and you have u^2 - 11u + 24 = 0 (u-8)(u-3) = 0 now plug back in for x: (x^2-2x-8)(x^2-2x-3) = 0 (x-4)(x+2)(x-3)(x+1) = 0 That help?
*Tuesday, September 18, 2012 at 3:14pm by Steve*

**math**

(3s-4)(3s+5) is correct. For the other, factor out the 2x first, leaving 2x(2x^2 - 5x - 25) 2x(2x+5)(x-5)
*Monday, March 25, 2013 at 8:58am by Steve*

**Algebra**

let x = width 2x + 4 = length A = lw A = x(2x + 4) 400 = x(2x + 4) 400 = 2x^2 + 4x 200 = x^2 + 2x x^2 + 2x = 200 do you know how to complete the square?
*Tuesday, April 9, 2013 at 4:22pm by rbowh*

**algebra**

in each case find the LCD , then multiply each term by that. I will do the #2 5/(2x-1) = 1 - (8x-16)/(10x-5) 5/(2x-1) = 1 - (8x-16)/(5(2x-1) ) LCD = 5(2x-1) 5/(2x-1)*5(2x-1) = 1*5(2x-1) - (8x-16)/(5(2x-1) )*5(2x-1) 25 = 10x - 5 - (8x-16) 14 = 2x x = 7
*Saturday, February 12, 2011 at 10:38am by Reiny*

**math-calculus**

i think i do something wrong again(2x+1)/(x-3)-(4x-1)/(2x-3) = 0 (2x-3)(2x+1)-(4x-1) = (0)(x-3)(2x-3) 4x^2+2x-6x-3-4x-1 = 2x^2-3x-6x+9 4x^2-4 = 2x^2-3x-6x+9 -2x^2-9x+13 i put this in quadratic formula and i think it wrong becaue i get 9/2 and 0for x and it have to be 2/3.
*Saturday, January 7, 2012 at 10:29pm by Shreya*

**Algebra**

h = x feet. B = 2X - 4. A = Bh/2 = (2X -4)X/2 = 15. Cross multiply: 2X^2 -4X = 30, X^2 -2X -15 = 0, Solve for X BY FACTORING: (X -5) (X + 3) = 0 X - 5 = 0, X = 5; X + 3 = 0, X = -3 Choose positive value: X = 5 = height. 2X -4 = 10 -4 = 6 = Base.
*Sunday, July 18, 2010 at 7:16pm by Henry*

**Maths**

I assume the left reads three times nine to the x power. 3*9^x=3*3^2x=3^(2x+1) 3^(2x+1)=81 take the log base three of both sides. 2x+1=4 2x=3 x=1.5 weird problem
*Sunday, December 1, 2013 at 5:41pm by bobpursley*

**mathematics**

y-2x-1=0 so y=2x+1 x^2+xy-5x-y+2=0 x^2 + x(2x+1) - 5x - (2x+1) + 2 = 0 x^2+2x^2+x-5x-2x-1+2=0 3x^2 - 6x + 1 = 0 and you can take it from there, I assume
*Tuesday, March 26, 2013 at 3:36am by Steve*

**inequality**

(x-4)/(2x+4) is greater than or equal to 1. How do i express solutions to inequalities in interval notation without using the calculator? please help. (x-4)/(2x+4) >=1 multiplying by 2x+4 1) when (2x+4) is +, x-4>=2x+4 or x-2x>=8 -x>=8 x<=-8 notice the ...
*Friday, June 16, 2006 at 10:54am by greg*

**Math CONTINUATION OF QUESTION FROM bobpursley**

I don't think bobpursley is on line, so permit me to continue ... from Bob, y' = -18x^2(1+2x^3)-2 so using the product rule and chain rule combination y'' = -18x^2)(-2)(1+2x^3)^-3(6x^2) + (1+2x^2)^-2(-36x) = 216x^4(1+2x^3)^-3 - 36x(1+2x^3)^-2 = 36x(1+2x^3)^-3 [ 6x^3 - (1+2x^3...
*Thursday, March 10, 2011 at 9:05pm by Reiny*

**math**

H=Hight W=Width L=Length V=Volume V=H*W*L V=(4x+1)*(x-3)*2x V=[4x*x + 1*x + 4x*(-3) +1*(-3)]*2x V=(4x^2 +x -12x -3)*2x V=(4x^2 -11x -3)*2x V=4x^2*2x - 11x*2x -3*2x V=8x^3 -22x^2 -6x
*Monday, May 23, 2011 at 10:02pm by Anonymous*

**math 2**

If you mean [6 + (2x-7)/2]/3 = 31 then just start unwinding the operations: 6 + (2x-7)/2 = 93 (2x-7)/2 = 87 2x-7 = 174 2x = 181 x = 181/2 If I got the expression wrong, fix it and do similar steps.
*Tuesday, June 12, 2012 at 3:33pm by Steve*

**8th grade math**

2(x-3)=-x+3+5x 2x-6=-x+3+5x -2x-6=-x+3-2x -6=-x+3+3x -6=+3+2x -3=-3+2x -9=2x _=_ 2 2 -4.5=x
*Tuesday, December 15, 2009 at 10:13pm by Michele*

**math**

2│2x-7│+11=25 |2x-7| = 7 then 2x-7 = 7 or -2x + 7 = 7 2x = 14 or 2x = 0 x = 7 or x = 0
*Monday, November 22, 2010 at 10:28pm by Reiny*

** a.p. calculus**

y = ln(1+e^2x)^3 = 3ln(ln(1+e^2x) y' = 3(2e^(2x))/(1 + e^(2x)) = 6e^(2x) /(1 + e^(2x))
*Sunday, December 12, 2010 at 4:16pm by Reiny*

**chemistry**

I assume this is Kc and knot Kp. Initial (CO2) = 2.00mol/5.0L = 0.40 ..........2CO2 ==> 2CO + O2 I.........0.40......0.....0 C.........-2x......2x......x E........0.4-2x...2x......x Substitute the E line of the ICE chart into the Kc expression and solve for x, 2x, and 04-2x...
*Monday, December 3, 2012 at 1:44am by DrBob222*

**math**

Assuming you mean: 4|3-2x^6|-3²=0 Then 4|3-2x^6|=3² |3-2x^6|=9/4 There are two cases, a. when 3-2x^6 >0 |3-2x^6|=9/4 => 3-2x^6=9/4 => 2x^6=3-9/4=3/4 => x^6=3/8 => x=±(3/8)^(1/6) (x∈ℝ, i.e. x is real) b. when 3-2x^6<0 |3-2x^6|=9/4...
*Monday, February 7, 2011 at 10:32pm by MathMate*

**College Algebra**

Here is the whole question... I have figured most of it out but I need help with part (b)(f) and (g)... Maybe you also check the work I have already done Given the following function, find: (a) vertex, (b) axis of symmetry, (c) intercepts, (d) domain, (e) range, (f) intervals ...
*Wednesday, January 4, 2012 at 5:43pm by Sunny*

**Calculus Help Please!!!**

just use the quotient rule: y" = (2xy^2 - 2x^2y y')/y^4 = 2x(y-xy')/y^3 = 2x(y-x(x^2/y^2))/y^3 = 2x(y^3-x^3)/y^5 or, use implicit differentiation twice: x^3-y^3 = 3 3x^2 - 3y^2 y' = 0 6x - 6y(y')^2 - 3y^2y" = 0 y^2 y" = 2x - 2y(y')^2 = 2(x -y(x^2/y^2)^2) = 2(xy^4-x^4y)/y^4 = ...
*Tuesday, April 8, 2014 at 1:44am by Steve*

**math**

sin(2x)+cos(2x)=tan(x) cos(2x) = sinx/cosx - 2sinxcosx cos(2x) = (sinx - 2sinxcos^2x)/cosx cos(2x) = sinx(1 - 2cos^2x)/cosx cos(2x) = tanx(-cos(2x)) -1 = tanx x = 135º or x = 315º or in radians x = 3pi/4 or 7pi/4
*Thursday, May 28, 2009 at 10:07pm by Reiny*

**11th grade Algebra II**

original garden is (x)(2x) = 2x^2 new garden = (x+2)(2x+2) = so difference = (x+2)(2x+2) - x(2x)
*Tuesday, March 31, 2009 at 10:00pm by Reiny*

**Math**

Note that (4-x) -3 = 1-x x^2 + 4x + 4 + 1 -2x + x^2 = 5 2x^2 + 2x + 4 = 5 2x^2 + 2x -1 = 0 x = [-2 +/-sqrt12]/4 = [-1 +/-sqrt3]/2 Check my math. I'm getting old
*Tuesday, May 12, 2009 at 1:05pm by drwls*

**calculus**

y = cot^2x - csc^2x = cot^2x - (1+cot^2x) = -1 so, y' = 0 Or, if you want to work it out, use the chain rule: y = cot^2x - csc^2x y' = 2cotx(-csc^2x) - 2cscx(-cscx cotx) = -2cotx csc^2x + 2cotx csc^2x = 0
*Tuesday, October 25, 2011 at 6:48pm by Steve*

**maths**

f'(x)=e^3x(3cos(2x)-2sin(2x))=0 3cos(2x)=2sin(2x) divide by 2cos(2x) tan(2x)=1.5 x=0.4914(rad) f(0.2618)=1.90 f(0.4914)=2.42 <--- f(0.7854)=0
*Wednesday, May 11, 2011 at 6:30pm by Mgraph*

**College Algebra**

1. fog(x) = 2 /[(7/x) - 3] simplifying, 2x/(7 - 3x) Domain: all real numbers except 3, 0 and 7/3 2. 2^(1-9x) = e^(2x) get ln of both sides: ln 2^(1-9x) = ln e^(2x) (1-9x)*(ln 2) = 2x (1-9x)/(2x) = 1/(ln 2) 1/(2x) - 9/2 = 1/(ln 2) 1/(2x) = 1/(ln 2) + 9/2 2x = 1/[1/(ln 2) + 4.5...
*Tuesday, August 14, 2012 at 1:45pm by Jai*

**math**

I'll use x for theta, for my convenience. (cos^4x - sin^4x)/sin^2x cos^2x = (cos^2x - sin^2x)(cos^2x + sin^2x)/sin^2x cos^2x = (cos^2x - sin^2x)/sin^2x * (cos^2x + sin^2x)/cos^2x = (cot^2x - 1)(1+tan^2x) = cot^2x - 1 + cot^2xtan^2x - tan^2x = cot^2x - 1 + 1 - tan^2x = cot^2x...
*Wednesday, March 14, 2012 at 11:30pm by Steve*

**functions**

I will assume you defined you first function as f(x) = x+1 also g(x) = x^2 + 2x + 1 the simple ones are where you do one of the arithmetic operations e.g. (f+g)(x) = (x+1) + (x^2 + 2x + 1) = x^2 + 3x + 2 similarly (g/f)(x) = (x^2+2x+1)/(x+1) etc the more complicated ones would...
*Friday, October 18, 2013 at 5:17pm by Reiny*

**math-logic(shsat)**

15 - 2x < 7 Add 2x to each side 15-2x+2x < 7+2x or 15 < 7+2x Switch sides (switch inequality sign also) 7+2x > 15 subtract 7 7-7+2x > 15-7 2x > 8 x > 4 So it's (c) What's important is when you multiply inequalities by -1 (probably what you did), you have ...
*Friday, September 23, 2011 at 9:06pm by MathMate*

**alegebra**

In this forum we type powers using the ^ to show exponents e.g 4x^2 , not 4x2 anyway 1. 4x^2 + 12x + 5 = (2x+1)(2x+5) 2. 2x^2 + x - 3 = (x-1)(2x+3) 3. ok 4. 16x^2 - 16x - 12 = 4(4x^2 - 4x -3) = 4(2x+1)(2x - 3) which is D) 5. = 3(x^2 + 7x - 8) = 3(x+8)(x-1) which is A) looks ...
*Tuesday, February 4, 2014 at 9:09pm by Reiny*

**Math**

(x/(x+1))+ ((2x-2)/x) = -1/(x+1) ((2x-2)/x) = -1/(x+1) -x/(x+1) ((2x-2)/x) = (-1-x)/(x+1) ((2x-2)/x) = -(x+1)/(x+1) ((2x-2)/x) = -1 2x-2 = -x 2x+x = 2 3x = 2 x = 2/3
*Sunday, August 22, 2010 at 10:09pm by Michael*

**pre-calculus**

LS = sin^2x/cos^2 - sin^2x = (sin^2x - sin^2cos^2x)/cos^2x = sin^2(1 - cos^2x)cos^2 = (sin^2x)(sin^2x)/cos^2x = (tan^2x((sin^2x) = RS
*Wednesday, February 3, 2010 at 6:29pm by Reiny*

**Algebra**

I am having trouble with this problem. I have done it several times and come up with two different answers. Can someone tell me what I am doing wrong? Write the quadratic function in the form a(x-h)^2+k y=x^2-2x-9 (x^2-2x)-9 (x^2+2x+1-1)-9 (x^2+2x+1) answers (x+1)^2 or x^2-2x-9=y
*Wednesday, April 2, 2008 at 4:08pm by Cynthia*

**calculus**

first of all simplify it a bit, multiplying top and bottom by e^x to get y = (e^(2x) -1)/(e^(2x) + 1) dy/dx = [(e^(2x) + 1)(2e^(2x)) - (e^(2x) -1)(2e^(2x)) )/(e^(2x) + 1)^2 = 2e^(2x)(e^(2x) + 1 + e^(2x) - 1)/(e^(2x+1)^2 = 4e^(2x)/(e^(2x)+1)^2 confirmed by Wolfram: http://www....
*Monday, November 5, 2012 at 8:51pm by Reiny*

**algebra**

I would multiply each term by 2x(x-2) the LCD to get 2(x-2)(x-2) + x(x-2) = 2x(x+1) 2x^2 - 8x + 8 + x^2 - 2x = 2x^2 + 2x x^2 - 12x +8 = 0 x = (12 ± √112)2 = (12 ± 4√7)/2 = 6 ± 2√7
*Monday, January 16, 2012 at 4:22pm by Reiny*

**equations trig**

very tricky 8sin^2x+sinxcosx+cos^2x=4 8sin^2x+sinxcosx+cos^2x=4(sin^2x + cos^2x) 8sin^2x+sinxcosx+cos^2x=4sin^2x + 4cos^2x 4sin^2x + sinxcosx - 3cos^2x = 0 (sinx + cosx)(4sinx - 3cosx) = 0 sinx = -cosx or sinx = (3/4)cosx sinx/cosx = -1 or sinx/cosx = 3/4 tanx = -1 or tanx = 3...
*Monday, May 3, 2010 at 11:19am by Reiny*

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