Sunday

January 25, 2015

January 25, 2015

Total # Posts: 28,321

**Math**

Since y = kx, y/x = k That is, pick any two known values of x and y. Their ratio is the constant k.
*November 9, 2014*

**College algebra**

f(x) has 3 changes, so 3 or 1 + roots f(-x) has no sign changes, so no - roots check: http://www.wolframalpha.com/input/?i=6x^5-6x^4%2B7x^3-8
*November 9, 2014*

**consumer math**

799.99(1-0.40) = 479.99 or, $480
*November 9, 2014*

**calculus**

y = log |2-x| y' = 1/|2-x| (-1) = 1/(x-2)
*November 9, 2014*

**Pre-trig/Calc**

1 positive, 2 or 0 negative check: http://www.wolframalpha.com/input/?i=x^3%2B3x^2-18x-40
*November 9, 2014*

**Math**

start with x. The value after each step is listed below x 2x 2x+6 (2x+6)/2 = x+3 x+3-3 = x you always get your original number.
*November 9, 2014*

**calculus**

s = t^9 ln|t| s' = 9t^8 ln|t| + t^9/t = t^8(9ln|t| + 1)
*November 9, 2014*

**Math**

15.3g + 2q + 3.79 Note that you don't need an equation, just an expression
*November 9, 2014*

**calculus**

y' = (26x^2-26)e^-3x - 3(13x^2-26x+26)e^-3x = -13(3x^2-8x+8)e^-3x
*November 9, 2014*

**Math**

#1 False A triangle has three angles, which add up to 180 degrees. The two right angles use all that up. #2 True #3 True #4 False Just draw any obtuse triangle. If two of the legs happen to be equal, stretch one of them a little. #5 True #6 False. Draw a tall skinny isosceles ...
*November 9, 2014*

**Math -word problem**

Consider the room. The front wall is square ABCD, and the back wall is EFGH. The front outlet (P) is a foot from the floor (AB), and the back outlet (Q) is a foot from the ceiling (GH). A line directly up and across and down is 20 feet. But, you can do the following: Unfold ...
*November 9, 2014*

**calculus**

just use the product rule and chain rule: y = uv y' = u'v + uv' where u=(2x+1)^3 v=(4x+1)^-4 y' = 3(2x+1)^2(2)*(4x+1)^-4 + (2x+1)^3*(-4)(4x+1)^-5(4) = -2(2x+1)^2(4x+5) --------------------- (4x+1)^5
*November 9, 2014*

**math pre algebra**

how about 5x+7y
*November 9, 2014*

**Algebra 2**

plug in your coefficients at http://www.gregthatcher.com/Mathematics/GaussJordan.aspx to see all the details Or, you can try this: http://www.wolframalpha.com/input/?i=solve+{{3%2C4%2C1}%2C{1%2C3%2C-2}%2C{-2%2C1%2C-2}}*{{x}%2C{y}%2C{z}}%3D{{1}%2C{-2}%2C{-5}}
*November 9, 2014*

**Algebra 2**

(2x-1)^2 = x^2 + 33 4x^2-4x+1 = x^2+33 3x^2-4x-32 = 0 (3x+8)(x-4) = 0 ...
*November 9, 2014*

**math**

315 = 3^2 5 7 882 = 2 3^2 7^2 So, GCF = 3^2 7 = 63
*November 9, 2014*

**math count warm up**

just start writing down the facts. If the 4 digits, left to right are a,b,c,d, then: a+b=c b+c=d c+d=10a+b a+b=c d-b=c 10a+b-d=c a+b=d-b a+2b = d 10a+b-(a+2b) = a+b 8a = 2b b = 4a so, c=5a and, d=9a Since a,b,c,d are single digits, a must be 1. So, we get 1459
*November 9, 2014*

**algebra**

w(3w-2) = 16 w = 8/3 p = 2(8/3 + 6) = ?
*November 9, 2014*

**math**

6 - 4(2/3) = ?
*November 9, 2014*

**projectives**

time to hit: 4.9t^2 = 90 t = 4.29 s 4.29s * 50 m/s = 214.28 m
*November 9, 2014*

**calculus**

see related questions below
*November 9, 2014*

**Math**

Drop a perpendicular from G on CD to F. Now you have a rectangle GDEF with sides 3.5 and 4.5 Also, triangle CGF where CG = 3.5/√3 CF = 7.0/√3 Now the perimeter is easy - add up three sides of the rectangle, and the triangle.
*November 8, 2014*

**math**

I don't see any fractions If you meant (11+x)/(13+x) = 10/11 then x=9
*November 8, 2014*

**Calculus**

Do you mean ln√x or √lnx ?
*November 8, 2014*

**trig**

recall your sum of angles formula for tan recall that tan(180) = 0
*November 8, 2014*

**Math Question**

QR = 7.1 tan58° Area = (1/2)(PR)(QR)
*November 8, 2014*

**math**

your math is ok, freddy, but the spelling, not so much...
*November 8, 2014*

**math**

Hmmm. How can someone savvy enough to use a computer, not know what 1+1 is? I suspect either a test posting, or someone really bored.
*November 8, 2014*

**Algebra 2**

You always want to set things equal to zero: 6x^2-13x+6 = 0 Now, just have to have some practice factoring to get to (3x-2)(2x-3) = 0 Now, since if the product of two numbers is zero, one of the numbers must be zero. So, either 3x-2 = 0 and x = 2/3 or 2x-3 = 0 and x = 3/2 You ...
*November 8, 2014*

**Calculus (math)**

Draw a diagram, looking from the side. If the observer is at distance x, then The angle Ø from the eye to the bottom of the picture is such that tanØ = d/x The angle θ subtending the picture is tan(Ø+θ) = (d+h)/x We want to determine x for ...
*November 8, 2014*

**math**

(3x-2)/8 + (2-x)/ 4 = - 1/2 easy way: clear fractions by multiplying by 8: (3x-2) + 2(2-x) = (-1/2)(4) 3x-2+4-2x = -2 x+2 = -2 x = -6 check: (3(-6)-2)/8 + (2-(-6))/4 = -20/8 + 8/4 = -5/2 + 2 = -1/2
*November 8, 2014*

**mathematics**

a + a+2d = 12 (a)(a+d) = 24 a=4, d=2 check: 4,6,8,10,... 4+8=12 4*6=24
*November 8, 2014*

**Mustapha**

ever try google? Here's one discussion: http://www.meritnation.com/ask-answer/question/determine-the-oxidation-number-of-cr-in-k2cr2o7/redox-reactions/1515787
*November 8, 2014*

**algebra**

#1 If you mean f(n) = 3(n-3) then f(5) = 3(5-3) = 6 f(n) = 3^(n-3) then f(5) = 3^2 = 9 f(n) = 3^n - 3 then f(5) = 3^5-3 = 240 How did you come up with 279? #2 f(n) = 2f(n-1)+2 f(1) = 4 f(2) = 2(4)+2 = 10 f(3) = 2(10)+2 = 22 #3 f(n) = (4-n)/(n+3) f(3) = (4-3)/(3+3) = 1/6 #4 I ...
*November 8, 2014*

**Maths**

C = pi * d so, the turns is 44/C where C is in meters
*November 8, 2014*

**math - typo**

3:1:15 requires three quantities.
*November 8, 2014*

**calculus**

d/dx u^n = nu^(n-1) u' d/dx arctan(u) = 1/(1+u^2) u' d/dx ln(u) = 1/u u' So, we have y' = 2(3/2) √(22x+31) * (22) + 1/(1+(22x+21)) * (1/2) * 1/√(22x+21) * 22 + 1/(11x+16) * 11 - 22 = 66/√(22x+31) + 11/((22x+22)√(22x+21)) + 11/(11x+16) - 22
*November 8, 2014*

**calculus**

That would be 4(cos(6x-1) - 6xsin(6x-1)) ------------------------------ 3cos(3x) - 3 at x=0 that is 4(cos(-1)-0) ---------------- 3-3 -> -∞
*November 8, 2014*

**Calculus**

See the related questions below
*November 8, 2014*

**Trig**

Extend AB to meet the vertical from C at D. It is easy to determine that in triangle ABC ∠C = 47° ∠B = 88° Now just use the law of sines to get AB.
*November 8, 2014*

**Intermediate Algebra**

well, let's just assume that there is a t. After all, if there is none, the B is just a constant: 604.47 So, since polynomials are normally written in descending power order, I'd guess B = 18.75t + 585.72 So, you want to solve 18.75t + 585.72 > 1050 18.75t > 464....
*November 8, 2014*

**Intermediate Algebra**

B=18.75+585.72 where is the t?
*November 8, 2014*

**Calculus (math)**

See the related questions below
*November 8, 2014*

**Math**

cos is negative in QII, QIII You reference angle is 0.927 So, x = pi±0.927
*November 8, 2014*

**calculus**

V = IR dV/dt = R dI/dt + I dR/dt Now just plug in your numbers.
*November 8, 2014*

**Calculus**

I guess you should have checked my math. The actual formula for the volume is (π/3)((18+y)(3+y/6)^2-18*3^2) = π(y^3/108 + y^2/2 + 9y) because the water is in the bottom of the cone dv/dt = π/36(y+18)^2 dy/dt 4.9 = π/36*20^2 dy/dt dy/dt = .14 m/min or 14.0 ...
*November 7, 2014*

**Calculus**

When the water has depth y, the radius of the surface is 3 + (y/12)(5-3) = 3 + y/6 Its volume at that point is v = 1/3 (R^2-r^2) h = (1/3)((3+y/6)^2-3^2)y = y^3/108 + y^2/3 Now we are ready to begin. When y=2, v = 8/108 + 4/3 = 38/27 dv/dt = (y^2/36 + 2y/3) dy/dt 4.9 = (4/36...
*November 7, 2014*

**Calculus (math)**

as usual, draw a diagram. The distance z between the boats at time t hrs after 2:00 is z^2 = (15t)^2 + (20-20t)^2 2z dz/dt = 2(15t)(15) + 2(20-20t)(-20) z dz/dt = 1250t - 800 clearly dz/dt=0 when t=8/1250 hours Now just change that to a time of day.
*November 7, 2014*

**AP CALC. AB**

the .75 comes from the fact that the diameter is increasing 1t 1.5in/min. The radius is half that. and yes, he meant to solve for dh/dt, since that was the question asked. Part 1 just uses the fact that the volume of a box is length*width*height. can you say duh?
*November 7, 2014*

**helpppppppp**

7c+3d = 4m 1m = 1c+1d So, 4m = 4c+4d 7c+3d = 4c+4d 3c = d
*November 7, 2014*

**BRAINMATH**

If the family has b boys and g girls, b = g+4 So, b = (g-1)+5 so, each sister has 5 more brothers than sisters
*November 7, 2014*

**Pre-Calc/Trig**

numerator will be 1 or 2 denominator will be 1 or 3 all can be positive or negative
*November 7, 2014*

**Math**

cos pi/3 = 1/2 so, t = pi/3 That's the reference angle. cos is positive in QI and QIV, so the other solution is t = 2pi - pi/3
*November 7, 2014*

**Physics**

the vertical speed is 200 sin30 = 100 m/s So, the height is h(t) = 100t - 4.9t^2 As you recall from algebra I, the vertex of the parabola is at t = -b/2a, which in this case is 100/9.8 So, plug that in to get the max height. The ball is in the air twice that long, and you know...
*November 7, 2014*

**Algebra**

I'll do one, and you try the other. 2x^2+7x = 4 (x^2 + 7/2 x) = 2 (x^2 + 7/2 x + (7/4)^2) = 2 + (7/4)^2 (x + 7/4)^2 = 81/16 x + 7/4 = ±9/4 x = -7/4 ± 9/4 x = -4 or 1/2 Just recall that (x+a)^2 = x^2 + 2ax + a^2 That's why I added (7/4)^2 to both sides. ...
*November 7, 2014*

**science Help 911!!**

Looks good to me.
*November 7, 2014*

**ALakh narayan sing high school Ekma saran**

1/6(2x+3y)+x/3=8 1/2(7y-3x)-y=11 Rearranging things a bit to get rid of the bothersome fractions, we have 4x+3y = 48 -3x+5y = 22 Since y = (48-4x)/3, -3x+5(48-4x)/3 = 22 Or, -9x+240-20x = 66 29x = 174 x = 6 Now you can easily find y.
*November 7, 2014*

**algebra - incomplete**

I think you are missing a t in the formula.
*November 7, 2014*

**multiplying integers**

10^2 = 10*10 = 100 So, -10^2 = -100 Powers are done before multiplication. SO, the expressions is evaluated as -(10^2), NOT (-10)^2, which is (-10)(-10) = +100
*November 7, 2014*

**algebra**

no, it would never be positive. multiplying two negatives always gives a positive. That third negative then yields a negative product.
*November 7, 2014*

**MATH**

#1 -15 - (-5.6) = -9.4 #2 ...
*November 6, 2014*

**Programming (Integer generator)**

Ignoring the syntax error of the missing close brace, all you do is set the text to the latest string displayed. All the other strings are replaced. You need to append a new line for each value of i, not just do a clean reset of the text area each time.
*November 6, 2014*

**algebra help!**

The domain is the set of 1st values from the pairs: (b) In #2 you are correct.
*November 6, 2014*

**Algebra**

The slope of the line is 2, so y-a^2 = 2(x-a) y = 2x+(a^2-2a) 2(-3/2) + a^2-a = 0 a^2-a-2 = 0 2(0) + b^2-b = 3 b^2-b-3 = 0 Now you can find a and b, and pick the ones that work with the line.
*November 6, 2014*

**Algebra**

well, that would be where x^3-x^2+x+1 = x^3+x^2+x-1 2x^2-2 = 0 x = ±1 See http://www.wolframalpha.com/input/?i=solve+x^3-x^2%2Bx%2B1+%3D+%28+x^3%2Bx^2%2Bx-1%29
*November 6, 2014*

**algebra**

you have slope=3, so using the point-slope form, the line can be written as y-3 = 3(x-0) or, y = 3x+3
*November 6, 2014*

**Calculus**

If v is viewed as a piecewise linear function, then the slopes of the various segments are 154/7 = 21 (374-154)/(17-7) = 22 (418-374)/(19-17) = 22 (770-418)/(35-19) = 22 (1320-770)/(60-35) = 22 (1508-1320)/(64-60) = 47 (3905-1508)/(115-64) = 47 Looks like the 2nd stage kicked ...
*November 6, 2014*

**Intermediate Algebra**

If you are talking about powers, 9^(1/2) = 3 x^(1/2) = √x so you are talking about square roots. (-9)^(1/2) is not real, because negative numbers do not have real square roots. That is because if x = √(-9) then x^2 = -9. But all numbers are positive when squared.
*November 6, 2014*

**Algebra help**

ok ok v = 243 - 0.2s
*November 6, 2014*

**Physics**

Consider m3 to be at (0,0) m1 exerts its force at an angle of 30° m2 exerts its force at an angle of -30° As you say, each force is F=GMm/r^2 The resultant force will be at 0°, with a magnitude of F√3, the sum of the x-components of the two forces.
*November 6, 2014*

**Calculus**

y = (3x^2+2√x)/x = 3x + 2/√x y' = 3 - 1/x^3/2 You are correct
*November 6, 2014*

**Trigonometry**

secx = 2.1754 x = 0.4776 SInce secx is positive in QI and QIV, the other angle is 2π-0.4776 arcsin(2x) makes no sense, since x is an angle. That's if we are still referring to the first question. Now, in general, if sinθ = 2x, then cosθ = √(1-sin^2&#...
*November 6, 2014*

**Trigonometry**

c = 2πr, so one revolution of the plane travels 4π meters So, now we know that 5 min * 60s/min * 0.4rev/s * 4πm/rev = (5*60*4π)/(0.4) = 300π meters
*November 6, 2014*

**Math**

There is a good discussion of this problem at http://mathforum.org/library/drmath/view/65003.html This is, of course, a case involving Fermat's Last Theorem.
*November 6, 2014*

**Math- Had Sub teacher, need help**

apparently you stopped reading before finishing things. Read on past where it says So, the revenue will be ...
*November 6, 2014*

**Math- Had Sub teacher, need help**

revenue = rent * apts if there are x $7 increases, then we have rent = 294+7x units rented = 120-x So, the revenue will be (294+7x)(120-x) = -7x^2 + 546x + 35280 Recall that the vertex of a parabola is at x = -b/2a, which in this case is x = 546/14 = 39 So, with 39 rent ...
*November 6, 2014*

**Algebra**

recall that the vertex of a parabola is at x = -b/2a, which in this case is x = 1/.0064 = 156.25 so, just find f(156.25)
*November 6, 2014*

**Math - incomplete**

better explain some more about the diagram.
*November 6, 2014*

**Math help!**

m=r+4 m-10 = 3(r-10) Miriam is 16 and Ricardo is 12
*November 6, 2014*

**Calculus**

in that case, z^2 = (x+3)^2 + (5(x+3)/x)^2 z = (x+3)√(x^2+25) / x dz/dx = (x^3-75) / x^2√(x^2+25) and x = ∛75
*November 6, 2014*

**Calculus**

If the foot of the ladder is x feet from the fence, and it reaches to a height h on the wall, then using similar triangles, x/3 = (x+7)/h h = 3(x+7)/x the ladder's length z is z^2 = (x+7)^2 + h^2 = (x+7)^2 + (3(x+7)/x)^2 z = (x+7)√(x^2+9) / x So, for minimum z, we ...
*November 6, 2014*

**CAL**

#1 correct #2 is only partially complete f'(x) or df/dx y' or dy/dx
*November 6, 2014*

**CAL**

The _____________ of a function is a formula for the slope of the tangent line to that function at any point x. When the function is defined as f(x), the derivative will be written as __________or ________. When the function is written in the form of y =, the derivative is ...
*November 6, 2014*

**Calculus**

ohhh.. I think I know what I did wrong.. I did not even realize that I was doing something completely different
*November 6, 2014*

**Calculus**

wow - remember the chain rule if u = cos(x^4) then we have y = u^4 y' = 4u^3 u' y' = 4 cos^3(x^4) * (-sin(x^4)(4x^3) = -16x^3 sin(x^4) cos^3(x^4) y = sinx/(1+cos^2(x)) y' = [(cosx)(1+cos^2(x)) - (sinx)(2cosx)(-sinx)]/(1+cos^2(x))^2 y = sinx(sinx+cosx) y' = ...
*November 6, 2014*

**Calculus**

Find the derivatives of the following 18. y=〖cos〗^4 x^4 ANSWER: tanx2 19. y= sinx/(1+ 〖cos〗^2 x) ANSWER: 3sinx 20. y=sinx(sinx+cosx) ANSWER: sinx Can you check my answers?
*November 6, 2014*

**Math Help**

since loga + logb = log(ab) we have 4(ln(z(z+5)) = ln (z(z+5))^4 2ln(z-5) = ln (z-5)^2 Now, since loga - logb = log(a/b), we have ln ( (z(z+5))^4 / (z-5)^2 )
*November 6, 2014*

**Calculus**

Looks good to me. Nice work.
*November 6, 2014*

**Calculus**

3G 4H 5F 6B (though I never heard of such a term) 7E 8C 9A 10D Wow - either your order is mixed up, or you have some serious reviewing to do...
*November 6, 2014*

**Calculus URGENT**

f(x) = √x f(0) = 0 f(9) = 3 So, the slope of the secant is 1/3 f'(x) = 1 / 2√x So, we want to find c such that 1 / 2√c = 1/3 √c = 3/2 c = 9/4 So, c is in the interval [0,9] just to check, f(9/4) = 3/2 So, the equation of the tangent line at c is y...
*November 6, 2014*

**Algebra 2**

192x^2y + 72x^3 = 24x^2(8y+3x) 24rxy - 9rx^2 = 3rx(8y-3x) The only common factor I can see is 3x, so we have 3x(64xy+24x^2-8ry+3rx) Not sure just where you're trying to go with this Now, if the original had been without the parentheses, (192x^2y+72x^3)-24rxy-9rx^2 then we ...
*November 6, 2014*

**matg**

20 years is 4 half-lives, so there would be 300 * (1/2)^4
*November 6, 2014*

**Science**

PE = mgh = (45)(9.81)(2) J
*November 6, 2014*

**Math**

I get z(2.75,0) = 22
*November 6, 2014*

**physics**

recall your equation of motion. In this case, the height, as a function of t is h(t) = 2.54t - 4.9t^2 This is just a parabola, with the vertex at t = -b/2a = 2.54/9.8 = 0.26 So, just plug in t=0.26 to find the maximum height, h(0.26)
*November 6, 2014*

**algebra**

It's easy to construct such a polynomial with zeroes that are very difficult to find: 3x^5 - 12x^4 + 9x^3 + x^2 - 12 But, if you start out with the roots, it's easy to build the polynomial: (x-3)(2x+5)(3x-7)(x^2+4) = 6x^5 - 17x^4 - 14x^3 + 37x^2 - 152x + 420
*November 6, 2014*

**geometry**

draw a line mark point A. mark point B at a distance of 8.5 units away from A. Now, draw a circle of radius 6.2, with center at B. Point C can be anywhere on the circle just drawn.
*November 6, 2014*

**Math**

.815 = 815/1000 .021 = 21/1000 So, she added thusly: 5 815/1000 + 6 21/1000 = 11 876/1000 = 11 219/250
*November 6, 2014*

**math**

1.2 - (0.75 + 0.125 + 0.09) = 1.2 - 0.965 = 0.235
*November 6, 2014*

**math**

recall the slope-intercept form of a line: y = mx+b
*November 6, 2014*

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