Thursday

December 18, 2014

December 18, 2014

Total # Posts: 27,588

**Calculus**

f(x) = √(x^3+4x) f'(x) = (3x^2+4) / 2√(x^3+4x) f'(2) = 16/8 = 2 So, now you have a point and a slope. The line is y-4 = 2(x-2) y = 2x See the graphs at http://www.wolframalpha.com/input/?i=plot+y%3D%E2%88%9A%28x^3%2B4x%29%2C+y%3D2x+for+x%3D-1..3
*November 14, 2014*

**math**

32/100 That can, of course, be reduced as desired.
*November 14, 2014*

**sixth grade math**

do you mean 24% ? If so, then 360 = .24x x = 360/.24 = 1500
*November 14, 2014*

**sixth grade math**

I'm having trouble with the English: What is the "of" doing there?
*November 13, 2014*

**Calculus Please Help**

F(x,y) = e^(2x+y) Fx = 2e^(2x+y) Fy = e^(2x+y) Fx(1,-1) = 2e^(2-1) = 2e Fy(1,-1) = e^(2-1) = e F(x, y, z) = xy + xz -yz Fx = y+z Fy = x-z Fz = x-y Fx(0,-1,1) = 0 Fy(0,-1,1) = -1 Fz(0,-1,1) = 1 F(x,y,z) = 2x^0.2 y^0.8 + z^2 Fx = 0.4x^-0.8 y^0.8 Fy = 1.6x^0.2 y^-0.2 Fz = 2z Fx(0...
*November 13, 2014*

**Math**

multiply top and bottom by -9-9i, Note that (-9+9i)(-9-9i) = 9^2+9^2 = 162 So, you have (-10-6i)(-9-9i)/162 Now just do the multiplication and put into standard form.
*November 13, 2014*

**Trigonometry**

well, just plug in the numbers: cot x = (3.5)(9.8)/5.4^2
*November 13, 2014*

**Algebra**

(a) f and g are both degree 4, so their sum cannot be of higher degree. (b) 1/2 g(x) = x^4-3x^2+x - 1/2 f - 1/2 g = -x + 5/2 Looks like 1 is the smallest possible degree.
*November 13, 2014*

**Math**

You might take all 6 red balls, all 5 green balls first. So, you might have to take out 6+5+3=14 balls to be sure you also have 3 yellows. Extra credit: how many do you need to be sure you have 3 balls of some color? Hint: There are 3 colors.
*November 13, 2014*

**Math**

Thanks alot Reiny that was super helpful. I have a math test tomrrow!
*November 13, 2014*

**Math**

I'm sorry but this does not help. WHat is D?
*November 13, 2014*

**Math**

Can anyone show me how to solve this? .1g+3.75m=7.36 and 8.2g+21m=73 Thank you
*November 13, 2014*

**Math help please!!!!!!**

correct again
*November 13, 2014*

**Math (PLEASE!)**

But you are right. D is also true.
*November 13, 2014*

**Math (PLEASE!)**

well, 11 is greater than 10.7 11 > 10.7 11.1 is even greater 11.1 > 11 > 10.7 Can you use the number line? If, plot all those values. The ones on the right are greater than those to their left.
*November 13, 2014*

**Math (PLEASE!)**

I like C
*November 13, 2014*

**Math help please!!!!!!**

correct
*November 13, 2014*

**Algebra**

I get -15 6x^3*4 + 4x^2*3x + -7x*3x^2 + -5*6x^3 24+12-21-30 = -15
*November 13, 2014*

**Algebra**

h(x) = (3x^2-5x+3)-(9x^3-3x+1) so, just subtract tghe coefficients of like powers.
*November 13, 2014*

**math take 2**

fcator out 4tan^2θ to get 4tan^2θ(2sinθ+1) = 0 so, tanθ = 0 or sinθ = -1/2 θ=0 or θ = 7π/6 or θ = 11π/6 You didn't specify the domain, so I'll let you worry about that. If you want all solutions, then that would be &#...
*November 13, 2014*

**math**

fcator out 4tan^2θ to get 4tan^2θ(2sinθ+1) = 0 so, tanθ = 0 or sinθ = -1/2 θ=0 or θ = -π/6 You didn't specify the domain, so I'll let you worry about that. If you want all solutions, then that would be θ = πn θ...
*November 13, 2014*

**science**

the same
*November 13, 2014*

**Math *checking answers*?**

Nope - my bad. B,C,D are all the same You want 9.5/3 = x/7 but B,C,D are all 9.5/3 = 7/x
*November 13, 2014*

**Math *checking answers*?**

I like C for #2. It's the same as 9.5/3 = x/7
*November 13, 2014*

**calc**

I think you want g(x) = 3x√(x+5) on [-5,0] g(-5) = 0 g(0) = 0 So, we want c somewhere in (-5,0) such that f'(c) = 0 f'(x) = 3(3x+10) / 2√(x+5) f' = 0 when 3x+10 = 0 So, c = -10/3 which is in the domain.
*November 13, 2014*

**calc**

In that case, g(0) = 0 g(5) = 15√10 Again Rolle's Theorem does not apply. Try again.
*November 13, 2014*

**calc - eh?**

g(0) is not defined, since sqrt(-5) is not real. So, g(x) does not satisfy the condition that g(0) = 0 You want to reconsider any part of the question?
*November 13, 2014*

**MATH HELP PLEASE**

(13/4)(2x + 3/4) (13/4)(2x) + 13/4 * 3/4 (13/2)x + 39/16
*November 13, 2014*

**Math: Chi Squared**

Boy, you don't give much context, do you?
*November 13, 2014*

**Intermediate Algebra**

clearly the domain is not (-oo,100) negative percent makes no sense. I'd say the domain is [0,100) C(X) -> oo as x->100, meaning that the harder you try to squeeze out that last bit of pollution, the morse it costs. So, just evaluate C(90). Then double that, since we...
*November 13, 2014*

**Algebra**

-3x^5 + 24x^2 factor out -3x^2 and you have (-3x^2)(x^3) + (-3x^2)(-8) -3x^2 (x^3-8) Now recall the factorization of the difference of two cubes, and you have -3x^2(x-2)(x^2+2x+4)
*November 13, 2014*

**Science**

F = ma so, a = F/m the x-component is F cos60° and sine for y
*November 13, 2014*

**Molecular Calculations**

convert g of Sn to moles. You will need that many moles of O2. Then find the mass for that many moles of O2.
*November 13, 2014*

**Algebra 2**

7x^2 = 21x 7x^2-21x = 0 7x(x-3) = 0 This is why we always set stuff to zero. If the product of factors is zero, one of the factors must be zero. So, we have 7=0: never x=0: x=0 x-3 = 0: x=3 Those are the solutions: x = 0 or 3
*November 13, 2014*

**math**

0.8 = 8/9 So, you have 10.1 + 8/90 = 101/10 + 8/90 = 917/90
*November 13, 2014*

**Calculus**

yes
*November 12, 2014*

**math**

at any depth y, the radius of the surface of the water is y/2. So, v = pi/3 (y/2)^2 y = pi/12 y^3 dv/dt = pi/4 y^2 dy/dt Now plug in your numbers.
*November 12, 2014*

**Calculus - PLEASE HELP!!**

Looks good to me
*November 12, 2014*

**math**

(6.0-0.3)/4
*November 12, 2014*

**Inverses of functions**

f^-1(a-1) = -3 a-1 = f(-3) a-1 = 2 a = 3 Not sure just what you're asking
*November 12, 2014*

**math**

b/g = 8/5 b = g+135 now just solve for g
*November 12, 2014*

**Pre-Calculus**

Look what happens when x gets huge. The equation is basically 5y^2 = 20x^2 y^2 = 4x^2 so the slopes are 2 and -2
*November 12, 2014*

**Math**

x-25.6 = 84.3
*November 12, 2014*

**Math**

m-4=11 m+u=53 now just find u
*November 12, 2014*

**Math**

(100-50)/(3-1.5)
*November 12, 2014*

**Science**

you are correct
*November 12, 2014*

**math**

s=m m = 4(s-18) Each had 24
*November 12, 2014*

**Pre-Calc/Trig**

range is all positive >2, as well as all negative On the interval (-1,1) the numerator is positive and the denominator is negative.
*November 12, 2014*

**Math HELP!**

x-10 = 47 x-25.6 = 84.3
*November 12, 2014*

**Algebra HELP!**

If the population is P, then 14/50 * P = 30
*November 12, 2014*

**Math - eh?**

eh? try that in English
*November 12, 2014*

**Arithmetic Progression**

a + a+d = 4 a+9d = 19 a=1 d=2 So, now add T5 and T6
*November 12, 2014*

**Algebra**

the two roots have product c and sum 7 Since the coefficient of x is 1, all the rational roots are integers. (x-1)(x-6) = x^2-7x+6 (x-2)(x-5) = x^2-7x+10 ... I think you can see that there are only a few values for c.
*November 12, 2014*

**Calculus**

∫v(v^3+5)^2 dv = ∫v(v^6 + 10v^3 + 25) dv = ∫v^7 + 10v^4 + 25v Now just integrate each term.
*November 12, 2014*

**caculus**

f"(x) = -4sin(2x) f'(x) = 2cos(2x) + c1 f(x) = sin(2x) + c1*x + c2 Now plug in the initial conditions to find c1 and c2
*November 12, 2014*

**calculus**

a(t) = 36t+10 v(t) = 18t^2+10t+c v(0)=9, so c=9 and v(t) = 18t^2+10t+9 s(t) = 6t^3+5t^2+9t+c s(0) = 16, so c=16 and s(t) = 6t^3+5t^2+9t+16 So, find s(11)
*November 12, 2014*

**Calculus**

Try this site: http://mathworld.wolfram.com/RiemannSum.html
*November 12, 2014*

**alg 2**

think back to algebra I. The QF is x = [-b +/- sqrt(b^2-4a)]/2a So, just plug in your number. The first is x = [-4 +/- sqrt(16-4(1)(-2)]/2 = [-4 +/- sqrt(16+8)]/2 = -4 +/- sqrt(24)]/2 = -2 +/- sqrt(6) Do the same with the others. First step is to get everything on the left ...
*November 12, 2014*

**Algebra 2 HELP plz**

t=2 is the correct time. So, plug in t=2 for your formula for h to get the max height. to see when it hits the ground, just set h=0: -16t^2+64t+80 = 0 -16(t+1)(t-5) = 0 so, h=0 at t=5
*November 12, 2014*

**Pre-Calc/Trig...**

2(x+5)/(x-3)
*November 12, 2014*

**College Algebra**

This is just a parabola, with its vertex at h = -b/2a, or in this case h = 3.817/0.14
*November 12, 2014*

**Math**

w(3w-5) = 12 ...
*November 12, 2014*

**College Algebra**

y = a(x+2)(x-8) The vertex is midway between the roots, at x=3, so y(3) = 5 a(5)(-5) = 5 a = -1/5 y = -1/5 (x+2)(x-8) See http://www.wolframalpha.com/input/?i=-1%2F5+%28x%2B2%29%28x-8%29
*November 12, 2014*

**College Algebra**

usually college algebra students haven't learned about derivatives. However, s(t) is just a parabola, with its vertex at t = 112/32 so, plug that into s(t) to find the max height Note that you got your derivative wrong, anyway.
*November 12, 2014*

**algebra**

if x at 2%, then the rest (10000-x) at 5%. So, 10000-x = 3x so solve for x, and the interest earned is .02x + .10(3x) = ?
*November 12, 2014*

**algebra**

y = ab^x a*b^2 = 3 a*b^9 = 33 a = 3/b^2, so a*b^9 = 3/b^2 * b^9 = 3b^7 = 33 b = 11^(1/7) a = 3/11^(2/7) y = 3/11^(2/7) * 11^(x/7) = 3*11^((x-2)/7) y(15) = 3*11^(13/7) = k
*November 12, 2014*

**math olympiad**

there are many such rectangles. 1 x √88 2 x √85 ... √44 x √45
*November 12, 2014*

**intermediate algebra**

the line through (2,3) and (2,0) is x = 2 That is a vertical line. So, the perpendicular line through (2,3) is y = 3
*November 12, 2014*

**Maths**

h/10 = 1.8/1.2
*November 12, 2014*

**St Ans School**

75-x = x-45 ...
*November 12, 2014*

**Math**

How about writing that as 3x - 8y - 4 = 0
*November 12, 2014*

**Math Help**

ln√x-8=5 ln√x=13 √x = e^13 x = e^26 log2x+log2(x+2)=log2(x+6) log2(x(x+2)) = log2(x+6) x(x+2) = x+6 x^2+2x = x+6 x^2+x-6 = 0 (x-2)(x+3) = 0 x = 2 or -3 But, log2(x) is not defined for x <= 0, so x=2 is the only solution check: log2(2)+log2(2+2) = log2(2+6...
*November 12, 2014*

**Math**

if course not "all reals" plug in just some value and see whether the equation balances. In particular, try x=0: log(4) =?= log(2) I don't think so. What you need to realize is that if 2^x = 2^y or log2(x) = log2(y) or 2x = 2y or 2+x = 2+y or in general, for any ...
*November 12, 2014*

**Maths**

1.3/1.8 = h/(7+1.8)
*November 12, 2014*

**Math**

you want e^.12x = 3
*November 12, 2014*

**maths**

a:b = 3:5 = 12:20 b:c = 4:7 = 20:35 a:b:c = 12:20:35
*November 12, 2014*

**math algebra**

Hard to tell, but it appears that you have (x-3)/4 - (y-2)/5 = -1/4 (4-x)/3 + (3-y)/2 = -13/20 which is the same as 5x - 4y = 2 20x + 30y = 209 20x-16y = 8 20x+30y = 209 46y = 201 so, x = 448/115 y = 201/46 Hmmm. Maybe you had x - 3/4 - y - 2/5 = -1/4 4 - x/3 + 3 - y/2 = -13/...
*November 12, 2014*

**Pre-Calculus**

recall that cos 2x = cos^2 x - sin^2 x
*November 12, 2014*

**Pre-Calculus**

7/12 = 1/3 + 1/4 now review your addition formula for tan(x+y)
*November 12, 2014*

**Pre-Calculus**

recall that cos(x-y) = cosxcosy+sinxsiny hence the name cosine -- sine of complementary angle
*November 12, 2014*

**maths**

we need (1+x)(0.9) = 1.08 x = 0.20 = 20% check: on $1.00 cost, mark up to $1.20 discount = $0.12, so the final price is $1.08, or 8% profit
*November 12, 2014*

**Pre-Calculus**

cos(cos^2+sin^2) don't forget your Algebra I now that you're taking trig!
*November 12, 2014*

**Math**

surely the explanation in your text will be more thorough than any offered here. Also, google is your friend...
*November 12, 2014*

**Geometry**

s = rθ, so θ = s/r
*November 11, 2014*

**math**

12 and 15
*November 11, 2014*

**math**

yes, you did
*November 11, 2014*

**Calculus**

just use the power rule: d/dx(x^n) = n x^(n-1) so, f'(x) = 3x^2 + 5/x^2
*November 11, 2014*

**Geometry**

No, it's November 11. I think you need to provide a little more context. p -> q where p=It's snowing today and q=It is December
*November 11, 2014*

**math**

Assuming that's $2.90 each, then 2*2.9 = 5.80 So, now, which has cost more for two packs of batteries?
*November 11, 2014*

**calculus related rates**

PV = k V dP/dt + P dV/dt = 0 dP/dt = -(P/V) dV/dt Now just plug in your numbers. By the way, this is true only if the temperature remains constant. More generally, PV = kT
*November 11, 2014*

**calculus inverted cone**

when the contents have depth y, the radius of the surface of the liquid is y/2 So, v = pi/3 r^2 y = pi/3 (y/2)^2 y = pi/12 y^3 dv/dt = pi/4 r^2 dy/dt Now just plug in your numbers.
*November 11, 2014*

**Math**

e+s = 80 e-17 = 2(2/3 s) Now just solve for e and s, and then get e-s.
*November 11, 2014*

**Math**

looks ok. But that depends on what x represents...
*November 11, 2014*

**Math**

9^2+35
*November 11, 2014*

**Math**

5 * 50/(5+8+7) = 12.5 m
*November 11, 2014*

**College Algebra**

there are lots of good graphing sites. Try here: http://www.wolframalpha.com/input/?i=plot+y%3D1%2F2x-3 The two equations describe the same line.
*November 11, 2014*

**Advanced Algebra**

3x=4(4-2x)=11 I suspect a typo
*November 11, 2014*

**Pre-Calc/Trig**

extrema where 12x^2 - 24x = 0 12x(x-2) = 0 Now apply the 2nd derivative test, and what you know about the general shape of cubics.
*November 11, 2014*

**math**

I suspect that V = XYZ So, just multiply the three values
*November 11, 2014*

**Calculus**

Go to http://www.wolframalpha.com/input/?i=8ln%28sec%28x%29%29+ and scroll down a bit.
*November 11, 2014*

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