Monday

April 21, 2014

April 21, 2014

Total # Posts: 21,808

**5TH GRADE MATH**

If you mean that the sum of the three perimeters is 140 yards, then you can divide 140 up any way you like. If you pick, say, 30,50,60 yards as your three perimeters, then you can divide the values up as you will. For example perimeter 30: 7x8 yards perimeter 50: 25 ft by 50 f...

**Mathematics**

P is in QI, so if cosP = 5/13, tanP=12/5 Q is in QIII, so if tanQ = 8/15, sinQ = -8/17 17tanPsinQ = 17(12/5)(-8/17) = -96/5

**trigonometry**

h/63 = sin 45

**calculus help**

There's a nice discussion here: http://math.stackexchange.com/questions/126755/lhospitals-rule-with-indeterminate-powers

**Math**

to figure a percentage, just divide and multiply by 100. 2023.60/55953.77 = 0.0361 = 3.61%

**Calculus Help!!!!!**

should be no problem. That's just Algebra II. Anyway, y = x^3/(x+1)^2 y' = x^2(x+3)/(x+1)^3 y" = 6x/(x+1)^4 see the graph and some analysis here: http://www.wolframalpha.com/input/?i=x^3%2F%28x%2B1%29^2

**Precalculus**

tan does not become 2tan expand (a+b-1)^2 and you will see the 2's, just as (a+b)^2 = a^2 + 2ab + b^2

**Precalculus**

You have (tan+sec-1)/(tan-(sec-1)) multiply top and bottom by tan+sec-1 and you have (tan+sec-1)^2/(tan^2-(sec-1)^2) tan^2+sec^2+1+2tan*sec-2tan-2sec)/(tan^2-sec^2+2sec-1) (sec^2+tan*sec-tan-sec)/(sec-1) (tan+sec)(sec-1)/(sec-1) tan+sec

**Calculus Help Please!!!**

As mentioned, let BC = x distance over water: √(x^2+49) distance over land: 11-x (a) energy expended: y = 1.3√(x^2+49) + 11-x dy/dx = 1.3x/√(x^2+49) - 1 y'=0 at x = 70/√69 what do you get for the other parts?

**algebra**

you just have a right triangle with legs 5 and 8. So, the rope, forming the hypotenuse is √(25+64) = √89

**Calc 2**

and the (-2/3)^k part is just a geometric series S = (-2/3)/(1+2/3) = -2/5

**geometry**

x^2 + 6371^2 = (6371+353)^2 x = 2150

**Algebra**

2 hours

**Math**

√(25^2 + 10^2)

**math**

(P2-P1)/P1 * 100

**Calc**

dQ/dt = kQ (if you please)

**CHEM**

since PV/T is constant, if P is held constant, then V/T is constant. if T is increased by a factor of 296/90, then V is increased by a factor of 296/90. So, multiply 3.3 by that factor.

**math (Matrix algebra)**

Enter your data at http://www.gregthatcher.com/Mathematics/GaussJordan.aspx to see all the details

**math (Matrix algebra)**

assuming you mean a determinant, and calling your matrix M, I get |M| = 31-3x So, if 31-3x = 25, x=2 The inverse can be seen at http://www.wolframalpha.com/input/?i=inverse+{{-2%2C+1%2C+x%2B2}%2C{3%2C+x-4%2C+5}%2C{0%2C+1%2C+3}}+ substitute in x=2 if you just want the numbers.

**maths**

since one line is slanting, and the other is horizontal, I'd say they are not parallel. Even is there's an x missing from the 2nd line, it will still not be parallel.

**maths**

x y -4 13/2 -3 5 -2 7/2 ...

**Algebra**

since time = distance/speed, we have 5/x + 5/(2x) = 15/(2x)

**Math**

5% growth is just 1200000*1.05^n after n years for #c, we have 1200000*(1.05)^7*(0.92)^8 = 866,581

**Algebra**

x^2+8x+15 = (x+3)(x+5) so, you have (x+3)(x+5)/(x-4) * (x-4)(x+4)/2(x+3) the x+3 and x-4 cancel, leaving (x+5)(x+4)/2

**math**

out of 36 possible rolls, only 4 have the sum < 5: 12 13 21 22 What's the chance of no 6's? 5/6 * 5/6 so, take 1-P(~6,~6) = 11/36

**Algebra**

the point of Reiny's question is to remind you that division by zero is undefined. So, when b+4=0, the expression cannot be evaluated. That is, when b = -4, the expression is undefined.

**Math**

I get zero. ∫[-1,2] 1-x^2 dx = x - x^3/3 [-1,2] = (2 - 8/3) - (-1 + 1/3) = 0 yeah, I know, we have to divide by 3 ...

**Trig - heh**

now, just change all those 225's to 255's and redo the calculations.

**Math**

PR/QR = sin(Q) PR/120 = sin(10.5)

**Calculus Help Please Urgent!!!**

1/3 pi r^2 h = 30, so h = 90/(pi r^2) surface area = 2 pi r s where r^2+h^2 = s^2, so a = 2 pi r √(r^2 + (90/(pi r^2))^2) = 2/r √(pi^2 r^6 + 8100) for minimum paper, we need da/dr = 0, so, as wolframalpha so ably shows at http://www.wolframalpha.com/input/?i=2+pi+r...

**mathematics for management**

5^x 25^2y = 1 5^x 5^4y = 1 5^(x+4y) = 1 x+4y = 0 3^5x 9^y = 1/9 3^5x 2^2y = 3^-2 5x+2y = -2 x = -4/9 y = 1/9

**math**

Assuming you mean (8^y)^y, we have 8^y^2 = 2^3y^2 1/32^y = 2^-5y 4 = 2^2 3y^2-5y = 2 y = 2 or -1/3

**mathematics for management**

2log(a+b) = log (a+b)^2 = log (a^2+2ab+b^2) 2log a + log(1 + 2b/a + b^2/a^2) = log (a^2 (a^2+2ab+b^2)/a^2) looks the same to me

**math**

Assuming a 360-day year, 8000 * .095 * 1/3 = 253.33

**math**

ignoring all those trailing zeros, 58/6 = 9.67

**calculus**

1/10 ∫[0,10] 3/(x+1) dx = 3/10 ln(11)

**math**

the bisector BD and half the chord CD form a right triangle with the radius BC BC^2 = 3^2+5^2

**math**

the one with an x term, but no higher powers.

**calculus**

So, the height is 40/x^2 c(x) = 2x^2*5 + 4x(40/x^2)*2 = 10x^2 + 320/x set the derivative to zero, and you find minimum cost at x = 2∛2

**Precalculus**

for #2c, there is a second point at about (-1.38,3) see the graph at http://www.wolframalpha.com/input/?i=solve+5x^3-3x^5%2B1+%3D+3

**cal**

false. These are formulas for interest compounded k times per year.

**math**

have you tried your calculator?

**Math**

850000(1+.06/4)^(4*3.5)/4.75 = 220,419 850000(1+.06/4)^(4*5)/5.10 = 224,476

**statistics**

well, who has a negative z score?

**Math**

As I showed you earlier, it is ∫[0,8] 7/(x+1) dx ---------------------- (8-0) = 1/8 (7log(x+1) [0,8]) = 7/8 (log9-log1) = 7/8 log9 what were your solution steps?

**Algebra**

(r+2)/(r+4) - 3/(r+1) = (r^2-10)/(r^2+5r+4)

**math**

5n + 25q = 325 q = n-5 now just plug in q and solve for n.

**Algebra**

1/(g+2) + 3/(g+1) = (4g+7)/(g^2+3g+2)

**Math**

using FOIL, you get (2n)(6n) + (2n)(1) + (2)(6n) + (2)(1) = 12n^2 + 2n + 12n + 2 = 12n^2 + 14n + 2

**math trigonometry**

the law of cosines says 68^2 = 53^2 + 71^2 - 2*53*72 cosθ so, cosθ = (68^2-53^2-71^2)/(-2*53*72) = 0.42269 θ = 65°

**Algebra**

the common denominator is 2b, so you have 1/2b + b^2/2b = (1+b^2)/2b

**Math**

and there's your answer.

**Math**

17 is 17*1 so, what is 17*23.125?

**Mean, Median and Mode HELP!**

sort the data to get 4 6 7 9 10 11 11 12 15 16 Now it is easy to see mode = 11 median = 10.5 mean = 10.1

**Math**

distance is ∫[0,8] 3t√(640t^2) dt = 512

**Math**

that would be ∫[0,3] 300t^2/(t^3+32) + 5 dt let u = t^3 + 32 du = 3t^2 dt and now you have ∫[32,59] 100/u du + ∫[0,3] 5 dt = 15 + 100log(59/32)

**Math**

∫[-1,2] 1-x^2 dx ----------------------- (2-(-1))

**Math**

∫[0,8] 7/(x+1) dx ---------------------- (8-0)

**Math**

that would be (∫[0,8] 5e^-x dx)/(8-0) = 5/8 (1-e^-8)

**Algebra**

since the denominators are equal, just subtract the numerators: ((5x-2)-(x-2))/(4x) = (4x)/(4x) = 1

**trying to understand math**

in March, total sales were 9.00 * 125 * 31 = 34875.00 In April, (assuming 30 days, 15 days per half-month) 1st half: 9.00*125 * 15 = 16875.00 2nd half: 11.00*125 * 15 = 20625.00 total: 37500 avg at 11.50 = 43125 so she falls 5625 short. Two items need clarification... There...

**math**

(pi * .8^2 * .5 m^3) ----------------------- = 32 s (pi * .1^2 * 1 m^3/s)

**Algebra**

just solve your equation: x^2-81 = 115 x^2 = 196 x = 14 so, the base is 14+9 = 23 height is 14-9 = 5

**Algebra**

√(5x^2)+7x+2-√(4x^2+7x+18) = x-4 √(5x^2)-√(4x^2+7x+18) = -6x-2 now square both sides to get (5x-2) - 2√(5x^2)(4x^2+7x+18) + (4x^2+7x+18) = (6x+2)^2 4x^2+12x+16 - 2√(20x^4+35x^3+90x^2) = 36x^2 + 24x + 4 √(20x^4+35x^3+90x^2) = -16x^2-6x+...

**algebra 1**

both correct However, I'd have written #2d as (5x+2 + 3x-1)/2 just to show it's the average of two numbers, rather than half of each.

**algebra1**

looks like A to me

**College Cal 1**

if the 2-strand dimension is x, then we have f(x) = 2x + x + 2(15000/x) df/dx = 3 - 30000/x^2 min fence when df/dx = 0, at x = 100 so, the field is 100x150 as usual, the fencing is divided equally among the dimensions. ** f(x) = fencing(x) !

**alg**

since f(1) = 5, f-1(5) = 1

**Algebra**

probably not any better than the user's manual that came with it.

**Alg.**

well, just plug in the values g(-4+h) = (-4+h)^2 - 3 = 13-8h+h^2 g(-4) = (-4)^2 - 3 = 13 so, the numerator is -8h+h^2 divide that by h and you wind up with -8 + h

**math**

v1 - 3v2 = (1+2j)-3(1+j) = 1+2j-3-3j = -2-j v1*v2 = (1+2j)(1+j) = 1+3j+2j^2 = -1+3j v1*v1 = (1+2j)(1+2j) = -3+4j v1/v2 = (1+2j)/(1+j) * (1-j)/(1-j) = (3+j)/2 you can practice you skills at wolframalpha.com, but use i instead of j.

**trigonometry**

since 10*468 = 13*360, if we can find all the values of n < 10, then just adding 10 to those values will produce all the possibilities. Finding n<10 should not take too long. For example n=1 works, since 468 = 108(mod 360)

**caculus**

from the product rule, d(uv) = v du + u dv so, u dv = d(uv) - v du now integrate

**MATH HELP**

2 kg is about 4.5 pounds So, what do you think?

**Calculus Help Please Urgent!!!**

Hmmm. we all know that tan x -> x as x->0 so, cot x -> 1/x as x->0 so the limit ought to be zero.

**M A T H**

no, cost = 4800 + 1400x

**Algebra**

.04x + .30(200-x) = .08(200) x = 169.2 lbs whole milk so, 30.8 lbs of cream.

**Math**

Hmmm. we have ∫[0,2] 0.18t^2 + 0.16t + 2.64 = 0.06t^3 + 0.08t^2 + 2.64t [0,2] = .48 + .32 + 5.28 = 6.08 What did you do to arrive at your answer? 1.04 is obviously way off, since f(0) = 2.64 just to start with.

**trig/ vectors**

he needs to head upstream at an angle θ (measured from the line CD), such that tan θ = 4/12 so, θ = 18.4° The speed required is √(16+144) = 12.65 mph

**Math**

call the 2 piles x and y P(x) = 3/5 P(y) = 2/(b+2) 3/5 * 2/(b+2) = 6/45 6 / 5(b+2) = 6/45 5(b+2) = 45 b+2 = 9 b = 7

**Calc III**

well, just plug them in ∂x/∂u = 0 ∂x/∂v = 7 ∂x/∂w = 14w ∂y/∂u = 18u ∂y/∂v = 0 ∂y/∂w = 9 ∂z/∂u = 2 ∂z/∂v = 4v ∂z/∂w = 0 So J = (0 7 14w) (18u 0 9) (2 4v 0) |J| = 1008uvw ...

**maths**

the area of a sector is 1/2 r^2 θ So, the volume of the cake slice of height h is 1/2 r^2 θh subtract that from the volume of the cake, which is pi r^2 h v = r^2 h (pi-θ/2)

**college algebra**

If the hole has depth h, then √h/4 + h/1100 = 3.0 h = 132'8"

**math trigonometry**

if the two towers have heights a and b, then (b-a)/31 = tan 25° a/31 = tan 31° Now just solve for a and b.

**math**

sinA = 1/2, so tanA = 1/√3 tan(A-B) = (tanA-tanB)/(1+tanAtanB) = (1/√3 - 3/4)/(1+(1/√3)(3/4)) = (25√3 - 48)/39

**math**

I get AC+B = 3x^^3+14x^2+6x-6 AC has to be 3rd degree, since A is 1 and C is 2 (A+B)(B-C) = -3x^3-5x^2-6x+24 Somewhere you're messing up. Visit calc101.com and click on the "long multiplication" link, and it will show all the details of polynomial multiplication.

**math**

just substitute in. For example, ABC = (3x+4)(x^2+2)(x^2+3x-2) = 3x^5+13x^4+12x^3+18x^2+12x-16 what do you get for the others?

**college algebra**

max height when t = 1 h(1) = 576 -16t^2+32t+560 = 0 when t = 7

**algebra**

recall that lna-lnb = ln a/b, so ln (x+3)/(x-5) = ln 7 (x+3)/(x-5) = 7 x+3 = 7x-35 x = 19/3

**algebra**

643 e^10k = 813 e^10k = 813/643 = 1.2644 10k = ln 1.2644 = 0.2346 k = 0.02346 now just solve for t in 643 e^0.02346t = 2x10^9

**science**

well, 1 mole = 22.4 L, so you have 1.15/22.4 moles. that should help.

**math**

common ratio, so ...

**math**

identify the sequence 1.6, .8, .4, .2 as arithmetic, geometric, or neither and why.

**math**

C(g) = 2.19g

**math**

a gas station charges $2.19 per gallon of gas. use function notation to describe the relationship between the total cost C(g) and the number of gallons purchased g.

**math**

looks like 8 11 14 17

**math**

find the 1st 4 numbers of the sequence 3 n + 5

**math**

625cm/2500cm = 1:4 so, what's 5000/4 ?

**Math**

since you give no information as to their directions, I will assume they are traveling in opposite directions. So, if the .1kg mass is moving in the + direction, we have 0.1*0.4 - 0.2*1.0 = (0.1+0.2)v v = -8/15

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