Wednesday

April 23, 2014

April 23, 2014

Total # Posts: 21,876

**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

**science**

why do people have a hard time moving a large rock out of the road

**science**

what animal would have less momentum then a 52 kg cheetah running at 10 m/s.

**science**

what is the momentum of a 20kg scooter traveling at 5.o m/s.?

**math**

this is bogus. Maybe X,Y,Z are all disjoint. So, Z – (X U Y) = Z maybe Z⊆(XUY) so Z – (X U Y) = Ø Looks like you're on your own for this one.

**math**

well, duh: (2,-3)

**Math**

w(2w+4) = 400 w = √201 - 1

**linear algebra**

3/4 as big as or 25% smaller than

**Math**

assuming you mean a 1 ft cube, then it has 4^3 = 64 smaller cubes. The 8 corners have 3 faces yellow The 12 edges have 2 blocks each with 2 faces yellow, making 24 The 6 faces have a 2x2=4 set of cubes with 1 face yellow, or 24 That leaves the interior 2x2x2=8 cubes all green

**Math 10**

the graph is a parabola, with vertex at t=1 h(1) = 20 see a picture and analysis at http://www.wolframalpha.com/input/?i=-5t^2+%2B+10t+%2B+15

**Math**

x+y = 14 30x+60y = 660 6 @ 30" 8 @ 60"

**Algebra**

w=t+11 2w+t=64 w = 25 t = 14

**physics grade 10 icse**

v is the same forces in ration 1:2

**calculus**

Hmmm. If the population is p, then we have dp/dt = kp dp/p = k dt ln(p) = kt + c p = c e^kt Now we know that p triples in 14 days, so we can forgo using e, and just use 3 as our base: p(t) = Po * 3^(t/14) so, starting with 100, and gaining 15 per day and losing 23 each day, ma...

**calculus**

if the printed area has width and height x and y, then the total area is a = (x+8)(y+12) and xy=384 a = (x+8)(384/x + 12) = 12x + 3072/x + 480 that is minimum when da/dx = 0, so we need 12 - 3072/x^2 = 0 x = 16 so the poster is (16+8) by (24+6) or 24x30

**math**

a pyramid has a height of 5 inches and a surface area of 90 inches squared. A similar pyramid has a height of 10 inches what is it's surface area?

**Math**

n(n-1)/2

**Algebra**

x/(2x-5) - 5/(6x-15) 3x / 3(2x-5) - 5 / 3(2x-5) (3x-5) / 3(2x-5) or 1/2 + 5 / 6(2x-5)

**math**

I have 7 different pairs of jeans and want to take 2 with me on vacation. how many different ways can you pick 2 pairs of jeans to take on vacation. sorry for not proofing before

**math**

pairs of jeans and want to take 2 with you. how many different ways can you pick 2 pairs of jeans.

**Math**

Use partial fractions to see that you have -1/x^2 + 2/x + 3/(x+2) Now it's a piece of cake, right?

**math**

well, how many numbers are there? 10 how many are divisible by 2? 5

**math**

you have a spinner labeled 1 through 10. you spin once. what is p (divisimble by 2)

**Calculus**

y = xe^(-4x) y' = (1-4x)e^(-4x) y" = 8(2x-1)e^(-4x) since e^(-4x) > 0 for all x, y" < 0 when 2x-1 < 0. see the graph at http://www.wolframalpha.com/input/?i=+xe^%28-4x%29+for+0+%3C%3D+x+%3C%3D+1

**maths**

substitute y=7-2x and you have x^2 - x(7-2x) = 6 x^2 - 7x + 2x^2 - 6 = 0 3x^2 - 7x - 6 = 0 (3x+2)(x-3) = 0 x = 3 or -2/3 so, y = 1 or 25/3

**CALCULUS**

algebraically, since both functions are symmetric about the origin, the area is zero. geometrically, using symmetry, you have the area is 2∫[0,pi/9] (tan3x - 2sin3x) dx = 2/3 (2cos 3x - log cos 3x)[0,pi/9] = 2/3 (log2 - 1)

**CALCULUS PLEASE HELP**

seems pretty straightforward. We have ∫[0,10] 2300e^(.024t)-1450e^(.018t) dt = 10107.8 see http://www.wolframalpha.com/input/?i=%E2%88%AB[0%2C10]+2300e^%28.024t%29-1450e^%28.018t%29+dt

**CALCULUS**

If I read it correctly, you have y=4/x, y=16x, and y=x/16 for x>0 You need to break the area into two regions, because the two top curves intersect at x = 1/2 So the area is ∫[0,1/2] 16x - x/16 dx + ∫[1/2,8] 4/x - x/16 dx = (8x^2-x^2/32)[0,1/2] + (4logx-x^2/32)[...

**math**

7/10 * 6/9

**algebra**

(1+.09/12)^(12x) = 500000/25000 x = 33.411 years I figure you can calculate the required day

**algebra**

just solve for t in (1/2)^(t/5715) = 0.76 t/5715 = log(.76)/log(.5) t = 2262 makes sense, since 76% is about half way to 50%.

**math**

take a look here: http://www.wolframalpha.com/input/?i=plot+x%2B2y%3D-2%2C+-2x%2By%3D14+%2C+x%3D0%2C+y%3D0

**birzit**

find t with 5t^2 = 20 then use that t to find the distance d in d = 25t

**maths**

Since the formula says BAC% = ... I'd say just set it up as 0.1 = (5.14)(4*16*.04) / (.73)(12.5*14) - 0.015H H = 0.2 hours

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