Wednesday

May 27, 2015

May 27, 2015

Total # Posts: 31,412

**math**

8 cubes of side 2 have surface 8(6*2^2) one cube of side 4 has surface 6*4^2
*March 10, 2015*

**math**

it appears you want 2^2010 + 2^2012 + 10 --------------------------- 2^2009 + 1 (2^2009*2 + 2^2009*8 + 10)/(2^2009+1) (2^2009*10 + 10)/(2^2009+1) 10(2^2009+1)/(2^2009+1) 10
*March 10, 2015*

**trapezoidal method**

there are many trapezoid rule calculators online, which explain the method and do the math.
*March 10, 2015*

**College Algebra**

if there is x at 10%, the remainder (12000-x) is at 3%. So, add up the interest: .10x + .03(12000-x) = .06(12000)
*March 10, 2015*

**math**

2 min = 8*15sec so, it will double 8 times. 10 * 2^8 = ?
*March 10, 2015*

**Algebra 2**

If you mean the nth term is -4 (1/3)^(n-1) then clearly you have a GP where a = -4 r = 1/3 The terms are -4, -4/3, -4/9, ... The sum is thus a/(1-r) = -4/(2/3) = -6
*March 10, 2015*

**Geometry**

75 at W20°S = <-70.48,-25.65> 125 at N10°W = <-21.71,123.10> Add them up to get the resultant: <-92.19,97.45> Now just convert that back to distance and heading.
*March 10, 2015*

**Elective mathematics**

since there is a common ratio, (2k-4)/(k-3) = (4k-3)/(2k-4) (2k-4)^2 = (4k-3)(k-3) 4k^2 - 16k + 16 = 4k^2 - 15k + 9 k = 7 Now just plug and chug.
*March 10, 2015*

**Analytic geometry**

I don't know what you mean by necessary, but the hyperbola is 9x^2+18x - 4y^2+8y = -41 9(x^2+2x+1) - 4(y^2-2y+1) = -41 + 9*1 - 4*1 9(x+1)^2 - 4(y-1)^2 = -36 or, 4(y-1)^2 - 9(x+1)^2 = 36 (y-1)^2/9 - (x+1)^2/4 = 1 That should tell you what you need to know.
*March 10, 2015*

**Calculus**

exponentials grow faster than any power. So, x^5/e^x -> 0 consider using l'Hospital's Rule. After 5 times, the limit is 120/e^x -> 0
*March 10, 2015*

**math**

Or, if the order is important, 5P3 = 5*4*3 = 60
*March 10, 2015*

**math**

5C3 = 5C2 = 5*4 / 1*2 = 10
*March 10, 2015*

**Please Check My Math Answers. Thanks!**

They all are correct. Good work.
*March 10, 2015*

**Math**

7/8 * pi * 7^2
*March 10, 2015*

**precalculus**

ln2 + lny - ln(2y) ln(2 * y / (2y)) ln(1) 0 If I misread the cryptic 2lN then fix it and follow the rules to get the right answer.
*March 9, 2015*

**precalculus**

decrease of 1.8% is a remaining portion of 98.2% So, the population after t years is 89000 * 0.982^t now just find t when the function = 50000
*March 9, 2015*

**math**

it is the set of products, where each value in each pair is an element of the set described above, with no duplications: {7*8,...9*11}
*March 9, 2015*

**math**

hard to say, without knowing the relationship of the angles. In a triangle? Between parallel lines?
*March 9, 2015*

**precalc**

x^2+3x-10 = (x+5)(x-2) x^2-8x+15 = (x-5)(x-3) So, you want (x+5)(x-2) / (x-5)(x-3) * R = (x-2)/(x-3) So, you have R = (x-2)/(x-3) * (x-5)(x-3) / (x+5)(x-2) R = (x-5)/(x+5)
*March 9, 2015*

**3rd grade fraction**

you sure there are 15 coins?
*March 9, 2015*

**Vectors**

The vector equation for the line is r(t) = x(t) i + y(t) j = (-2-t)i + (4+2t)j r(0) = -2i+4j r(t) = -2i+4j + t(-i+2j)
*March 9, 2015*

**math**

π(2.4^2-2.2^2)cm^2*350cm * 10g/cm^3 = 10.12 kg
*March 9, 2015*

**MATH**

40mm * 25mm * 2mm = 2000mm^3 2000mm^3 * .0005 oz/mm^3 = 1 oz now, why you tell us about a bar of soap and then about the weight of gold, I have no clue...
*March 9, 2015*

**Algebra 1**

since y=2x-1, 2x + 2(2x-1) = 22 6x = 24 x = 4 so, y = 2x-1 = 7 one solution.
*March 9, 2015*

**Math**

you want two partitions of 36 that multiply to between 300 and 350 18x18 = 324 17x19 = 323 16x20 = 320 15x21 = 315 14x22 = 308
*March 9, 2015*

**Trig verifying identities**

oops. That is csc^2(x) - 1 = cot^2(x)
*March 9, 2015*

**Trig verifying identities**

sec(pi/2 - x) = csc x sec^2(x) - 1 = cot^2(x) remember that the co- in cosine, cotan, cosec means the complementary angle. cos(x) = sin(pi/2 - x) and so on
*March 9, 2015*

**preCal help please**

can't figure what you want. 5√-3 / (√5-√3) ? multiply top and bottom by √5+√3
*March 9, 2015*

**Math**

all you have is the difference. You need one more piece of information. normal - storm > 121
*March 9, 2015*

**Math help and check**

1 and 2 are ok 3 is eponential as in (1,2) (2,4) (3,8) (4,16) ... where the common ratio is 2 y = 2^x
*March 9, 2015*

**Math**

correct 80 < t or t > 80
*March 9, 2015*

**FRACTIONS (simplify)**

2m^2-8 = 2(m-2)(m+2) so, that is the common denominator. You end up with (m+2)(m-2)(m+2) - m(2)(m-2) + 2m^2-m^3 ---------------------------------------- 2(m-2)(m+2) m^3+2m^2-4m-8 - (2m^2-4m) + (2m^2-m^3) ---------------------------------------- 2(m-2)(m+2) 2m^2-8 ------------ ...
*March 9, 2015*

**PreCal**

(√2 + 1)/i (√2 + 1)i/i^2 -(√2 + 1)i
*March 9, 2015*

**Calculus 2**

no, that is point-slope form. To me, it's just as good as the slop-intercept form, since it contains all the necessary info. For this kind of problem, you have a point on the curve, and the slope of the tangent. The point-slope form is ideal. x = tcost dx/dt = cost - tsint...
*March 9, 2015*

**MATH**

1.8E-5 = 18E-6 so, sqrt will be √18 E -3 You can do that by hand, or just do some estimating 4^2 = 16 4.2^2 = 17.64 4.3^2 = 18.49 So, you can see it is between 4.3 and 4.3
*March 9, 2015*

**geometry**

no diagram. more data.
*March 9, 2015*

**math**

v = pi r^2 h = 100pi h dv/dt = 100pi dh/dt Now just plug in your numbers to get dv/dt. Note that it does not matter what the height is. Since the cylinder has a constant radius, its volume is directly proportional to its height.
*March 9, 2015*

**math**

the magnitude is clearly √(16^2+18^2) now just figure the angle. Draw a diagram.
*March 9, 2015*

**math**

just use the product rule v = x^2 h dv/dt = 2xh dx/dt + x^2 dh/dt Now just plug in the given values.
*March 9, 2015*

**Math**

avg cost is c(x)/x = .1x + 36 + 90/x now just find x when dc/dx = 0.
*March 9, 2015*

**Trigonometry**

draw a diagram. Use the law of cosines, since you have two sides and the angle between them.
*March 9, 2015*

**calc**

the distance z between the ships after t hours, is given by z^2 = (150-35t)^2 + (30t)^2 at t=4, z=30√5 So, plug in your values, using 2z dz/dt = -70(150-35t) + 1800t
*March 9, 2015*

**Precal**

http://www.wolframalpha.com/input/?i=plot&a=*C.plot-_*Calculator.dflt-&a=FSelect_**Plot-.dflt-&f3=2x%5E2%2F((x-2)(x%2B4))&f=Plot.plotfunction%5Cu005f2x%5E2%2F((x-2)(x%2B4))&f4=x&f=Plot.plotvariable%5Cu005fx&f5=-6&f=Plot.plotlowerrange%5Cu005f-6&f6=6&f=Plot.plotupperrange_6&f7...
*March 9, 2015*

**Functions - Please Help**

the amplitude is (15.28-9.08)/2 = 3.1 Let's say Jun21 is day 172 Dec21 is day 355 The period is clearly 365 days, so y = 3.1cos(2π/365 (x-172)) Now just find x when y=13.5 and figure the dates.
*March 9, 2015*

**math**

f(z) clearly has poles of order 4 at z = 3 order 2 at z = ±√3 i
*March 9, 2015*

**Trigonometry (Identities)**

(cos-(sin-1))/(cos+(sin-1)) (cos-(sin-1))^2/(cos+(sin-1))(cos-(sin-1)) cos^2 - 2cos(sin-1) + (sin-1)^2 ------------------------------------- cos^2 - (sin-1)^2 cos^2 - 2sin*cos + 2cos + sin^2 - 2sin + 1 ----------------------------------------- cos^2 - sin^2 + 2sin - 1 2cos - ...
*March 9, 2015*

**Maths**

can't tell whether you mean 2^x - 1 or 2^(x-1) In either case, see http://www.wolframalpha.com/input/?i=plot+2^x-1%2C+2^%28x-1%29
*March 8, 2015*

**math**

the easy ones are (0,-10/3) and (2,0) Other than that, pick any value for x or y, and then calculate the other.
*March 8, 2015*

**math**

it's already two decimal places. If you mean two significant figures, then that is 7.4
*March 8, 2015*

**maths**

clearly zero
*March 8, 2015*

**Ms Sue I need you MATH!**

if the pool is circular, then pi r^2 = 25.75, so its radius r = 2.86 m. Adding a meter all around means its new area is pi * 4.86^2 = 74.20 m^2 So, the difference is 48.45 m^2 The volume of concrete is thus 4.845 m^3 If the pool is not round, adjust the figures to allow for ...
*March 8, 2015*

**math**

4y+8y+16 12y+16 4(3y+4)
*March 8, 2015*

**Math ASAP**

Since Sprint requires a 2-year (24-month) contract, for M months and m non-weekend minutes per month, that is (70-30) + 24(30+.029m) = 760 + .696m Verizon for 24 months with m non-network minutes minutes each, charges 24(28+.3m) = 672 + 7.2m All that stuff about weekend ...
*March 7, 2015*

**Math 112**

suppose 1/0 = x That means that 1 = 0*x But zero times anything is zero. There is no number x which can be defined as 1/0. 0! is useful because n! is a recursive definition. That is, n! = n * (n-1)! 3! = 3 * 2! 2! = 2 * 1! 1! = 1 * 0! But, n! is also defined as the product of ...
*March 7, 2015*

**addmath A.P.**

a = T1 = 3*1-2 d = T2-T1 = (3*2-2) - (3*1-2) = 3 T5 = 3*5-2
*March 7, 2015*

**math**

If you mean 1/x^2 e^-x dx, there is no integral using elementary functions. Maybe I misread it. http://www.wolframalpha.com/input/?i=integral+1%2Fx^2+e^-x+dx
*March 7, 2015*

**math**

If you graph it, you find that x = .19 or 1.28 so, now you can figure cos and tan
*March 7, 2015*

**Math**

looks like 2160 pi to me.
*March 7, 2015*

**Math**

V = pi r^2 h also, there are 5 tanks, not just 1. So, we need 5 * pi * 36 * 12 = 2160 pi ft^3
*March 7, 2015*

**algebra - Oops**

3/4 = h/15 This assumes that she is 5 feet up the ladder, or 4 feet closer along the ground. If she is in fact 5 feet closer along the ground, then the original answer is good. The wording of the problem is unclear.
*March 7, 2015*

**algebra**

Assuming that she is 5 feet along the ladder, then using similar triangles, 3/5 = h/15
*March 7, 2015*

**math**

They said the speed of the bus is x. time = distance/speed. So, 280/x = 280/(x+20) + 7/6
*March 7, 2015*

**physics**

since all the velocities are constant, the only force acting on the skaters is gravity. (neglecting friction, as usual)
*March 7, 2015*

**Math - Oops**

x = 15000√3 = 25981 4/3 * 0.0384 = -15000/25981^2 dx/dt dx/dt = -2304 ft/s My bad.
*March 7, 2015*

**Math**

As far as I can tell, the time of day makes no difference. So, as usual, draw a diagram. If the plane is at distance x from the spot directly overhead, its speed is dx/dt, and tanθ = 15000/x sec^2θ dθ/dt = -15000/x^2 dx/dt when θ = 30°, secθ = 2/&#...
*March 7, 2015*

**Math**

1/R = 1/R1 + 1/R2 -1/R^2 dR/dt = -1/R1^2 dR1/dt - 1/R2^2 dR2/dt using the two given resistances, 1/R = 1/3 + 1/4 = 7/12 (7/12)^2 dR/dt = (1/3)^2(.04) + (1/4)^2(-.03) dR/dt = 37/4900 = 0.0076
*March 7, 2015*

**indian ridian ridge**

1/∛x = x^(-1/3) (x^(-1/3))^-6 = x^((-1/3)(-6)) = x^2
*March 7, 2015*

**maths**

104 = 8*13 96 = 8*12 That should help
*March 7, 2015*

**Maths**

3d = 3-192 = -189 d = -63 192;129;66;3
*March 7, 2015*

**physics**

Draw a diagram. You have a 3-4-5 triangle. well, 60-80-100, but it's the same ratio of sides.
*March 7, 2015*

**Math**

Looks good to me
*March 7, 2015*

**math**

If the price was p, then 1800000/(p-4000) = 1800000/p +5 p = 40000 so, at $4000 each he could buy 45 at 3600 each, he could buy 50 he sold 48 computers for 1800000 * 1.15 . . .
*March 7, 2015*

**math**

see related questions below
*March 7, 2015*

**math**

I'll use h for height, because l for length looks like the number 1. the surface area is two semi-circular ends: 2*(πr^2/2) = πr^2 the rectangular cut surface: 2rh the curved surface: πrh Now just plug and chug
*March 7, 2015*

**math**

T = A+(B-A)/2 = (A+B)/2 B = 2T-A
*March 7, 2015*

**math**

If KN=x, and Z has speed b-20 Z flew for 3.5 hr B flew for 1.5 hr (b-20)(3.5) = 340+x b(1.5) = x (b-20)(3.5) = 340+b(1.5) 2b = 340+70 b = 205 check: K is 307.5 KM west of N So, K is 647.5 km west of G 647.5/185 = 3.5 307.5/205 = 1.5
*March 6, 2015*

**math**

clearly, 23 is not the largest. If it is the smallest, then 23 + 23+d + (23+d)+d = 180 3d + 69 = 180 3d = 111 d = 37 and the angles are 23,60,97
*March 6, 2015*

**math**

s(t)equals = 5cos(t) minus- sin(3t) ??? why all the words? Just write s(t) = 5cos(t) - sin(3t) v(t) = -5sin(t) - 3cos(3t) a(t) = -5cos(t) + 9sin(3t) a(π) = -5(-1) + 9(0) = 5 you are correct.
*March 6, 2015*

**SAT Math**

those two terms are 6 apart, so their difference is 6*4 = 24
*March 6, 2015*

**Math**

after a couple of others just like this, do you have some work of your own to show?
*March 6, 2015*

**Pre-Calc.**

just set y=0. x = ±3 So, the vertices are at (±3,0) You know that it has to be y=0, because if x=0, -y^2/4 = 1, meaning y^2 has to be negative.
*March 6, 2015*

**Math**

The domain is the set of first numbers of the pairs: {0,-2,2,1,-5} the range is the set of second numbers: {2,4,8,6,0} Looks like D to me.
*March 6, 2015*

**math, physics**

the initial velocity is 60 ft/s s = 60t - 1/2 at^2 So, a = 3 ft/s^2 60/3 = 20 seconds I wouldn't call that a sudden stop!
*March 6, 2015*

**MaTH**

meet 1: 7'46" meet 2: 7'46" - 42" = 7'04" next meet the reduction was 2*42 = 84 seconds meet 3: 7'04" - 84" = 5'40"
*March 6, 2015*

**math**

(i) 50(0.035) + 30(0.0475) = 3.175 (ii) 3.175/(50+30) = 0.03968 = 3.97% .035x + .0475(50-x) >= 0.04*50 x <= 30 That is, with 30kg from A and 20kg from B, the result is exactly 4% fat Since A's content is less than 4%, using any more of A will reduce the result below ...
*March 6, 2015*

**College Algebra**

y = x^2-13x+36 clearly the y-intercept is (0,36) y = x^2 - 13x + (13/2)^2 + 36 - (13/2)^2 y = (x - 13/2)^2 - 25/4 So, the vertex is at (13/2,-25/4) y = x^2-13x+36 y = (x-4)(x-9) The x-intercepts are at (0,4),(0,9) The coefficient of x^2 is positive, so it opens up. (C) No idea...
*March 6, 2015*

**algebra**

12t^2 + 17t - 40 = 0 (4t-5)(3t+8) = 0 t = 5/4 or -8/3
*March 6, 2015*

**Pre-Calc.**

you have to complete the squares. Rearrange things a bit and you have x^2+6x + 4y^2-8y = -9 x^2+6x + 4(y^2-2y) = -9 Now complete the squares, and be sure to make the same changes to both sides of the equation: x^2+6x+9 + 4(y^2-2y+1) = -9+9+4 (x+3)^2 + 4(y-1)^2 = 4 (x+3)^2/4...
*March 6, 2015*

**algebra**

F = Av^2/400 v^2 = 400F/A v = 20√(F/A)
*March 6, 2015*

**calculus**

if the base has width 2x, then the area is a = 2xy = 2x(27-x^2) = 54x - 2x^3 da/dx = 54 - 6x^2 = 6(9-x^2) da/dx=0 when x=3 So, the rectangle is 6 by 18 with area=108
*March 6, 2015*

**Math Check**

#1 C #2 B 4x + 2z = 10, so z = 5-2x That gives 2x + y - 3(5-2x) = 4 -2x + 3y - 13(5-2x) = -8 or 8x + y = 19 24x + 3y = 57 so, y = 19-8x That is, whatever x you choose, y and z can be found.
*March 6, 2015*

**Pre-Calc**

as you know, the hyperbola x^2/a^2 - y^2/b^2 = 1 has asymptotes of y = ± b/a x You have a=2, b=1, so the asymptotes have slope ±1/2. Also, your hyperbola is shifted, so its center is not at (0,0). Thus, your asymptotes are y-2 = 1/2 (x+1) y-2 = -1/2 (x+1) or, y...
*March 6, 2015*

**College Algebra**

4x+y-5=0 has slope -4 So, now you have a point and a slope: y-3 = -4(x-3)
*March 6, 2015*

**College Algebra**

just plug and chug. No mysteries here: f(g(2)) = f(1) = 0 g(f(-1)) = g(4) = 3 f(2)-g(3) = 1-2 = -1 f(-1)*g(-2) = 4(-3) = -12 f(x)/g(x) = (x-1)^2 / (x-1) = x-1 for x≠1
*March 6, 2015*

**Math**

sorry, bub. No diagrams here.
*March 6, 2015*

**Math**

the horizontal sections are the same shape as the base of the pyramid, but of different sizes. What do you think of the vertical sections? As with the cube, I assume the vertical sections are parallel to a side of the base.
*March 6, 2015*

**Maths**

come on, guy. For the first equation, pick any two values. Make them easy ones. y=-1, x=4 x=-3, y=4 So, plot (4,-1) and (-3,4) and draw the line. Do the same for the other one.
*March 6, 2015*

**Maths**

c'mon. Can you plot points? Plot two points for each equation. Connect the pairs into lines. See where the lines intersect.
*March 6, 2015*

**math**

see http://www.wolframalpha.com/input/?i=plot+x-y%2F3%3D2%2C+x-5y%3D0+ or use the graphing utility of your choice. See where the lines intersect.
*March 6, 2015*

**math**

ar^5 - ar^2 = 3(ar^6-ar^5) r^3-1 = 3r^4 - 3r^3 3r^4 - 4r^3 + 1 = 0 The only real root of this equation is r=1. Is there a typo? Can you find an error I made?
*March 6, 2015*