Sunday

March 29, 2015

March 29, 2015

Total # Posts: 30,170

**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*

**Pre-calc/trig**

Both a and b are not truly periodic, since the oscillation gradually has lower and lower amplitude. But, all other things being equal, I'd say the pendulum keeps its amplitude much better than the ball, so I'd have to go with (a) Even (c), if the plane is always on ...
*March 6, 2015*

**Math**

using substitution, z = -3, so 3y-8(-3) = -9 3y = -33 y = -11 so, x-(-11)+2(-3) = 22 x+11-6 = 22 x = 17 So the solution is (17,-11,-3) The other is not quite so simple, but since z = 1-2x-4y, we have x-2y-3(1-2x-4y) = 2 x+y-(1-2x-4y) = -1 or, 7x+10y = 5 3x+5y = 0 So, since 5y...
*March 6, 2015*

**Pre-Calculus**

the vertices are at the ends of the major axis, and the covertices are on the minor axis. So, since 25 > 4, the vertices are on the y-axis. when x=0, y=5 when y=0, x=2 So, the vertices are at (0,±5) the covertices are at (±2,0) Looks like you need to review ...
*March 6, 2015*

**College Algebra**

the two-point form is y-3 = (-4-3)/(4-1) (x-1) and you can massage that into what you need.
*March 6, 2015*

**College Algebra**

just use the distance formula: d = √((-1-2)^2 + (-1-3)^2) = √(9+16) = 5
*March 6, 2015*

**College Algebra**

the domain is all reals except where the denominator is zero. see more information at http://www.wolframalpha.com/input/?i=%28x%2B4%29%2F%28x^2-1%29
*March 6, 2015*

**calculus**

#1 I think you have a typo here. Since the r = d/2, the radius always grows half as fast as the diameter. #2 If the base is x from the wall, and the top is y in height, then you have x^2+y^2 = 25^2 so, 2x dx/dt + 2y dy/dt = 0 or, getting rid of the 2's, x dx/dt + y dy/dt...
*March 6, 2015*

**College Algebra**

since body/snout is constant, body/3.5 = 7/2
*March 6, 2015*

**College Algebra**

680(1+.04/12)^(12*17)
*March 6, 2015*

**Business Math GBM 200**

cost is 300*.24 = 72.00 desired end revenue is 72*2 = 144.00 But only 79% = 237 lbs get sold $144/237lb = .6076 = $0.61/lb
*March 6, 2015*

**Physics**

how long does it take to fall 0.75m? 4.9t^2 = 0.75 the speed is 2/t m/s I don't think the above answer is correct.
*March 6, 2015*

**MATHS**

Looks to me as though there are in fact many possible solutions. Maybe the "correct" one is the smallest total. Nope. 1244 +125 ------- 1369
*March 6, 2015*

**o level mathematics**

Draw your Venn diagram. The intersection of all 3 circles is 2 The other 3 numbers are placed outside any intersections. So, that means we have accounted for 20 of the 35 students. The other 15 can play any combination of two of the games, or just one game. So, there could be ...
*March 6, 2015*

**AHOOOGAAH! HOMEWORK DUMP!**

...
*March 6, 2015*

**computer programming**

you're going to have to explain those lengths a bit better than that.
*March 5, 2015*

**Pre-Calc**

sinB/b = sinA/a so, sinB = 108*sin21°/76 = 0.50925 so, B = 30.6° or 149.4° Since A+B+C = 180, that means C is 128.4° or 9.6° So, since c/sinC = a/sinA there are two choices for c, as well.
*March 5, 2015*

**Math**

assume an infinite number of hops. In practice, of course, the ball stops earlier than that, but you have a geometric progression where a = 20 r = 4/5 So, the initial drop is 20, and each bounce is a round trip 4/5 as high as the one before, so the total distance after n hops ...
*March 5, 2015*

**maths**

12% off means 88% of the price. So, .88p = 1.21c p = 1.375c so, the marked price is 37.5% above cost
*March 5, 2015*

**math**

c+p+s = 100 .5c + 2p + 5s = 100 There are lots of answers, as long as there are an even number of chicks (so you have whole dollars for their cost) For example, getting the most sheep possible, you have 2 chicks, 4 pigs, 19 sheep Now, you can start converting each pig into 4 ...
*March 5, 2015*

**Directions**

Face North, and turn 70° eastward. Due east is N90°E NE is N45°E
*March 5, 2015*

**math**

the more negative the balance, the greater the debt.
*March 5, 2015*

**math**

3 1/3 = 10/3 1 1/4 = 5/4 10/3 * 5/4 = 50/12 = 25/6 = 4 1/6 miles
*March 5, 2015*

**Calculus**

dx/dt = (2-x)√(1-x) dx/(2-x)√(1-x) = dt Note that 2-x = 1 + 1-x = 1 + √(1-x)^2 2 arctan √(1-x) = t + c √(1-x) = tan(t/2 + c) = tan(t/2) + c 1-x = (tan(t/2) + c)^2 x = 1 - (tan(t/2)+c)^2
*March 5, 2015*

**Calculus**

what, have you forgotten your algebra I? y' = (y^2 + y^2 cosx)^2 y' = y^4 (1+cosx)^2 y'/y^4 = (1+cosx)^2 dy/y^4 = (1+cosx)^2 dx -1/3 y^-3 = 1/4 (6x+8sinx+sin2x) + c y = -∛(4/(3(6x+8sinx+sin2x) + c))
*March 5, 2015*

**algebra**

y = x^2-2x-8 = x^2-2x+1 - 9 = (x-1)^2 - 9
*March 5, 2015*

**math 5th grade**

8/(10+8) this assumes the socks feel identical, and they are stacked in a messy pile.
*March 5, 2015*

**math**

evidently you have not spent the intervening time studying the problems. #1 A flat rate is just that: a constant amount, say $25 A commission is a certain amount, say $50 for every sale. So, for x sales, he would earn 50x so, his total earnings are 25 + 50x #3 is just like #1...
*March 5, 2015*

**math**

If (x) = -2 and (y) = 2, then everything is false.
*March 5, 2015*

**math**

I assume (x) means |x| = absolute value? well, you have |x| = 2 |y| = 2 Looks like (c) is false. Can't tell with (a), since "1 (y)" eludes me.
*March 5, 2015*

**Math**

2x+y = 2400 so, y = 2400-2x a = xy = x(2400-2x) = 2400x-2x^2 That is just a parabola, with vertex at x=600 So, the field is 600 x 1200 As usual, these problems achieve maximum area when the perimeter is divided equally among lengths and widths.
*March 5, 2015*

**Math**

If the rectangle has width 2x, then the area is a = 2xy = 2x(12-x^2) da/dx = 24 - 6x^2 = 6(4-x^2) da/dx=0 when x=2 So, the largest area is 4(12-2^2) = 32
*March 5, 2015*

**Math**

how about 100 and 100 99*101 = 9999 100*100 = 10,000
*March 5, 2015*

**Acceleration**

SInce 1 mi = 1,6 km, 1 mi/hr/s = 1.61km/hr/s So, 23.3 mi/hr/s = 37.5 km/hr/s So, now you can compare them.
*March 5, 2015*

**math - typos**

Holy crap! Is ! the NOT operator? If so, we have !(!((5-10)*(4-(9/2)) > 60) has unbalanced parentheses (((5*7)/(4/(5+3)))=15)) has unbalanced parentheses In both cases there are not enough ('s Better go over it and make sure things balance out
*March 5, 2015*

**math: algebra**

so, who asked anything about the volume?
*March 5, 2015*

**math: algebra**

Drop an altitude from the vertex to the center of the base. Now draw a line from there to the center of a side of the base. Seen from the side, you now have a right triangle with legs 4 and 22. The hypotenuse is the slant height of the pyramid, and thus the altitude of one of ...
*March 5, 2015*

**Math**

there is only one statement, and it is true. why?
*March 5, 2015*

Pages: <<Prev | 1 | 2 | 3 | 4 | 5 | 6 | 7 | **8** | 9 | 10 | 11 | 12 | 13 | 14 | 15 | Next>>