Friday

April 29, 2016
Total # Posts: 40,000

**maths**

do you recognize powers of 3?
*April 26, 2016*

**physics**

Let the speed of sound be s. Then the time taken for the splash to be heard is 45/s seconds. That means that the stone fell for 3.12 - 45/s seconds. You know that stone fell 45m, so 5(3.12-45/s)^2 = 45 s = 375 m/s not too close to the expected value of 343 m/s...
*April 26, 2016*

**Calc 2**

an+1/an = e^-3 (1 + 2/n + 1/n^2) that ratio is less than 1, so the series converges. see http://www.wolframalpha.com/input/?i=sum+n^2%2Fe^%283n%29
*April 26, 2016*

**Math**

44 - 0.95 * (54-34) = 25
*April 26, 2016*

**Math**

7+9=16 80/16 = 5 7:9 times 5 is 35:45
*April 26, 2016*

**math**

well, 3000 ml is 3 liters.
*April 26, 2016*

**College Algebra**

just find t where 15 e^(-0.051t) = 1
*April 25, 2016*

**algrea brs so hard**

your parentheses do not balance. is m7 m*7 or m^7 ? if all you want is to expand it, use the distributive property.
*April 25, 2016*

**math**

1/3 for each son. if each ate 1/2 of his share, then 1/2 of the pie was eaten, leaving 1/2 uneaten. You can see that the number of slices does not matter here.
*April 25, 2016*

**Maths Urgent please**

well, assuming you meant g(x) = (1/16 + x)^(-5/4) g'(x) = (-5/4)(1/16 + x)^(-9/4) using the chain rule for g(x) = u^n and u = (1/16 + x) do that again for g", and evaluate the fractions.
*April 25, 2016*

**Maths Urgent please**

so, what do you get for g'(x)?
*April 25, 2016*

**Calculus II**

Looks good to me.
*April 25, 2016*

**Calculus**

since the radius is 3/4 the height, The volume of the pile is thus v = π/3 r^2 h = π/3 (3h/4)^2 h = 3π/16 h^3 dv/dt = 9π/16 h^2 dh/dt now, plugging in your numbers, at t=3, v = 3/2, so 3π/16 h^3 = 3/2 h = 2/∛π so, 9π/16 4/∛π^...
*April 25, 2016*

**math**

huh? How much do you make on regular days? It will be 1.5 times that much.
*April 25, 2016*

**Math**

the given line has slope 1/3 so, perpendicular lines will have slope -3 The line you want is thus (using the point-slope form): y+2 = -3(x-6) or, if you prefer, y = -3x + 16
*April 25, 2016*

**Calculus II**

no, but x^(4n+2) will work. duh.
*April 25, 2016*

**Calculus II**

squaring the entire series would be (sin x)^2, not sin(x^2) Just replace all the x's with x^2. ∑(-1)^n[(x^2)^(2n+1)]/(2n+1)!
*April 25, 2016*

**Length of similar shapes**

25/10 = 5/2 all of the other dimensions are multiplied by that same ratio.
*April 25, 2016*

**Calculus**

the separation distance z after t hours is found using z^2 = (200t)^2 + (150t)^2 so, find dz/dt when t=2
*April 25, 2016*

**Math**

The medians of the triangle all pass through the center of the circle. The medians intersect 2/3 of the way to the opposite side. The altitude of the triangle is also one of the medians, and it has length 10√3. So, r = (2/3)10√3 = 20/√3 For an equilateral ...
*April 25, 2016*

**Roots, complex numbers**

huh? negatives are just numbers, like positives. z^5 = -1 = 1 cisπ z = 1^(1/5) cis(π/5) but since cisπ = cis3π = cis5π, etc., to get all the values from 0 to 2π, z = 1 cis(π/5 k) where k = 1,3,5,7,9 Reiny went over this with you. There are 5 ...
*April 25, 2016*

**Math/Precal**

sorry. I meant logy(8y-7)
*April 25, 2016*

**Math/Precal**

depends on what you mean by "solve" It is log7(8y-7)
*April 25, 2016*

**math**

depends on the table.
*April 25, 2016*

**Math**

Does 34 mean 3/4? How can Ava give 920 cranes, when she only made 100? anyway, start writing down the amounts, adding and subtracting as required. At the end, subtract Ava's from Brittany's.
*April 25, 2016*

**Math**

yep
*April 25, 2016*

**Calculus**

v = 4π/3 r^3 dv/dt = 4πr^2 dr/dt Now just plug in your numbers.
*April 25, 2016*

**Calculus**

plug and chug x^2+y^2 = 25 2x dx/dt + 2y dy/dt = 0 Now you have x, y, dx/dt, so find dy/dt
*April 25, 2016*

**Calculus**

If his distance from the pole is x and the length of the shadow is s, then s/2 = (s+x)/10 1/2 ds/dt = 1/10 (ds/dt + dx/dt) 4 ds/dt = dx/dt So, the length of the shadow is growing 1/4 as fast as the man's speed. But that's not how fast the tip of the shadow is moving. ...
*April 25, 2016*

**PHYSICS! HELIUM BALLON**

the balloon weighs 277*9.8 = 2714.6 N add to that the weight of the helium enclosed. (volume * density * 9.8) so, the volume of air displaced must weigh the same amount. Now just plug in the numbers and crank it out.
*April 25, 2016*

**Calculus**

see the other problams - they are all the same thing.
*April 25, 2016*

**Calculus**

No, they said that the radius is increasing at a constant rate of 2 cm per minute That means dr/dt = 2. dv/dt = 4πr^2 dr/dt da/dt = 8πr dr/dt You have r and dr/dt, so crank it out.
*April 25, 2016*

**Calculus**

the separation distance z after t hours is found using z^2 = (200t)^2 + (150t)^2 so, find dz/dt when t=2
*April 25, 2016*

**Trig**

for Lions, assuming that all points lie on the curve, k + Asin(0+C) = 1272 k + Asin(2B+C) = 1523 k + Asin(4B+C) = 1152 k + Asin(6B+C) = 891 I'd start by noting that sin(C) = (1272-k)/A Now you can find cos(C) and then use the sum formulas to expand the other sines.
*April 25, 2016*

**O.D.E**

If someone can help me with this ODE I would greatly appreciate it. Thank you in advance! ------ Consider the differential equation dx/dt = 1/2x This is a separable O.D.E., so we know how to find all of its solutions: they are of the form x(t) = sqrt(t+c) where C is a constant...
*April 25, 2016*

**CALCULUS**

the small tank has a volume of 96π ft^3 its depth decreases at a rate of (1/2)/(16π) = 1/(32π) ft/s So, A: v(t) = 96π - 1/2 t B: duh C: the large tank's area is 4 times as big, so its depth increases 1/4 as fast, or 1/8 ft/s. D: 6/(192π)
*April 25, 2016*

**Brokport**

If the amounts invested are x,y,z then they have told us that x+y+z = 180000 .08x + .06y + .09z = 8520 .08x = 6 * .06y I suspect a typo, since even if the whole amount were invested at 6%, the interest would be 10,800. In other words, there's no way to solve these ...
*April 25, 2016*

**Algebra SOLVING LINEAR EQUATIONS AND INEQUALITIES**

#1 you can solve for 1/x and 1/y in the usual ways: elimination: double the 1st and subtract 6/x - 4/y = 28 6/x + 3/y = 7 ------------------- 7/y = -21 1/y = -3 then, 1/x = 8/3 or, y = -1/3 and x = 3/8 using substitution, 3/x = 14+2/y 2(14+2/y) + 3/y = 7 28 + 4/y + 3/y = 7 7/y...
*April 25, 2016*

**Math**

(2)(33)(19)[2(33)(19)-33-19] (2)(33)(19)[1254-33-19] (2)(33)(19)[1202] 1,507,308
*April 25, 2016*

**geometry**

Find the volume in terms of pi of a sphere with a surface area of 9 pi sq ft
*April 24, 2016*

**Math2**

f(x) = 3x^2 - 12x + 2 the y-intercept is clearly at 2. x-intercepts? where y=0. Use the quadratic formula to get x = 2±√(10/3)
*April 24, 2016*

**Math**

3 : 2 = 1 : 2/3 = 5 : 10/3
*April 24, 2016*

**Algebra**

slope = 15/9 = 5/3 y-4 = 5/3 (x-6) y = 5/3 x - 6 The lines are not perpendicular, because the slopes are not negative reciprocals. The line through the points is perpendicular to y = -3/5 x - 6 However, they do both have a y-intercept of -6.
*April 24, 2016*

**Math**

using the vertex form, y = a(x-3)^2 - 6 using the point given, 10 = a(-1-3)^2 - 6 10 = 16a-6 a = 1 y = (x-3)^2 - 6 = x^2-6x+3
*April 24, 2016*

**math**

clearly getting a 44 (lower than any of her scores so far) will not raise her grade!! Add up the total points she has so far. A "B" average for 7 tests needs at least 80*7 = 560 points. So, how many does she need to get there?
*April 24, 2016*

**Math**

The sequence is not geometric! 17111/19 = 900.5789 1539121/17111 = 89.9492 the ratio is not constant. Fix it, find the common ratio r, and then a6 = 19*r^5
*April 24, 2016*

**Calculus quick question pls**

huh? huh? How can you say that? x^2 is always positive, right? At least it is never negative. So, x^2+1 is always positive, and always at least 1. So, dividing by x^2+1 will never yield a vertical asymptote.
*April 24, 2016*

**Calculus quick question pls**

vertical asymptotes occur when you try to divide by zero. x^2+1 is never zero. You solved for x^2-1 = 0
*April 24, 2016*

**Math**

surely you can do the math. ∫[2,3] (3x^2)-(x^4-10x^2+36) dx = ∫[2,3] -x^4 + 13x^2 - 36 dx = -x^5/5 + 13/3 x^3 - 36x [2,3] = (-243/5 + 13*27/3 - 36*3)-(-32/5 + 13/3 * 8 - 36*2) = 62/15 double that for 124/15 Now you need to ask your teacher how 1436/15 can be right...
*April 24, 2016*

**Math**

what's the trouble? You are adding up a bunch of thin rectangles, with width dx and height the distance between the curves. The curves intersect at (±2,12) and (±3,27). Since both functions are even, we can use symmetry and use a = 2∫[2,3] (3x^2)-(x^4-...
*April 24, 2016*

**Absolute and relative change**

Nope. 1.07x = 16.05 the 16.05 includes the tax, right?
*April 24, 2016*

**math**

30P4 = 30*29*28*27
*April 24, 2016*

**math**

4! = 24
*April 24, 2016*

**Maths**

words, words, words. This is math, so use math 4x^2-5x+7 terms are separated by + and - signs, so there are three terms
*April 24, 2016*

**math(counting)**

if all you want is permutations of the letters, MAIRE: 5! = 120 OISEAU: 6! = 720 If you want your anagrams to be actual words, then of course there will be a lot fewer. Better get a good French dictionary.
*April 24, 2016*

**math(counting)**

what does "near" mean? There are only 5 kids, after all.
*April 24, 2016*

**Math**

If she started out with x, and was left with 15 more than half of what she started with, then x - x/8 - x/5 - x/10 = x/2 + 15 x = 200
*April 24, 2016*

**Ic**

I assume you can handle the java syntax, but the flow of logic is integer a,b,c,option,err print "Enter two integers: " read a,b print "Enter option: 1: add 2: subtract 3: multiply 4: divide " read option err=0 case option { 1: c=a+b 2: c=a-b 3: c=a*b 4: if...
*April 24, 2016*

**math**

233489
*April 24, 2016*

**precalc**

oops. tanϕ = -1/√15 tan(θ+ϕ) = (√8 - 1/√15)/(1-(√8)(-1/√15)) = (32√2 - 9√15)/7 = 1.48
*April 24, 2016*

**precalc**

cos(θ) = −1/3 QIII, so tanθ = √8 sin(ϕ) = 1/4 in QII, so tanϕ = -√15/4 tan(θ+ϕ) = (tanθ + tanϕ)/(1-tanθ tanϕ) = (√8 - √15/4)/(1-(√8)(-√15/4)) = (18√15 - 31√2)/52
*April 24, 2016*

**Absolute and relative change**

6%
*April 23, 2016*

**Precal/Math**

log(3-2x)-log(x+24)=0 log (3-2x)/(x+24) = 0 (3-2x)/(x+24) = 1 3-2x = x+24 x = -7
*April 23, 2016*

**geometry**

Thye area of a regular hexagon is 38 cm^2. What is the area of a regular hexagon with sides 4 times as long?
*April 23, 2016*

**Calculus!!**

dy/dx = xy/2 dy/y = x/2 dx ln y = 1/4 x^2 + c y = c e^(x^2/4)
*April 23, 2016*

**AP Calculus**

A yes, if f"=0 and f'≠0 B find g'(0) Then the tangent line is y-5 = g'(0) (x-0)
*April 23, 2016*

**Calculus**

The large cylinder has a cross-section 4 times that of the smaller one, so its water level rises 1/4 as fast as it falls in the smaller one.
*April 23, 2016*

**Calculus**

A a = ∫[0,5] 3/e^x dx = 3 - 3/e^5 B now you want c such that ∫[0,c] 3/e^x dx = ∫[c,5] 3/e^x dx 3 - 3/e^c = 3/e^c - 3/e^5 6/e^c = 3 + 3/e^5 e^c = 2/(1+e^-5) c = ln 2/(1+e^-5) C each semicircle has diameter equal to y. Adding up all those thin slices of ...
*April 23, 2016*

**math**

not sure what H= 10 inches from measuring the perimeter of the base is 2 inches means
*April 23, 2016*

**AP Calculus**

A (1+x)y^3 + (x^4)y - 85 = 0 y^3 + 3(1+x)y^2y' + 4x^3y + x^4y' = 0 y' = -(y^3+4x^3y)/(3(1+x)+x^4) = -(y^3+4x^3y)/(x^4+3x+3) B y'(3) = -109/93 So, the tangent line is y-1 = -109/93 (x-3) C y(3) = 1, so g(1) = 3 g'(1) = 1/y'(3) = -93/109
*April 23, 2016*

**GEOMETRY HELP PLEASE**

well, maybe not. It just might mean that AB is shorter than CD. In that case, h = 3√5 and x = -2, so CD=8 and AB=4, making the area (8+4)/2 (3√5) = 18√5
*April 23, 2016*

**GEOMETRY HELP PLEASE**

Drop altitudes CE and DF The trapezoid now can be seen to be rectangle CDEF and two right triangles of height h and base x. h^2+x^2 = 7^2 h^2 + (8+x)^2 = 9^2 Hmmm. I get x = -2 I guess I have drawn the figure incorrectly. Maybe you can use my ideas in your own drawing.
*April 23, 2016*

**math**

L = K∛M 15 = K∛125 15 = K*5 3 = K
*April 23, 2016*

**converting parabolic equations**

x = 8(y-1)^2-15 x = 8(y^2-2y+1)-15 x = 8y^2-16y-7
*April 23, 2016*

**physics**

So, if their ratio is 1.1, you have determined that Ve/Vb = 1.1 since distance = speed * time 3*10^-9 Ve = 1.1*3*10^-9 Vb = 3.3*10^9 Vb So, De-Db = 0.3*10^-9 Vb
*April 23, 2016*

**Calculus**

c'mon, you can do this. dm/dt = -0.22m dm/m = -0.22 dt ln(m) = -0.22t + c m = c e^(-0.22t) c is the initial amount, so m(t) = 20 e^(-0.22t) I'm sure you can find the half-life now, ok?
*April 23, 2016*

**Calculus**

we know the slope is dy/dx, so at (1,2), the slope is -3. The tangent line is thus y-2 = -3(x-1) y dy = -6x^2 dx 1/2 y^2 = -2x^3 + c y = √2 √(c-2x^3) you know that the domain of √ is non-negative numbers, so c-2x^3 >= 0 2x^3 <= c x <= ∛(c/2)
*April 23, 2016*

**Calculus**

recall that for parametric curves, the curvature is x'y" - x"y' ----------------- (x'^2 + y'^2)^(3/2) so, for your function, that is x' = 2t x" = 2 y' = 2t^2 y" = 4t (2t)(4t)-(2)(2t^2) ----------------------- = 4t^2/(8t^3(1+t^2)) = 1...
*April 23, 2016*

**Pre-Calc**

the domain of log(u) is u>0, regardless of the base of the logs. So, you need 3x-8 >0 x > 8/3 or, (8/3,∞)
*April 23, 2016*

**maths**

also, we have (x+y)(x-y) = 12 The pairs of factors of 12 are 1,12 2,6 3,4 x+y=6 x-y=2 x=4,y=2
*April 23, 2016*

**math**

is the light source between the center of the circle and the object's line, or on the far side of the line? That will affect where the shadow falls. Also, you have not indicated the speed of the object, dx/dt. In either case, label the light source L, the object P and the ...
*April 23, 2016*

**geometry (check answers)**

how can a rhombus be sometimes a square, but never a rectangle? a square is always a rhombus a rhombus is sometimes a square so, a rhombus is sometimes a rectangle, since a square is always a rectangle. as for trapezoids, some places define them as a quadrilateral with at ...
*April 22, 2016*

**Calculus I**

yes, so using symmetry, the area is a = 4∫[0,1] √(1-(x-1)^2) dx = π
*April 22, 2016*

**HELP ME MATH**

the perimeter is just the sum of the sides. So, that is large: 4x+2 + 7x+7 + 5x-4 = 16x+5 small: x+3 + 2x-5 + x+7 = 4x+5 large-small = (16x+5)-(4x+5) = 12x
*April 22, 2016*

**Precal**

but -2 is not a solution, since the domain of log(x) is x>0.
*April 22, 2016*

**Program**

assuming positive numbers, largest=0 for i=1 to 10 read val if (val > largest) largest=val end for print largest so, what change is needed if positive and negative numbers are allowed?
*April 22, 2016*

**Calculus I**

using shells of thickness dx, v = ∫[1,2] 2π(2-x)(x^3-1) dx using discs of thickness dy, v = ∫[1,8] π(2-∛y)^2 dy
*April 22, 2016*

**Calculus I**

each square of thickness dx has side 2y, so its area is 4y^2. Adding up all the thin squares, and using symmetry, v = 2∫[0,3] 4(9-x^2) dx
*April 22, 2016*

**Calculus I**

using shells of thickness dy, v = ∫[1,3]2πrh dy where r = y and h = x = (y-1)^2 v = ∫[1,3] 2πy(y-1)^2 dy = 40π/3 using discs (washers) of thickness dx, v = ∫[0,4]π(R^2-r^2) dx where R=3 and r=y=1+√x v = ∫[0,4]π(9-(1+√...
*April 22, 2016*

**math**

y = k/x^2 That is, x^2y = k is constant. So, you want y such that 19^2 * 9 = 17^2 * y
*April 22, 2016*

**Math**

come on. the angles of any quadrilateral sum to 360
*April 22, 2016*

**math**

what, can you not answer any of the guiding questions? I'll do #1 For x GB of data, the costs are Runfast: 25+10x BA&D: 18+15x now crank it out
*April 22, 2016*

**Algebra**

well, how many of the 52 cards are diamonds?
*April 22, 2016*

**Math**

since the opposite of "greater than or equal" is "less than", ~p = ~(x-2 >= 3) = x-2 < 3
*April 22, 2016*

**Trigonometry**

sinθ = tanθ cosθ, so tanθ - sinθ = tanθ (1-cosθ) squared that is tan^2θ (1-cosθ)^2 now we can add up the left side: tan^2θ(1-cosθ)^2 + (1-cosθ)^2 = (tan^2θ + 1)(1-cosθ)^2 = sec^2θ (1-cosθ)^2 on the...
*April 22, 2016*

**maths**

she walks on a heading, not a bearing. 50@25° = (50sin25°,50cos25°) = (21.13,45.32) add to that (200,0) and you end up at (221.13,45.32) the distance moved is thus 225.73m
*April 22, 2016*

**Math**

the sides are in the ratio 1:√3:2 multiply all those by 2.5 and you have a perimeter of (5/2)(3+√3)
*April 22, 2016*

**Math**

of the girls, brown: 4/5 long brown: 3/4 * 4/5 = 3/5
*April 22, 2016*