Thursday

September 3, 2015
Total # Posts: 33,290

**math**

Since a = 1 r = 2 S4 = 1(2^4-1)/(2-1) = 15 S15 = 2^15-1 = 32768 So, the sum from S5 to S15 = S15-S4 = 32753
*June 26, 2015*

**math**

So far, 70% of your score is allocated as follows: .15*74 + .15*96 + .40*60.5 = 49.7 Since that is only 70% of your score, it's equivalent to 49.7/.70 = 70.1%
*June 26, 2015*

**physics**

since the sled is not accelerating, the net force acting on it must be zero. So, since the friction is .15*(30*9.8), that is equal to the force acting on the sled.
*June 26, 2015*

**algebra**

If there are x fill-in y multiple-choice z long-answer they have given us the following facts: x+y+z = 24 2x+6y+5z = 100 x+y = z Now you can solve for x,y,z
*June 26, 2015*

**math**

400$ * 3tickets/5$ = 1200/5 = 240 tickets
*June 26, 2015*

**math**

The curves intersect at (-2,2) and (1,-1) so, the area is the sum of all the little horizontal strips between the line and the parabola: ∫[-1,2] (2-y^2)-(-y) dy = ∫[-1,2] 2+y-y^2 dy = 9/2
*June 26, 2015*

**math**

x' = y√(y^2+2) so, the arc length s = ∫[0,1] √(1+x'^2) dy = ∫[0,1] √(1+(y^2)(y^2+2)) dy = ∫[0,1] √(y^4+2y^2+1) dy = ∫[0,1] √(y^2+1)^2 dy = ∫[0,1] y^2+1 dy = 4/3
*June 26, 2015*

**physics**

the mass of the rocket does not matter. Given a liftoff speed of v, the height h(t) = vt - 4.9t^2 The max height is reached when t = v/9.8, so we have v(v/9.8) - 4.9(v/9.8)^2 = 100 v = 44.27 m/s Or, you could just use the formula for max height h = v^2/2g and get v^2/19.6 = ...
*June 25, 2015*

**Probability**

Assuming a uniform population, the 1st person has a birthday. The 2nd has the same birthday with probability 1/365 same for the 3rd. So, ...
*June 25, 2015*

**DTM**

you never have said how many gallons (or liters) of gas (or petrol) have been used. Tough to figure mileage (or kmage) without that.
*June 25, 2015*

**DTM**

.50$/km * 100km/day = 50$/day not sure just what it is you want to know.
*June 25, 2015*

**Algebra**

If there were x more meets to go, and they won them all, then, since the wins = the losses 2+x = 10-2 take it from there. Note that the value of x is not the answer.
*June 25, 2015*

**Math**

I already see one error: gt = ln(v+1)
*June 25, 2015*

**Math**

dv/dt = g(1+v) dv/(1+v) = g dt ln(1+v) = gt 1+v = e^(gt) + c v = e^(gt) + c (diferent c) v(0) = 0, so c = -1 v = e^(gt)-1 now, v = ds/dt, so s = 1/g e^(gt) - t + c now, e^(gt) = v-1, so s = 1/g (v-1) - t + c since v = e^(gt)-1, ln(v)+1 = gt t = 1/g (ln(v)+1)) and so s = 1/g (v...
*June 25, 2015*

**Math**

.9*77 + .1*67 = 76 Look like it won't hurt too much. 1 point.
*June 25, 2015*

**Algebra**

revenue is price * quantity, so R(x) = x*p(x) = x(-.16x+320) = -.16x^2 + 320x Now just find the vertex of that parabola to locate maximum revenue.
*June 25, 2015*

**Algebra**

I think you are missing a minus sign on the x^2 term, since g = -32 ft/s^2 h(x) = 1/42 (-16/42 x^2 + 72.7x) = x/42 (-16/42 x + 72.7) h=0 when x=0 (ball is hit), and when x = 190.8 (ball lands) The symmetry of a parabola means that the ball is at its highest midway between the ...
*June 25, 2015*

**math gr. 10**

Two good solutions are provided by Reiny and MathMate in the related questions below.
*June 25, 2015*

**maths**

well, if y = sum(a_n x^n) y' = sum(n*a_n x^(n-1)) y" = sum(n(n-1)*a_n x^(n-2)) so, from the DE, we have sum(n(n-1)*a_n x^(n-2)) + 3x^2*sum(n*a_n x^(n-1)) - x*sum(a_n x^n) sum(n(n-1)*a_n x^(n-2)) + 3sum(n*a_n x^(n+1)) - sum(a_n x^(n+1)) sum(n(n-1)*a_n x^(n-2)) + sum((...
*June 25, 2015*

**Math**

polar can get kind of complicated, but this one's a piece of cake: y = t+4, so t = y-4 x = (y-4)^2 + 4(y-4) + 8 Now expand it out and note that it's just a parabola.
*June 25, 2015*

**Math: Compounded Interest**

just use your formula. For quarterly, that is 200(1+.05/4)^(4*1/4) = 202.5
*June 25, 2015*

**Math - thx Reiny**

Good catch. I felt vaguely uneasy with the way things worked out, but was in a hurry. No doubt Justine caught the slip on her own...
*June 25, 2015*

**Math**

r = 1/(3-4sinθ) (a) see http://www.wolframalpha.com/input/?i=r+%3D+1%2F%283-4sin%CE%B8%29 (b) r = 1/(3-4sinθ) r(3-4sinθ) = 1 3r - 4rsinθ = 1 3√(x^2+y^2) - 4y = 1 3√(x^2+y^2) = 4y+1 3(x^2+y^2) = 16y^2+8y+1 3x^2 - 13y^2 - 8y = 1 3x^2 - 13(y^2 - 8/...
*June 25, 2015*

**math**

maybe you should lie on the floor and kick your feet, too. Beats me what would come next. What kind of material are you studying, that they'd drop a sequence like that on you? Or are you just wasting everybody's time with lottery numbers?
*June 25, 2015*

**physics**

You know, when you log on here you see all these posts with subjects like math, science, psychology, English, etc. Why do you think that Cowdray Park will be interpreted as a school subject? work = force * distance So, just plug in your numbers.
*June 25, 2015*

**advanced algebra**

how about you provide the function, instead of naming a file we cannot see? Is it p(t) = a*e^(kt) with suitable values of a and k?
*June 24, 2015*

**Math**

the properties of logs are the same as those of exponents. √2 = 2^1/2 because √2 * √2 = 2 and 2^a * 2^b = 2^(a+b) 2^1/2 * 2^1/2 = 2^1 = 2 And, 2^-n = 1/2^n so, 1/√2 = 2^-1/2
*June 24, 2015*

**Math**

4x - e^(-(4x-y+3)) = -e^(-2) e^(-(4x-y+3))= 4x + e^(-2) -(4x-y+3) = ln(4x + e^-2) 4x-y+3 = -ln(4x + e^-2) y = 4x+3 + ln(4x + e^-2)
*June 24, 2015*

**Math**

since y(2) = 2, e^(2-2) = -2+c 1 = -2+c 3 = c So far, so good. e^(x-y) = -x+3 x-y = ln(-x+3) y = x-ln(-x+3) Somehow you got an extra "-" sign in there
*June 24, 2015*

**6TH GRADE MATH**

assuming in and out mean up and down, then I get -2-19+7 = -14 You have the bait 10 feet above the water!
*June 24, 2015*

**math**

θ=42°
*June 24, 2015*

**Math - oops**

Oops. I added, instead of subtracting. The final answer, of course, is -1/2, as shown above.
*June 24, 2015*

**Math**

log(√x) = 1/2 log(x) since log(x^n) = n log(x) So, what you have is √36 = 6, so √72 = 6√2 So, assuming by log2 you mean log2, log26 + log26 + log22 but log22 = 1, so that is = 2log26 + 1/2 Now, what is log26? You don't have a log2 button on your ...
*June 24, 2015*

**physics**

Unless the pilot can track a speeding bullet, I don't see how he can take any of those measurements. I guess he could turn the gun backwards and fire it when passing over an observer. If the observer sees the bullets come straight down, the speeds are ok.
*June 24, 2015*

**physic pls help**

how long does it take to fall 45m? 4.9t^2 = 45 t = 3.03 seconds So, how fast do they have to be moving to cover the 15m in 3 sec? 5m/s How fast is 5m/s in terms we can really relate to? That's 100m in 20 sec. Great athletes cover 100m in about 10 seconds, so it's only ...
*June 24, 2015*

**statistics**

ummm that would be .3 + .12 * 1.3 = ?
*June 24, 2015*

**physics**

just solve for t in 24 + 24.5t - 4.9t^2 = 0
*June 24, 2015*

**Physics**

the max height is (v sinθ)^2/19.6 the speed at time t is √(Vx^2+Vy^2) = √((v cosθ)^2 + (v sinθ - 9.8t)^2) The speed at max height is just Vx = v cosθ So, find t when h(t) is half the max height. That is, when vsinθ t - 4.9t^2 = (v sinθ...
*June 24, 2015*

**Algebra**

Suppose there are x price decreases. so, price is 400-x membership is 300+3x revenue is thus (400-x)(300+3x) = -3x^2+900x+120000 So, just find the vertex of the parabola to determine price and max revenue.
*June 24, 2015*

**math**

friction coefficients? mass of man? force of man's steps?
*June 23, 2015*

**English**

2 and 4 are the same 1 and 3 are almost self-contradictory. If he kept in touch, that means he got in touch on a regular, continuing basis. You can't do that every day. So, 1 and 3 are really not good English.
*June 23, 2015*

**algebra**

if x is at 5%, the rest (30000-x) is at 9%. So, you want x in .05x + .09(30000-x) = .08(30000)
*June 23, 2015*

**algebra**

x^4 + 2x^3 - x^2 - 2x x(x^3 + 2x^2 - x - 2) x(x^2(x+2)-(x+2)) x(x^2-1)(x+2) x(x-1)(x+1)(x+2) I expect you can take it from there, yeah?
*June 23, 2015*

**maths**

if cot x = 3/4 sec x = 5/3 csc x = 5/4 So, plug in the values and see whether it's true.
*June 23, 2015*

**maths**

So, is it ever true? tanθ/(1+tan^2θ) = sinθ/cscθ tanθ/sec^2θ = sin^2θ sinθ cosθ = sin^2θ sinθ(cosθ-sinθ) = 0 θ = π/4
*June 23, 2015*

**maths**

If you can't spell theta you might as well just use x. secθ = 5/4 so, sinθ = 3/5 cscθ = 5/3 tanθ = 3/4 tanθ/(1+tan^2θ) = sinθ/cscθ (3/4)/(1+9/16) = (3/5)/(5/3) (3/4)/(25/16) = 9/25 3/4 * 16/25 = 9/25 12/25 = 9/25 Hmmm. I suspect a ...
*June 23, 2015*

**Guru found Singh**

If I said A product of two integers is 10. If one number is 5, find the other number. could you find the other? Follow the same step here.
*June 23, 2015*

**Science**

a = v^2/r, so .18 = v^2/5 Now just find v.
*June 23, 2015*

**Differential Equation**

as you know, |ai+bj| is √(a^2+b^2) and ai+bj + ci+dj = (a+c)i+(b+d)j so what do you get?
*June 23, 2015*

**vectors**

two sides are <8,6,-4> and <9,10,14> The cross product is <124,-148,26> Its magnitude 194.823 is twice the area of the triangle.
*June 23, 2015*

**maths**

well, all four angles add up to 360, so 30+50+d+d = 360 . . .
*June 23, 2015*

**statiSTIcs**

52 is 2 std above the mean of 44. So, look up Z=2 in your table, and subtract that from 1. Or, mosey on over to http://davidmlane.com/hyperstat/z_table.html and enter 2 in the "Above" box and it will calculate the Z value and show the tail of the graph.
*June 22, 2015*

**Calculus**

well, recall what you were told: The price(in dollars) p,and the quantity sold x obey the demand equation p=-1/10x+150 You found that maximum revenue R occurs at x=750, so p = -75+150 = 75
*June 22, 2015*

**algebra**

c = k-4 c+2 = (k+2)/2 + 11 c = 24
*June 22, 2015*

**math**

AB = 12, BC = 4, CD = 7, and DA = 15 Start off my assuming that the largest number is the whole line. That might not be so, but it's a good place to start. B must be to the right of C, since BC < CD That means we have ACBD. Check: AC=8 CB=4 BD=3 You just have to play ...
*June 22, 2015*

**algebra**

just find t when 5422(1.037^t) = 29000000 (I dropped 3 zeros on each side) 1.037^t = 5348.57 t = log(5348.57)/log(1.037) It'll be a while
*June 22, 2015*

**algebra**

You don't say whether the dividing fence is to be parallel to the river or not. If not, then let x be the length of the side parallel to the river. Then you have x+3y = 4700 so, x = 4700-3y The area is a = xy = (4700-3y)y = 4700y - 3y^2 The vertex of this parabola is at (...
*June 22, 2015*

**algebra**

same as the others. Give it a try. period
*June 22, 2015*

**algebra**

this is just like the river field. Why do you insist on saying "comma" and then writing the number with the comma? question mark
*June 22, 2015*

**algebra**

if the side parallel to the river is x, and the other two sides are y, you have x+2y = 9000 the area a = xy = (9000-2y)(y) = 9000y-2y^2 The vertex of this parabola is where you have maximum area. That is at y = 9000/4 You have y, so you can get x, and can find the area.
*June 22, 2015*

**Algebra**

why all the words, when you have the symbols? demand quantity at price x is p(x) = 1/40 x + 19000 The cost is C(X) = 15450x + 20000 (a) So, the revenue (price * quantity) is R(X) = xp(x) = 1/40 x^2 + 19000x (b) P(x) = R(x)-C(x) = (1/40 x^2 + 19000x)-(15450x + 20000) = 1/40 x^2...
*June 22, 2015*

**Geometry**

42 u2004-42-42-42-00_files/i0060042.jpg
*June 22, 2015*

**Math**

see related questions below
*June 22, 2015*

**math**

your answer is correct. Why all the nonsense below it?
*June 22, 2015*

**Math**

If the number of 37,50,58-in tv's are x,y,z, then x = 2(y+z) y = z 1800x+2000y+4800z = 168000 Now just solve the three equations.
*June 22, 2015*

**physics**

F = ma = 1.2*10^3kg * (20m/s)/(5s) = 4.8*10^3 kg-m/s^2 = 4.8*10^3 N
*June 22, 2015*

**Algebra**

If the number of 37,50,58-in tv's are x,y,z, then x = 2(y+z) z = ? (which are equal?) 1800x+2000y+4800z = 168000 Fix the second condition, and then solve the three equations.
*June 22, 2015*

**algebra**

√42 By the way, your files cannot be seen anywhere except on your computer.
*June 22, 2015*

**programming ( qbasic)**

I did google searches for qbasic input to find how to do input and print qbasic math to see how math was done. It also showed the DIM statements and a FOR loop qbasic strings to find string functions like Val and CHR and ASC
*June 22, 2015*

**programming ( qbasic)**

DIM WORD AS STRING DIM SUM AS INTEGER INPUT "Enter a word: ", WORD$ SUM = 0 FOR N=1 to LEN(WORD$) SUM = SUM + ASC(MID$(WORD$,N,1)) NEXT N PRINT "Word value: ",SUM google is your friend. I have never used qbasic, but in about five minutes I found enough code...
*June 22, 2015*

**Physics**

you assumed g/min was gallon/minute Not so. It is grams/minute So, keeping track of the units, you want to convert volume to time: 196gal * 231 in^3/gal * (2.54cm/in)^3 * 1g/cm^3 * 1min/g = 741940 min = 1.41 yr Hmmm. check my math or your supposed answer
*June 22, 2015*

**Math - not**

I'm sure an acronym is not what you are after. An acronym is a word whose letters come from the words of a phrase represented, such as LASER - Light Amplification by Stimulated Emission of Radiation Maybe you are supposed to write a list using the letters of the word as in...
*June 22, 2015*

**mathematics**

type that into google or, what the heck -- use your calculator stop wasting time here with such nonsense.
*June 22, 2015*

**Math velocity**

The second ball hits when 60-4.9(t-2)^2 = 0, t = 5.5 v = 19.6(1 - 1/5.5) = 10.036 m/s So, you want v where 60 + 5.5v - 4.9*5.5^2 = 0 v = 16.04 m/s check:
*June 22, 2015*

**math conversion**

I don't know about Chinese, but from metric to English units, 1kg = 2.2046 lb so, now just multiply by the number of kg you have.
*June 21, 2015*

**Algebra II**

5x+12x+13x <= 120 30x <= 120 x <= 4 so, the shortest side can be no more than 20.
*June 21, 2015*

**math-urgent**

you have constant acceleration h" = -9.8 m/s^2 so, the speed is h' = -9.8t+Vo m/s Vo=20, so h' = -9.8t+20 h= -4.9t^2+20t+Ho Ho=20, so h = -4.9t^2+20t+30
*June 21, 2015*

**math-urgent**

the mass of the ball does not matter. The height as a function of time is h(t) = 30 + 20t - 4.9t^2 so, just find the vertex of the parabola, and solve for t when h=0. Just good old Algebra I.
*June 21, 2015*

**calculus**

each triangle has a base 2y. So, since the area of an equilateral triangle of side s is √3/4 s^2, our triangles have area √3/4 (9-x^2) Now all we have to do is sum up all the triangular plates of thickness dx, from -3<=x<=3. Or, because of symmetry, twice the...
*June 21, 2015*

**Calculus AB**

a) well, cosh(0) = 1 b) 211.49-20.96cosh(0.03291765x) = 70 20.96cosh(0.03291765x) = 141.49 cosh(0.03291765x) = 6.75 x = arccosh(6.75)/0.03291765 c) y' = -(20.96)(0.03291765x)sinh(0.03291765x) so, just plug in the values from (b). Use the graphs's symmetry to eliminate ...
*June 21, 2015*

**maths**

Angle RPQ is 75+30 = 105° Now just use the law of cosines to find the distance x from R to Q: x^2 = 32^2 + 24^2 - 2(32)(24)cos105°
*June 21, 2015*

**Calculus**

since the line 1-15x is the tangent line at (0,1), we could just find where |1/(1+3x)^5) - (1-15x)| < 0.01 -0.008353 < x < 0.008872 Using differentials, we have y(0) = 1 dy = -15/(1+3x)^6 dx So, if we want |dy| < 0.01, we need x such that 1/(1+3x)^5 - (1-15x) < ...
*June 21, 2015*

**Calculus**

However, given axis h, with 26 <= h <= 51 (why?) and angular speed ω, the height h(t) = h + 26sin(ωt) dh/dt = 26ω cos(ωt) so, solve for t when h=25, and then find dh/dt.
*June 21, 2015*

**Maths Problem**

their speed difference is 50m/min, so it will take Gerald 600/50 = 12 minutes to run the extra distance. So, how many laps does Gerald run in 12 minutes?
*June 21, 2015*

**gips ghoman**

fix the typo. 2nd equation has no "y" term. Then, note that 18 = 18x+18y and you can eliminate the 18x stuff.
*June 21, 2015*

**jdt trignometric**

using more conventional symbols, I assume we have tanθ = 7/24, in the 1st quadrant. So, cosθ = 24/25 I expect you can do the arithmetic now, eh?
*June 21, 2015*

**algebra**

If $x is invested at 2.3%, the rest (80000-x) is at 3.1% So, adding up the interest, we get .023x + .031(80000-x) = 2000
*June 20, 2015*

**ALGEBRA**

If his rowing speed is r, and the current speed is c, then (r+c)/4 = (r-c)/3 48/(r+c) + 48/(r-c) = 14 c = 1
*June 20, 2015*

**ALGEBRA**

Let's say the cop's leaps are 4 ft, while the snatcher's are 3 feet, and the cop's leaps take 6 seconds, while the snatcher's take 5 seconds. So, after x seconds, the distances and times must match up: (x/6)(4) = (x/5)(3) + 72*3 x = 3240 So, it will take ...
*June 20, 2015*

**pre-calculus**

http://www.wolframalpha.com/input/?i=x%2F%28x^2-100%29 as you can see from the graph, the concentration appears to be negative until x=10. Then it explodes and then drops down. I cannot think what the vertical asymptote can represent, unless there's some delayed-release ...
*June 20, 2015*

**Math**

P(x) = 750x(20-x) So, since x is always positive, where do you think P(x) is negative?
*June 20, 2015*

**Algebra**

each graph is a line. You don't need to use y=mx+b to "start." 18x+3y=12 or, 6x+y=12 If x=0, y=12 if y=0, x=2 So, plot the points (0,12) and (2,0) and join them in a line. Using the Ax+By=C (standard) form is very convenient, since you can easily get both ...
*June 20, 2015*

**Algebra**

You don't have to use any particular form of the equation. In fact, this problem asks you to graph the lines, and solve the system of equations using the graphs. For example, http://www.wolframalpha.com/input/?i=plot+6x+-+3y+%3D+24+%2C+18x+%2B+3y+%3D+12+
*June 20, 2015*

**pre-calculus**

The horizontal asymptote shows that the level of medicine approaches a constant level over a long time. A vertical asymptote? Got me. The level of medicine is always finite. Of course, it might be helpful if you actually provided the function under discussion. Just sayin'
*June 20, 2015*

**Geometry**

so, why not just follow the steps described, and see which way produces the desired result?
*June 20, 2015*

**algebra**

you know that the vertex of the parabola y = ax^2+bx+c is at x = -b/2a. So, your maximum height is at t = 90/32 Just find h(90/32) Or, rewrite the equation by completing the square: h(t) = –16t^2 + 90t + 15 = -16(t^2-90/16 t) + 15 = -16(t^2 - 90/16t + (90/32))^2 + 15 + 16...
*June 20, 2015*

**algebra - PS**

Or, you can think of it like this: all lines of the form 3x – 4y = C are parallel to the line 3x – 4y = 8 So, plug in your point (0,4) and you have 3(0)-4(4) = -16 So, your new line is 3x – 4y = -16 which is the same line derived above with the point-slope form.
*June 20, 2015*

**algebra**

3x-4y=8 has slope 3/4 So, you want the line through (0,4) with slope 3/4: y-4 = 3/4 (x-0)
*June 20, 2015*

**algebra**

If there were g general tickets, then the rest (416-g) were reserved. Now add up all the receipts: 8g+14(416-g) = 4840 Now just solve for g.
*June 20, 2015*

**ST MICHAEL NURSERY SECONDRY AND PRIMARY SCHOOL KONTAGORA**

force = mass * g work = force * distance power = work/time
*June 20, 2015*