Saturday

January 31, 2015

January 31, 2015

Total # Posts: 28,511

**Math**

a1+...+a9 = 5a + (0+2+4+6+8)d a2+...+a10 = 5a + (1+3+5+7+9)d so, 5a+20d = 17 5a+25d = 15 d = -2/5 a = 5
*December 15, 2014*

**Math Plz dont ignore Plz**

You just need to read through the words and write down the mathematical facts. They want to know how many adults and children were admitted. So, let's say there are a adults and c children. 1850 adults and children means a+c = 1850 $4.50 per child and $7.50 per adult and ...
*December 15, 2014*

**AP Calc BC**

y' = √x / -3y -3y dy = √x dx -3/2 y^2 = 2/3 x^(3/2) + c so, yes, you are correct y^2 = -4/9 x^(3/2) + c Not much to do with that. I guess you could use a new c. Name the old one k and you have y^2 = k - 4/9 x^(3/2) If k = 4/9 c then y^2 = 4/9 c - 4/9 x^(3/2) y...
*December 15, 2014*

**AP Calc B/C**

So, add up all the little wafers with diameter x^3, perpendicular to the x-y plane. They would have their centers at (x,y/2) with radius y/2 = x^3/2. The volume of a small semi-circular wafer of radius r and thickness dx is 1/2 πr^2 dx v = ∫[0,4] 1/2 π(x^3/2)^2...
*December 15, 2014*

**math**

well, since 27 = 4*6 + 3 what do you think?
*December 15, 2014*

**Math**

1/3 + √3/12 + 1/16 + √3/64 Using a common denominator of 64*3, we have (64 + 16√3 + 12 + 3√3)/192 (76+19√3)/192 or 19(4+√3)/192
*December 15, 2014*

**MATH**

-1+6 = 5, so -5(5) = ?
*December 15, 2014*

**math**

B:H = 4:7 H:J = 3:5 7*3 = 21, so we also have B:H = 12:21 H:J = 21:35 So, B:H:J = 12:21:35 12+21+35 = 68 136/68 = 2, so double everything so the sum is 136 B:H:J = 24:42:70 and those are the amounts for each girl.
*December 15, 2014*

**Algebra 1**

10.5^2
*December 15, 2014*

**Algebra 1 thx**

Does the coach count as one round? If so, the nth round has 3^n calls. If not, then the nth round has 3*3^n calls
*December 15, 2014*

**Algebra please help answer**

40/10 = 4, so it will triple 4 times. 3^4 = 81
*December 15, 2014*

**math**

If the cost is the same for x hours, then clearly 200+30x = 240+25x
*December 15, 2014*

**calc**

the tangent vector is ∇v = (dx/dt,dy/dt,dz/dt) ∇v = <4∛t,6t,-4/t^3> v(-1) = <3,-3,2>, so ∇v = <-4,6,4> Note that <-2,3,2> is parallel to ∇v(-1)
*December 15, 2014*

**math**

x=175, if that's what you are after
*December 15, 2014*

**Math**

since the ratio of the sides remains the same, x/5 = 5.1/3
*December 15, 2014*

**CAL**

y = 4/9 sinx - 1 y' = 4/9 cosx So, the slope of the tangent at x is 4/9 cosx The slope of the normal there is -1/y' = -1/(4/9 cosx) = -9/4 secx
*December 15, 2014*

**Math**

5000-7 = 4993 4993 = 5*998 + 3 7+4990 = 4997 The next number would have been 5002 Or, using modulo 5 math, 7 = 2(mod 5) 5000 = 0(mod 5) So, 5002 = 2 (mod 5)
*December 15, 2014*

**Math**

If the sequences are P and T, and their differences are Pd and Td, then T1 = -P1 2013/2 (2T1+2012Td) = 2013/2 (2P1 + 2012Pd) or, 2013P1 + 2025078Pd = 2013T1 + 2025078Td 4026P1 + 2025078(Pd-Td) = 0 We know that since T1 < 0 and the sum of T is the same as the sum of P, that ...
*December 15, 2014*

**calculus**

There must be problems on extrema. The extrema occur where f' = 0. Here, f(x) = x^(1/3) f'(x) = 1/3 x^(-2/3) f'(x) is never zero anywhere, especially in this interval. So, we just want the maximum and minimum values attained on the interval. Note that f' > 0...
*December 15, 2014*

**Algebra 1**

What is an equation of the line with slope 6 and y intercept -4?
*December 15, 2014*

**Math**

For all the garden subdivisions, 2 of 3 sections were planted. So, 2/3 of the garden space was planted. There were a lot of words there, obfuscating the simplicity of the problem. Now, if only some of the divided gardens had been planted, that would have changed things.
*December 15, 2014*

**Algebra**

Hmmm. I see h(t), but no x. I expect you meant to ask what the t-intercept means. Reiny's answer would be the one you want. Generally in problems like this, x represents the distance traveled horizontally. So, in that case, the x-intercept wold be how far away the ball hit...
*December 15, 2014*

**7th grade math - tsk**

#1 is * not ^ Looks ok to me.
*December 15, 2014*

**maths**

69 is not a fraction. d = n+4 (n+1)/(d+1) = 4/5 (n+1)/(n+4+1) = 4/5 (n+1)/(n+5) = 4/5 5(n+1) = 4(n+5) 5n+5 = 4n+20 n = 15 The fraction was 15/19
*December 15, 2014*

**Pre-Algebra**

draw a diagram. Using similar triangles, h/60 = 5/10
*December 15, 2014*

**Calculus**

f(1) = 2-3+1 = 0 df = (6x^2-3) dx So, starting from f(1), f(0.8) = f(1)+(6-3)(-0.2) = -1.8 The line is y-0 = 3(x-1) You can see the approximation at http://www.wolframalpha.com/input/?i=plot+2x^3-3x%2B1%2C3%28x-1%29 You could have started with a different value of x, say, x=0....
*December 15, 2014*

**Calculus (math)**

You want to integrate along x, so the shells have thickness dx. The volume is thus v = ∫[1,6] 2πrh dx where r = x+2 and h = y = 1/x^4 v = 2π∫[1,6] (x+2)/x^4 dx That's a nice easy integral to evaluate.
*December 15, 2014*

**Calculus (math)**

s(t) = 1/4 t^3 - 5/3 t^3 + c The displacement is s(6)-s(-1) The distance traveled is the arc length. So, what steps did you follow?
*December 14, 2014*

**math**

use the good old quadratic formula: x = (-6±√(36+96))/(-16)
*December 14, 2014*

**Calculus (math)**

dQ/dt = kQ dQ/Q = k dt lnQ = kt + c Q = c*e^(kt) since 2 = e^(ln2), that can of course be written by massaging the constants as Q(t) = c*2^(-t/k) where k=35 Q(t) = c*2^(-t/35) Since we are given no actual amounts, we can set c=1, so that as a fraction of the initial amount, Q(...
*December 14, 2014*

**Calculus (math)**

you can find useful formulas here. Pick the one for a paraboloid and use your numbers. http://www.had2know.com/academics/paraboloid-surface-area-volume-calculator.html I'm sure you can derive it by calculating the arc length and revolving the curve. The Theorem of Pappus ...
*December 14, 2014*

**Calculus (math)**

probably true. I'll show you mine if you show me yours...
*December 14, 2014*

**Math**

5 rad = 5(180/pi deg) 5pi/8 rad = 5pi/8(180/pi deg) I expect you can do the simplification.
*December 14, 2014*

**Math**

draw your triangle. The standard ones with integer sides include 8-15-17
*December 14, 2014*

**Calculus**

f(-1) = -2 f(4) = -2 f(x) is differentiable in the interval, so the conditions are met, and the theorem applies. Extra credit: We want to show that for c somewhere in [-1,4] f'(c) = 0. f'(x) = 3-2x f'(x) = 0 when x = 3/2 3/2 is within the interval [-1,4], so the ...
*December 14, 2014*

**Calculus**

y' = 15x^4 - 75x^2 + 60 So, is y'(1) = 0?
*December 14, 2014*

**calculus**

It looks like this very problem is discussed at http://tutorial.math.lamar.edu/Classes/CalcIII/StokesTheorem.aspx f(x,y,z) is different, but the surface is a paraboloid just like yours. It goes through all the steps to evaluate the integral.
*December 14, 2014*

**Math**

If you meant 36 / (2^5/8) + 7 * (3+11) = 107 then yes. The spacing helped determine what you wanted, but parentheses work even better.
*December 14, 2014*

**Any Math Tutor Online? Help**

Geez. Draw some diagrams, ok? The distance x is (a) x =12cot3° - 12cot6° (b) x =12cot3° + 12cot6° (c) x^2 = (12cot3°)^2 + (12cot3°)^2
*December 14, 2014*

**Math**

No. It is 2^3 * 11 All you want is the prime factors, and their powers.
*December 14, 2014*

**Math**

teacher tells 6 students each of those tells 6 more, for 36 brand-new people. But the total is now the original 6 plus the 36 new ones. And so on for each generation. Yiu are correct in saying that next time, there were 6^3 = 216 new people told, but the original 6 and the ...
*December 14, 2014*

**Math**

6+36+216 Don't forget the people doing the telling!
*December 14, 2014*

**calculus**

y' = 4x+(9x^2/(3x^3+1)^(3/2)) y = 2x^2 - 2/√(3x^3+1) + C Now plug in (0,2) to find C. integrate using u = 3x^3+1, so du = 9x^2 dx
*December 14, 2014*

**Analytic Geometry**

7x+y-7=0 intersects x+y+1=0 at (4/3,-7/3). 7x+y-7=0 makes an angle of 98.13° x+y+1=0 makes an angle of 135° with the x-axis. So, the bisector makes an angle of 116.56°, with slope -2 So, the equation of the line is y + 7/3 = -2(x - 4/3) 3y+7 = -2(3x-4) 6x+3y+15=0 ...
*December 14, 2014*

**Physics**

solve for t in 50+19.6t-4.9t^2 = 0
*December 14, 2014*

**physic 210**

the distance traveled is 17t + 1/2 at^2 with t=5 and a=2. So, divide that by the circumference of the wheel = 2pi*48
*December 14, 2014*

**easy geometry**

For the trapezoid, all we really need is the altitude. You can surely then find the area and perimeter. The long base is 10 cm longer than the shorter one. So, draw altitudes at the end of the short base and you will see two right triangles with base 5 and hypotenuse 10. So, ...
*December 14, 2014*

**Math**

AD = AE-DE BE = AE-AB AB=DE, so AD = AE-AB = BE You can reword that using all the "congruents" instead of = signs.
*December 14, 2014*

**Physics**

initially, all the energy is KE = 1/2 mv^2 At the peak, all the KE has become PE = mgh So, with 2v at first, there's 4 times the KE. That means 4 times the height at the peak.
*December 13, 2014*

**Calculus 1**

As usual, draw a diagram. If the angle subtended by the arc walked around the lake is θ, then the distance traveled on foot = 4θ by boat = √(32(1+cosθ)) = 8cos(θ/2) Use the law of cosines to get this. So, the total time t is t = 4θ/10 + 8/5 cos...
*December 13, 2014*

**algebra 2**

x^2+2x-35 x = (-2±√(4+140))/2 = (-2±12)/2 = 5,-7 Not sure how you got your answers, since you declined to show your work... x^2+10x+24 (x+4)(x+6) or, using the QF, (-10±√4)/2 x^2-11x+24 (x-8)(x-3) or, via QF, (11±√25)/2
*December 13, 2014*

**Algebra**

If a4a5 means a4*a5, then clearly a4=1 and a5=13, making d=12. So, a3a6 = (-11)(25) = -275
*December 13, 2014*

**algebra 2**

if you are aiming for a perfect square, then you are correct.
*December 13, 2014*

**Math**

Sorry, make that the long multiplication button.
*December 13, 2014*

**Math**

(2x-1)(3x-7) = 6x^2-17x+7 go to calc101.com and click on the long division button. Enter 6x^2-17x+7 and 4x+3 as your polynomials, and watch the fun! I'm sure you will find where you went wrong.
*December 13, 2014*

**precalculus**

(A) The denominator never zero, so the domain is all real numbers. So, of course F(x) is continuous, since there are no places where it is undefined. As x gets large, F(x) ≈ 3x^2/x^2 = 3. For the zeros of rational functions, they occur where the numerator is zero (as ...
*December 13, 2014*

**Math**

If the ladder of length z touches the ground x feet from the fence, and touches the building at height y, then z^2 = (x+2)^2 + y^2 Using similar triangles we see that x/12 = (x+2)/y y = 12(x+2)/x z^2 = (x+2)^2 + (12(x+2)/x)^2 So, find dz/dx and set it zero, to see that the ...
*December 13, 2014*

**physics**

the answer, of course, is 10.0-6.14 = 3.86 kg have been removed.
*December 13, 2014*

**physics**

since PV=nRT and RT and V are to remain constant, P/n = RT/V must be constant. n is usually in moles, but we can just as easily call it grams, since the ratio of moles to grams is constant for O2. We have reduced P by a factor of 26.4/43.0, so we must reduce the quantity of O2...
*December 13, 2014*

**physics**

1.8 (1 - 5.80/21.6) = 1.32 moles
*December 13, 2014*

**precalculus**

holes occur where the numerator and denominator are both zero. #16 x^2+2x+1 = (x+1)(x+1) 2x^2-x-3 = (x+1)(2x-3) So, f(x) = (x+1)/(2x-3) everywhere except at x = -1. At that point, f(x) = 0/0 which is not defined. The only vertical asymptote is at x = 3/2, and at x = -1 there ...
*December 13, 2014*

**Algebra 1**

2 units down sorry
*December 13, 2014*

**Algebra 1**

2 units down sorry
*December 13, 2014*

**Algebra 1**

what kind of translation? To shift the graph up k and right h, you need y-k = f(x-h) So, to shift f(x) = |x| up 2 and right 3, you have y-2 = |x-3| or, y = |x-3|+2 h and k may be negative, for shifting down and left.
*December 13, 2014*

**Algebra 1**

Write an equation for then translation of y = IXI
*December 13, 2014*

**algebra 2**

both correct
*December 13, 2014*

**Algebra**

⌊t⌋ = 2t+3 Clearly, t must be a multiple of 1/2 If an odd multiple, then we have t = (2k+1)/2 ⌊t⌋ = k 2t = 2k+1 k = 2k+1+3 k = -4 t = -7/2 If t is an even multiple of 1/2, then ⌊t⌋ = 2k/2 = k 2t = 2k k = 2k+3 k = -3 t = -3 See the graphs at ...
*December 13, 2014*

**Algebra 1**

Thank you from a 68 hear old Grandfather trying to help his Grandson
*December 13, 2014*

**Algebra 1**

What does the graph y = - absolute value of X -4 look like.?
*December 13, 2014*

**Algebra 1**

Yes I can. Thank you from a 68 year old Grandfather helping his Grandson
*December 13, 2014*

**Algebra 1**

What does the graph of y = absolute X-2 look like?
*December 13, 2014*

**Algebra 1**

Thank you -- A Grandfather trying to help his Grandson.
*December 13, 2014*

**Algebra 1**

Describe how the graphs of y = absolute value of x,and y = absolute value of X -15 are related. 1- Them graphs have the same shape. The y intercept of y = absolute x is 0, and them x intercept of the 2nd graph is -15. 2-m The graphs have the same y intercept. The second graph ...
*December 13, 2014*

**math**

Assuming the robot is completely submerged, then the water will rise by (20m^3)/(5m*4m) = 1m Doesn't really matter how tall the tank is, or how much water is already there, as long as the entire volume of the robot is submerged.
*December 13, 2014*

**math**

(-5/10)(-5/4) (-17/13)(-65/136)
*December 13, 2014*

**algebra**

A = Pe^(rt)
*December 13, 2014*

**algebra (please help)**

If $x is at 4.25%, the rest (22000-x) is at 5.75% So, just add up the interest received: .0425x + .0575(22000-x) = 1085
*December 13, 2014*

**algebra (please help)**

you are told that 2c+f = 1120 3c+2f = 1870 Now just solve for c and f.
*December 13, 2014*

**algebra**

the cost for x items is 4000+2.19x the income for x items is 4.59x Breakeven is where income and cost are equal: 4.59x = 4000 + 2.19x For 5500 profit, you need 4.59x = 4000 + 2.19x + 5500
*December 13, 2014*

**math**

πr^2h = 512 So, h = 512/(πr^2) area a = 2πr^2 + 2πrh = 2πr^2 + 2πr (512)/(πr^2) = 2πr^2 + 1024/r Assuming you have calculus, we see that da/dr = 4πr - 1024/r^2 da/dr=0 when r = 4∛(4/π) Now you can find h.
*December 13, 2014*

**Math**

f(x) = sin(x) df = cos(x) dx So, since sin(30°) = 1/2 sin 27° = 1/2 + (√3/2)(-π/60)
*December 12, 2014*

**Algebra**

can't be both. For floor, round down to next integer for ceiling, round up to next integer
*December 12, 2014*

**Algebra**

1 sometimes 0^0 is undefined 2 never 3 never 4 sometimes .3^1 < 1
*December 12, 2014*

**Algebra**

well, just solve 46 + .024x = 1558
*December 12, 2014*

**Math**

5r(13r+3s-5s^2) = 65r^2+15rs-25rs^2 -6s(13r+3s-5s^2) = -78rs-18s^2+30s^3 s^2(13r+3s-5s^2) = 13rs^2+3s^3-5s^4 Add them up to get 65r^2 - 63rs - 12rs^2 - 18s^2 + 33s^3 - 5s^4
*December 12, 2014*

**geometry**

google the number of segments determined by n points is and you will find many discussions of the topic. Find the one you like and use n=4.
*December 12, 2014*

**Grade 9 Science**

the orbit is not a circle, but an ellipse. The planet is not always exactly the same distance from the sun. Read up on Kepler's laws of planetary motion.
*December 12, 2014*

**physics 1**

This is just the same as figuring the time it takes for an object to fall 2cm on earth, except that the gravitation acceleration is not 9.81 m/s^2, but GM/r^2 where you have the values for G,M and r. Once you have that evaluated, s = 1/2 at^2 and you can just plug in s=0.02m ...
*December 12, 2014*

**Calculus - Correction**

Oops. The part from 0 to 0.0667 is not included, because it's all inside r=1. So, the real area is just 2∫[0.0667,π/2] 1/2 ((15sinθ)^2-1^2) dθ
*December 12, 2014*

**Calculus**

The area inside a polar curve is ∫1/2 r^2 dθ The circles intersect where 15sinθ = 1 θ = 0.0667 So, from 0 to 0.0667, we have r = 15sinθ From 0.0667 to π/2, we have to subtract the area inside r=1 from r=15sinθ, so the area is twice the area ...
*December 12, 2014*

**calculus (check)**

f(x) looks good No idea what F(n) is. If you mean the sequence of partial sums, it's just the normal sum formula for A.P.'s.
*December 12, 2014*

**calculus (Series)**

You just want the sum of the series where a = -2.02 d = -0.09 n = (-4 +2.02)/-.09 = 22 S22 = 22/2(-2.02 + -4) = -66.22
*December 12, 2014*

**Calculus**

You just want the sum of an arithmetic series where a = 4 d = 1/2 S28 = 28/2(2*4 + 27(1/2)) = 301
*December 12, 2014*

**Math**

That's what I get
*December 12, 2014*

**Physics**

Assume the heavier satellite is traveling in the +x direction. Conserve momentum, so the final velocity (xi+yj) is the solution to 1000*500i + 400*700(.707i+.707j) = (1000+700)(xi+yj) That will give the x- and y-components of the final velocity, which you can then express as ...
*December 12, 2014*

**math**

(n-2)180 / 360 = 4 So, find n to determine the number of sides, and that will let you find the angles.
*December 12, 2014*

**algebra**

I'd guess from previous posts that we want y > -5/3 x + 3 But what we are to do with it I don't know.
*December 12, 2014*

**algebra 2**

2√3+√27-√1/3 2√3+3√3-1/√3 5√3 - 1/√3 (5*3-1)/√3 14/√3 √1/2•(√8+√2) 1/√2 * 2√2 + 1/√2 * √2 2 + 1 3 √18-√3/√6 3√2 - √3/(√3√2) 3&#...
*December 12, 2014*

**physics**

BCE potential and kinetic energy depend on mass.
*December 12, 2014*

**Geometry**

tanθ = 1.89/50
*December 12, 2014*

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