Sunday

April 19, 2015

April 19, 2015

Total # Posts: 30,637

**Algebra**

Try using the symbols = < > and <= >=
*February 22, 2015*

**Math vectors**

clearly, AF = 1/2 AB AE = AF + FE = AB/2 + BC/2 BC = 2FE Now, since AF+FE=AE, FE=AE-AF BC = 2(AE-AF) = 2(EC-AF)
*February 22, 2015*

**Math**

the curves intersect at (5,2) and (5,-2) So, the volume can be done using shells or discs. Using shells, consider the volume as a bunch of nested shells, each of thickness dy, radius y+2 and height the distance between the curves: (9-y^2)-(y^2+1) v = ∫[0,2] 2πrh dy...
*February 22, 2015*

**Algebra**

since the remainders are equal, 2312-1417 = 895 2312-1059 = 1253 1417-1059 = 358 are all multiples of y. The GCF of those is 179. Divide any of the given numbers by 179 and the remainder is 164. So, y-x = 179-164 = 15
*February 22, 2015*

**Algebra**

T15-T5 = 10d = 40 so, d = 4 Now you can find T1 and T60 Then the sum is 30(T1+T60)
*February 22, 2015*

**math vectors**

well, you need a•b = 0, so (2)(4)+(2k)(-k) = 0 . . .
*February 22, 2015*

**Math vectors**

that would be a•b/|b| = (5)(1)+(-5)(2)+(3)(3) ----------------------------- √(1^2+2^2+3^2)
*February 22, 2015*

**Math**

For the consecutive number stuff, it is always best to recall that 1+2+3+...+n = n(n+1)/2 Then you can multiply that and shift it in various ways. Or, you can remember your arithmetic series stuff, and recall that the sum of the first n terms of an A.P. is n/2 (T1+Tn) = n/2 (...
*February 22, 2015*

**Math**

42000(1+.03)^5
*February 22, 2015*

**math**

could be zero. blue stripe: 110 oz red stripe: 110 oz white stripe: 110 oz pink stripe: 20 oz red and 40 oz white There are not enough constraints to find a unique solution.
*February 22, 2015*

**Math**

use what you know about arithmetic progressions. 10/2 (2a+9*2) = 200 2a+18 = 40 2a = 22 a = 11 So, the numbers are 11,13,...29 check: 5(11+29) = 200
*February 22, 2015*

**Math check**

#1 ok #2 ok #3 ok #4 ok #5 8(180) = 1440 each angle is 144, not the sum.
*February 22, 2015*

**Math**

#1 n(n+1)/2 >= 2000 n^2+n-4000 > 0 n > 62.7 So, 63 days #2 clearly the sum must be even. sum = n/2 (24+(n-1)(-2)) http://www.wolframalpha.com/input/?i=n%2F2+%2824%2B%28n-1%29%28-2%29%29+for+n%3D1+to+50 #3 1-digit #'s: 3 teens: 13 20's: 23 30's: 30-33
*February 22, 2015*

**math**

what do you mean by "solve"? As it stands, there's not much to be done except expand the polynomial. (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 so, you have (3x)^3 + 3(3x)^2(-2y) + 3(3x)(-2y)^2 + (-2y)^3 = 27x^3 - 54x^2y + 36xy^2 - 8y^3
*February 22, 2015*

**math**

if the small base has side 2x, then each face of the pyramid is a trapezoid with area a = (2x+10)/2 * √((5-x)^2 + 30^2) So, to find the minimum area, da/dx=0 Unfortunately, this never happens. Have I set it up wrong? Since the height of each face is constant, and the ...
*February 22, 2015*

**math**

In the real world, one would normally subtract a smaller volume from a larger one. So, the volume in question is v = πh(R^2-r^2) when h=30 and r=10, v = π(625-100)(30) = 15750π cm^3 since the volume of water does not change, dv/dt = 0. So, (R^2-r^2) dh/dt + h(-...
*February 22, 2015*

**Geometery**

beats me -- where is N?
*February 22, 2015*

**Math**

if the first odd integer is 2k+1, then the others are 2k+3,2k+5,2k+7 So, adding them up you get 8k+16=48 k=4. the numbers are 9,11,13,15 Your idea is also good, but to guarantee odd numbers, you need even numbers+1. In some situations you might get a solution for x but find ...
*February 22, 2015*

**AP CALC**

You have to take f(x/2), not f(x). Thus, the derivative is 1/2 e^(-(x/2)^2) = 1/2 e^(-x^2/4) So, at x=1, the slope is 1/2 e^(-1/4) = 0.3894 So, it looks like (b) is the choice. See the graphs here: http://www.wolframalpha.com/input/?i=plot+y%3D%E2%88%9A%CF%80%2F2+*+erf%28x%2F2...
*February 22, 2015*

**fractions**

the toppings overlapped.
*February 22, 2015*

**AP CALC**

Then you get 97/3
*February 21, 2015*

**AP CALC**

I also got (e). However, that is an algebraic area, which includes the sign. If you want the geometric area, you have to split into two parts, because it crosses the x-axis at x=0.
*February 21, 2015*

**calculus**

I assume you meant y(t) = 19(1+e^(-cos(t)) so, v(t) = 19sin(t) e^(-cos(t)) v(1) = 19 sin(1) e^-cos(1) = 9.31 v(4) = 19 sin(4) e^-cos(4) = -27.64
*February 21, 2015*

**math**

take a look here http://www.wolframalpha.com/input/?i=y%22%27+%2B+4y%22+%2B+4y%27+%3D+e^%28-2x%29
*February 21, 2015*

**Math**

ok up to the last line. x = -5/17
*February 21, 2015*

**math**

time to review proof by induction. Assume the sum is true for k=n. Show that if it is, then it is also true for k=n+1 1 = 1(2)/2 So, it is clear that the equation is true for k=1. Now, assume it is true for k=n. That is, 1+2+...+n = n(n+1)/2 Now, add n+1 to both sides: 1+2...
*February 21, 2015*

**math**

That is just 1+2+...+n I'm sure you have seen that before.
*February 21, 2015*

**Calculus**

or, minus all that.
*February 21, 2015*

**Calculus**

recall that the derivative of csc u = cscu cotu du
*February 21, 2015*

**Math**

find the slope of L2 The slope of L1 is -1/(slope of L2)
*February 21, 2015*

**Calculus**

x∛(x-2) dx let u^3 = x-2 x = u^3+2 3u^2 du = dx now the integrand is (u^3+2)(u)(3u^2 du) and that's just a b unch of powers.
*February 21, 2015*

**Math**

T15-T5 = 10d = 40 so, d = 4 Now you can find T1 and T60 Then the sum is 30(T1+T60)
*February 21, 2015*

**Math**

5x+3y+12=0 has slope -5/3 So, our desired line has slope 3/5 (-6-3r)/(2+r-4) = 3/5 Now find r.
*February 21, 2015*

**Trig application**

If the fire is labeled F, then in ΔABF, the angles are A = 14° B = 34° If the altitude from F to AB meets it at point P, then h/AP = sin 14° h/PB = sin 34° so, AP = h/sin14° PB = h/sin34° AP+PB=30 h/sin14° + h/sin34° = 30 h(1/sin14°+1/...
*February 21, 2015*

**Math**

you can see from #2 that x = 3y+6. Plug that into #1 and you have 2(3y+6)+5y=15 Now find y, then you can get x.
*February 21, 2015*

**Trig application**

(A) 428nm/20knot = 21.4 hr the direction does not matter. You have a distance and a speed. (B) E: 20*12 sin 1.4° S: 20*12 cos 1.4° (C) Surely the same heading (not bearing!) as the yacht, no? Ignoring water currents and winds aloft.
*February 21, 2015*

**maths**

2.63*10^11 0.978*10^11 1.789*10^11 -------------- 5.397*10^11
*February 21, 2015*

**physics**

since PE = mgh, ΔPE = mgΔh
*February 21, 2015*

**Math**

A^2/(√3/4 (A/2)^2) = 16/√3
*February 21, 2015*

**algebra**

5 tests at 82 pts each requires 410 points. So, remove the lowest score, add up the others, and see how much more he needs to get to 410.
*February 21, 2015*

**maths**

after n years, we have A: 4500*1.10^n B: 4500+525n so, plug in n=3 and see which is greater, and by how much.
*February 21, 2015*

**Math**

or, you can think of it as 5+10+...+50 = 5(1+2+...+10) = 5*55 = 275
*February 21, 2015*

**Math**

you know that y = -3x-2, so 2x+3(-3x-2) = 5 Now you can find x, and then y.
*February 21, 2015*

**maths**

whene revenue > cost, then there's a profit. So, for p pies, we need 8p > 200 + 3p
*February 21, 2015*

**physics**

each vector has two components, most commonly in the x and y directions. So, you have v = <31.8,0> + <0,-31.8> = <31.8,-31.8> |v| = 31.8√2 = 44.97 m/s in a southeast direction, since tanθ = -31.8/31.8 so, the momentum p = mv
*February 21, 2015*

**physics**

it takes 6.0/9.7 = 0.619 seconds to reach speed. so, s = 1/2 at^2 You have a, and now t.
*February 21, 2015*

**Calc**

I get 492.4 on a heading of 205°
*February 21, 2015*

**Pre ALGEBRA**

Looks like B to me. a cone and a pyramid have exactly one base. a sphere has 0 bases, which is also 1 or less. A cube has 6 faces, any of which could be called a base.
*February 21, 2015*

**math**

let the center of the earth be (0,0,0) find the coordinates of Sudbury. That will give you the tip of the vector pointing out. The negative of that is the vector pointing in. Take a look at spherical coordinates and their transformation to rectangular. You could start here: ...
*February 21, 2015*

**math**

when the water is y down from the top, the surface subtends an angle θ, so the cross-section has area 1/2 r^2(θ - sinθ) where cosθ = y/r So, plugging in our numbers, v = (400)(1/2)(20^2)(arccos(y/20)-√(20^2-y^2)) v = 800(arccos(y/20)-√(400-y^2...
*February 21, 2015*

**math**

if the square has side s and the triangle has side t (both in cm), then 4s+3t = 400 a = s^2 + √3/4 t^2 Now just express t or s in terms of the other, and you have a quadratic for the area. The vertex of the parabola will give the minimum area. The maximum will be ...
*February 21, 2015*

**maths**

since 9 = 3^2, 9^x = 3^2x So, you have 3^(x+1) = 3^(2x) x+1 = 2x x = 1 On the other one, I will guess you mean 3^2 * m^-1 = 243 m^-1 = 27 m = 1/27
*February 21, 2015*

**Maths**

2.500001 x 10^6
*February 21, 2015*

**math**

200000/2500 = 80 So, there were 120 to start with How far do you get on the rest?
*February 21, 2015*

**Math**

since the distance covered by constant acceleration a is s = 1/2 at^2, we have 1/2 a*10^2 = 500 a = 10 cm/s^2 50 m/s is the average speed during takeoff, not the acceleration.
*February 21, 2015*

**trigonometry- college**

I interpret your language to mean that BC is 20° south of west. I assume that the two-hour cruise from A actually arrived at port C, meaning AC is 100 mi. In problems like this, the first thing to do is always to draw a diagram. If you cannot do that, you have no hope of ...
*February 21, 2015*

**physics**

you have the numbers and the formula. What's the problem? All you need is the person's mass: 700/9.8 kg
*February 20, 2015*

**math**

s = (a+b+c)/2 If you found Heron's formula, why did you not read its explanation?
*February 20, 2015*

**Math/algebra**

If there are 52 bears, and one is left over after forming groups, then 51 must have divisors that are the # of groups and the # of bears in each group. 51 = 3*17, so there are either 17 groups of 3 or 3 groups of 17 After that, there is 1 bear left over, making 52 in all.
*February 20, 2015*

**grammer**

Reed is on the right track. I think it might have been better presented as Last night, my brother ________ me a secret. And you can prolly lose the comma.
*February 20, 2015*

**Calculus**

Take a look at this site. The very first problem is one just like yours, but with different numbers. http://tutorial.math.lamar.edu/Classes/CalcI/Work.aspx
*February 20, 2015*

**math**

that is just like the previous one. Just review your text section on linear DE's. There must be examples just like your problem.
*February 20, 2015*

**math**

y"' - 3y" + 2y' = e^(2x) D(D-1)(D-2)(y) = e^2x y = (1/4)(c1)e^x + (1/4)(c2)e^(2x) + (1/2)xe^(2x) + c3
*February 20, 2015*

**algebra**

v = 1/12 pi d^2 h now plug and chug...
*February 20, 2015*

**Math**

recall the dot product u•v = |u||v|cosθ 3*4 + 2*0 = √13 * 4 *cosθ cosθ = 12 / 4√13 θ = 33.7° Or, you could notice that v is just along the x-axis, so tanθ = 2/3 θ = 33.7
*February 20, 2015*

**Physics**

Just add/subtract the given acceleration to gravity. Then as usual, F=ma, where a is the net acceleration.
*February 20, 2015*

**AP calc**

Take a look here. The first problem is just like this. http://tutorial.math.lamar.edu/Classes/CalcI/Work.aspx
*February 20, 2015*

**Math**

(7+5+8)/(7+8+5+9+8+5)
*February 20, 2015*

**College Alegebra**

√(23^2-18^2)
*February 20, 2015*

**Trigonometry (Coterminal Angles)**

just keep adding 2π till you get a positive value. 2π = 6π/3, so you get -10π/3 + 6π/3 = -4π/3 + 6π/3 = 2π/3 and we have a winner!
*February 20, 2015*

**MATHS**

The three pumps work 3*8*2=48 hrs in 2 days, so one pump can empty 1/48 of the tank in an hour. 4 pumps can empty 4/48 = 1/12 of the tank in one hour. so, . . .
*February 20, 2015*

**College Algebra**

1000(1+.08)^3
*February 20, 2015*

**math**

On L1, as x increases by 3, y increases by 5*3, so (4,17) is on L1 The perpendicular line has slope -1/5, so L2 is y-17 = -1/5 (x-4) Now you can convert that to slope-intercept form.
*February 20, 2015*

**math**

the area will be 2.5 * 2.5
*February 20, 2015*

**Math**

40+.25m = 142.75
*February 20, 2015*

**ged**

google is your friend. You have not provided much information here to work with, I must say...
*February 20, 2015*

**trignometry**

Oops. I see where. Reiny did it right.
*February 20, 2015*

**trignometry**

1/(1-csc) - 1/(1+csc) = ((1+csc)-(1-csc))/(1-csc^2) = 2csc/cot^2 = 2csc tan^2 = 2/sin sin^2/cos^2 = 2sin/cos^2 = 2tan sec I seem to have lost a - sign. Hard to tell just what the original equation was.
*February 20, 2015*

**Math**

7% of the sample is white. So, what's 7% of 4150?
*February 20, 2015*

**Math**

false.
*February 20, 2015*

**maths**

for any rectangle, the area is length * width. So, get out your calculator and punch in your numbers. watch out for the units. you can't mix m and cm. For the cardboard with the border, the whole thing with border on 4 sides has area (120+10+10)^2 = 19600 cm^2 The inside ...
*February 19, 2015*

**Algebra 1**

assuming x is a function of y, note that x drops by 1/2 each time. Looks exponential
*February 19, 2015*

**Algebra 1**

If you had the following what model is most appropriate? y 3 1 0 -1 x 2 1 1/2, 0 quadratic linear exponential line
*February 19, 2015*

**MAth**

28 were boys, and each got x stickers. 28x + 14*2x = 840 x = 15 so, each girl got 30
*February 19, 2015*

**Ap calc**

the curves intersect at y=0,3 so the volume is v = ∫ 2πrh dy where r=y and h=4-(y-1)^2 - (3-y) v = ∫[0,3] 2πy(4-(y-1)^2 - (3-y)) dy = 27π/2
*February 19, 2015*

**AP calc**

each shell has volume 2πrh dx where r = x-2 h = y so, v = ∫[2,6]2π(x-2)(8x-x^2) dx
*February 19, 2015*

**AP calc**

oops. the shells have thickness dx. v = ∫[0,2] 2π (x+2)(13√x)dx = 1644√2/15 π
*February 19, 2015*

**AP calc**

the volume of a shell of radius r, height h and thickness dr is v = 2πrh dr So, add up your shells, where r = x+3 = (y/13)^2 h = y v = ∫[0,2] 2π ((y/13)^2+3) y dy = 2036π/169
*February 19, 2015*

**math**

πd*3 = 32 the area is 32*12 = 36πd clearly the cylinder's lateral area is 1/3 of the rectangle. Add on πd^2/2 to get the total area.
*February 19, 2015*

**Math**

#1 sometimes 6-2 = 4 > 0 2-6 = -4 < 0 #2 sometimes -3 - -5 = 2 > 0 -5 - -3 = -2 < 0 #3 always #4 never if by "the difference between a and b" you mean "a - b"
*February 19, 2015*

**Algebra 1**

water is 0.70 * 4*pi*(6.4*10^6)^2 for volume, multiply that by the depth.
*February 19, 2015*

**Calculus Physics**

F(t) = 5√t F=ma, so a(t) = 5/7 √t v(t) = 5/7 * (2/3) t^(3/2) + c v(0) = 0, so c=0 and v(t) = 10/21 t^(3/2) so, v(6.7) = 10/21 * 6.7^(3/2) = 8.26 m/s
*February 19, 2015*

**math**

3000(1+.09)^5
*February 19, 2015*

**Algebra 1**

the 1st x value is 1/2
*February 19, 2015*

**Algebra 1**

wait a minute. I think you have x and y switched. You cannot have two y values for a single x value (0).
*February 19, 2015*

**Algebra 1**

since it decreased, then came back to the last value, that indicates that it has started to rise again. Looks like a parabola (quadratic) to me.
*February 19, 2015*

**Algebra 1**

If you had the following what model is most appropriate? y 3 1 0 -1 x 2 1 0, 0 quadratic linear exponential line
*February 19, 2015*

**Math**

10*5x + 25*7x = 1575 x = 7 . . .
*February 19, 2015*