Friday

February 27, 2015

February 27, 2015

Total # Posts: 29,274

**3rd grade math**

I know. Maybe one of the band members is not a musician, eh? Probably the drummer. :-)
*February 26, 2015*

**3rd grade math**

since 3 does not divide 16, how can 2/3 of the 16 players be girls? 12/16 are brass
*February 26, 2015*

**algebra**

1695/(x + x+65) = 3
*February 26, 2015*

**Math**

v = 1/3 πr^2h 3v = πr^2 h h = 3v/(πr^2)
*February 26, 2015*

**math**

61.4% is the amount paid. That is not the discount (which is the amount not paid)
*February 26, 2015*

**Geometry**

90° is 1/4 of a circle. So, what's 1/4 of the circumference or area? c = πd, so d = c/π c = 2πr, so r = c/2π a = πr^2 = π*(c/2π)^2 = c^2/4π Now just plug in c where you need it.
*February 26, 2015*

**math**

not quite If they start out with o and c dollars, then o-40 = c+40 4(c-80) = o+80 rearrange things a bit and you have o-c = 80 o-4c = -400 3c = 480 c = 160 now find o
*February 26, 2015*

**Algebra II**

a parabola with focus at (0,p) and directrix at y = -p is x^2 = 4py The vertex is midway between the focus and directrix, so your parabola has p = 1 x^2 = 4y Now shift things so the vertex is at (-2,5) and you have (x+2)^2 = 4(y-5)
*February 26, 2015*

**math**

since Mickey had only gone 5/6 of the way, Dylan's speed was 6/5 of his. The last 1/6 (40 mi) of the trip took Mickey 48 minutes. That should get you started.
*February 26, 2015*

**Math**

you want to find factors of 81. They are 1,81 3,27 9,9 Each choice gives a different perimeter.
*February 26, 2015*

**math**

210/35 + 54/18
*February 26, 2015*

**Quick chemistry help**

each molecule is Cl2.
*February 26, 2015*

**math**

(5000-600)/4 grams per book
*February 26, 2015*

**math proportions**

since the ratio of height:shadow is constant, 5'1"/3'4" = h/12'10" That's kind of messy, so convert all the values to just inches. Find h, then convert it back to feet&inches.
*February 26, 2015*

**Math**

(a) is clearly explained at http://tinypic.com/view.php?pic=15qynhv&s=4#.VO_LNywYFvA You can probably find the others as well. Usually the Binomial Theorem is involved. For instance, (b) f(x+h) - f(x) = (x+h)^4 - x^4 = x^4+4x^3h+6x^2h^2+4xh^3+h^4 - x^4 = 4x^3h+6x^2h^2+4xh^3+h^...
*February 26, 2015*

**Math vectors**

recall that one factor in the result is the sine of the angle between the vectors.
*February 26, 2015*

**Math/Solve**

correct
*February 26, 2015*

**algebra**

x+y = 70 y = 2x+10 Now go for it.
*February 26, 2015*

**Math**

If the costs are x,y,z, respectively, then we have 2x+2y+z = 10.20 x = 2y z = x+3 You want to find z.
*February 26, 2015*

**linear algebra**

(a) Nope if f(t) >= 0 but c < 0, cf(t) is not a nonnegative function Looks to me like (b) and (c) are both subspaces. Do we have to choose only one?
*February 26, 2015*

**algebra**

3x means he withdrew x dollars each week. You want x to be the starting amount, so x - 3*14 > 65
*February 26, 2015*

**Calculus**

d/dx e^u = e^u du/dx
*February 26, 2015*

**Calculus**

well, y' = 4x^3 + 8x - 1 So, at (1,4), the slope of the tangent is 11. That means the line there is y-4 = 11(x-1)
*February 26, 2015*

**Calculus**

when a is a constant, g'(au) = a g' You know that if y=2x, y'=2(1) y=3x^7, y' = 3(7x^6) The constant is just a multiplier, and carries through. So, what you have is g = √2 u + √7√u
*February 26, 2015*

**math**

you need more than just the scale.
*February 26, 2015*

**Math**

the time in the air is t when h=0 again. So, just solve for h(t) = 0. the max height is the vertex of the parabola, clearly when t=4.
*February 26, 2015*

**math**

y = x/3 + 7 now just plug in x=18 and evaluate for y.
*February 26, 2015*

**Math**

You found the union, not the intersection. 3.3: You found A-B, not the union. 4.4: Again, you found the union. The intersection is the elements in all three sets: {1,5}
*February 26, 2015*

**math**

one recall that the roots are -b/2a ±√(discriminant) so, if the discriminant is zero, both roots are -b/2a
*February 26, 2015*

**math**

the lateral surface area is just the circumference times the height. So, a = 4*15 = 60 if you have to include the ends, then that is additional 2πr^2, where r = 4/2π
*February 26, 2015*

**MATH**

graphing the other methods are exact.
*February 26, 2015*

**Math**

(2/5)(1/5) + (3/5)(1/3)
*February 26, 2015*

**math**

there are several similar problems in the related questions below.
*February 26, 2015*

**Math!! Help ASAP**

11.4 is correct.
*February 26, 2015*

**maths**

3 5/9 = 32/9 40 1/2 = 81/2 So, the common ratio, r^6 = (32/9)/(81/2) = 64/729 r = 2/3 T5 = 32/9 * (2/3)^4
*February 26, 2015*

**maths**

ar = a+4 ar^2 = ar+9 Now just solve for a and r
*February 26, 2015*

**calculus**

no ideas on any of the questions? They gave you the function, so things should just fall right out.
*February 26, 2015*

**math**

y = (x^4+4)^2 (x^3+4)^4 ln y = 2ln(x^4+4) + 4ln(x^3+4) 1/y y' = 8x^3/(x^4+4) + 12x^2/(x^3+4) y' = [8x^3/(x^4+4) + 12x^2/(x^3+4)]*[(x^4+4)^2 (x^3+4)^4] Now you can massage that as you will. Eventually you can arrive at y' = 4x^2 (x^3+4)^3 (x^4+4) (5x^2+8x+12)
*February 26, 2015*

**Math**

By "side" I assume you mean one of the bases. If the other base is x, then we have (6+x)/2 * 13 = 91 (6+x)/2 = 7 6+x = 14 x = 8
*February 26, 2015*

**programming**

as usual, the average is the sum of all the ages, divided by the number of pupils. Count the pupils as they come in: num_pupils = sum_of_ages = 0 loop { read in years, months num_pupils++ age_in_months = years*12 + months sum_of_ages += age_in_months } average_age = ...
*February 26, 2015*

**math**

see related questions below
*February 26, 2015*

**maths**

the values are 1/√3, 1, √3, ... I think you can see what the constant ratio is.
*February 26, 2015*

**arithmetic**

the sides of such a hexagon are the same as the radius of the circle.
*February 26, 2015*

**Math**

since 3x^2+11x-20 = (3x-4)(x+5) you have log(3x-4)+log(x+5)-log(3x-4) = log(x+5) log3(27) = 3 log(18) = log(9) + log(2) So, you have 1/6 (3) + 2 + log3(2) - log3(2) = 5/2 625 = 5^4, so you have log5(5^(4/3)) = 4/3 surely you meant loga w. If so, you have log_a(x^6 y^3 z^4 w^3)
*February 26, 2015*

**Calculus**

That's what I get. For shells, ∫[2e^-6,2] 2π(4-y)(6+log(y/2)) dy comes out the same. Amazing!
*February 26, 2015*

**Calculus**

2e^-x has a y-intercept of 2, so you have a sort of triangular region with vertices at (0,2),(6,2),(6,2e^-6) using washers, each washer has an area π(R^2-r^2) where r = 4-2 = 2 and R = 4-y = 4-2e^-x Now the integral is just π∫[0,6] (4-2e^-x)^2-2^2) dx You can ...
*February 26, 2015*

**Math Help!**

each face of the pyramid is an isosceles triangle with base 8 and height √(22^2+4^2) To see this, view the pyramid from the side. Drop an altitude to the center of the base, and you have a right triangle whose hypotenuse is the height of the triangular face.
*February 26, 2015*

**math**

h = p-13 p = 29 so, h = 29-13 = 16 Alan is correct.
*February 26, 2015*

**math**

f = 2*5
*February 25, 2015*

**Calc**

#1 still does not contain the point (1,1) #2 I don't think there is such an A and B. There is a solution involving sin(√2 x), but that's not gonna fit the requirements. #3 y = e^x - e^2x y' = e^x - 2e^2x = e^x(1-2e^x) so, y'=0 when e^x=0 (never) 1-2e^x = ...
*February 25, 2015*

**Algebra 2**

(15^3-7^3)(15^3+7^3) (15-7)(225+105+49)(15+7)(225-105+49) (8)(379)(22)(169) (2^3)(379)(2 11)(13^2) 2^4 11 13^2 379
*February 25, 2015*

**science**

draw the batteries on the left, one above the other next over, draw two resistors, in line on the same wire. Connect the ends of the wire to the top and bottom of the batteries. Next over, draw two light bulbs beside each other. Connect the two top ends together, and the two ...
*February 25, 2015*

**math**

sure. You can label the axes any way you want. This is often useful when y varies by large amounts.
*February 25, 2015*

**Calculus Chain rule**

several ways expand and divide: 6/x * (x^2-1)^4 = 6/x (x^8 - 6x^6 + 4x^4 - 6x^2 + 1) 6(x^7 - 6x^5 + 4x^3 - 6x + 1/x) now take derivative: 6(7x^6 - 30x^4 + 12x^2 - 6 - 1/x^2) quotient rule: 6(4)(x^2-1)^3(2x)(x) - 6(x^2-1)^4(1) ----------------------------------- x^2 product ...
*February 25, 2015*

**maths**

no way. To plot a point, you have to know the coordinates, or have a way to calculate them.
*February 25, 2015*

**intermeidatealgebra**

you sure this isn't a trick question? Look at it carefully.
*February 25, 2015*

**intermeidatealgebra**

x+11/15
*February 25, 2015*

**Math**

it is C. Sorry, Tracy, not A.
*February 25, 2015*

**Math**

two complementary angles add up to 90 degrees. two supplementary angles add up to 180 degrees. So, what do you have to add to 47 to get those values?
*February 25, 2015*

**Geometry**

I like E, since you don't provide any clues as to the constraints on x and y.
*February 25, 2015*

**Math Alg. 2**

Well, let's see. #1: 1*1 2*1 4*3 8*7 16*17 Hmmm 2^1-1 2^2-2 2^4-4 2^6-8 2^8+16 Hmmm #2: 2 2*2+1 2*5+1 2*11+5 2*26+7 Hmmm 2^2-2 2^3-3 2^4-5 2^5-6 2^6-5 Hmmm. #3: clearly -3,-2,5 repeats as a sequence Sorry. I can't seem to get #1 and #2 right off. They almost follow a ...
*February 25, 2015*

**Math Algebra**

f(x0) = 2 f(x1) = f(2) = 2 f(x2) = f(2) must stay 2.
*February 25, 2015*

**Math**

look at the units 48 oz = 3 dollars 16 oz = 1 dollar 1 oz = 1/16 dollar so, 6.25 cents/oz you have 16oz/dollar, not 16cents/oz
*February 25, 2015*

**Math- Calc 1**

(1,1) is not on that curve. the other questions also have typos.
*February 24, 2015*

**Calculus**

d/du(tan u) = sec^2 u so, use that and apply the chain rule: sec^2(sec x) * secx tanx I'm sure you can do #2 y = sec^-1 x sec y = x secy tany y' = 1 y' = 1/(secy tany) y' = 1/(x √(x^2-1)) y = 2^(e^(sinx)) y' = ln2 2^(e^(sinx)) * e^(sinx) * cosx
*February 24, 2015*

**Homework Help (Math)**

Write your answer in <x,y> for the following question: If x=<6,-21>, then 1/3x.
*February 24, 2015*

**Pre-Calculus(Trigonometry)**

looks good to me
*February 24, 2015*

**math**

10x + 25x = 280
*February 24, 2015*

**math**

d/c = 2/7 c+15 = 2(d+15) now solve for c and d.
*February 24, 2015*

**Math**

6+9r-4(3r-7) 6+9r-12r+28 . . .
*February 24, 2015*

**math**

1/8 * 360 = ?
*February 24, 2015*

**honors chemistry**

the short answer is that the heavier molecules will require fewer to make up the 10 grams of mass.
*February 24, 2015*

**math**

lots of z table stuff here: http://davidmlane.com/hyperstat/z_table.html
*February 24, 2015*

**Motion in 2d**

well, y(t) = 39.11t - 4.9t^2 find t when y=0. Then multiply that by Vx. That's how far it went horizontally while going up and back down.
*February 24, 2015*

**algebraic connections**

if yu mean 12% per hour, then there will be 100*1.12^10
*February 24, 2015*

**Algebra 2**

just plug and chug. 4/(1-r) = 10 2/5 = 1-r r = 3/5 So, now you know the G.P.
*February 24, 2015*

**Math/ Compouded semiannually ?'s**

just refer to your compound interest formula. #1: solve for P where P(1+.10/2)^(2*1) = 30000 P = 27210.88 #2: same thing P(1+.12/2)^(2*2) = .25*175000
*February 24, 2015*

**math**

convert the two vectors p and w (fir plane and wind) to rectangular coordinates, and add them up. p+w is the resultant velocity on the ground. A vector w in the direction N63W, with magnitude w is <-0.454w,0.891w>.
*February 24, 2015*

**math**

Draw a diagram. Let the barge be in the canal, with Kim on the south shore, and Kent on the north shore, both pulling eastward. Then we have Kim: u = <50cos20,-50sin20> u = <46.98,-17.10> Kent: v = <kcos15,ksin15> = <0.966k,0.259k> You want the ...
*February 24, 2015*

**Calculus**

I suspect you meant p(x) = 10e^(-.1x) So, the mass will be ∫[0,a] 10e^(-0.1x) dx = -100e^(-.1x) [0,a] = 100 - 100e^(-a/10) = 100(1-e^(-a/10)) If that is 30, then a = 10 ln(10/7) So, the density decreases with increasing distance, meaning more of the mass is closer to x=0...
*February 24, 2015*

**math**

using the two-point form for line 1, it is y = 1/6 (x-1) + 1 Now do that for Line 2, and solve the two equations.
*February 24, 2015*

**ALGEBRA**

Still wrong.
*February 24, 2015*

**ALGEBRA**

#1: nope. That is f(2) -3(-2) = +6 #2 looks ok.
*February 24, 2015*

**math**

length is two sides of the rectangle, plus the circumference of the circle. area is the rectangle plus a whole circle.
*February 24, 2015*

**business math**

79.44/.8
*February 24, 2015*

**math**

so, just count up all of each type, and multiply by their weights.
*February 24, 2015*

**math**

13(h+3h) . . .
*February 24, 2015*

**math - oops**

make that +10.5
*February 24, 2015*

**math**

0.5(x^4 - 3)+12 0.5 x^4 - 1.5 + 12 0.5 x^4 - 10.5
*February 24, 2015*

**Math**

because log(0) is undefined, regardless of the base. There is no power of 3 or 5 which equals zero. Take a look at the graph of log x.
*February 24, 2015*

**math**

if there is $x at 4%, then the rest (15000-x) is at 5%. Now just add up the interest and solve for x: .04x + .05(15000-x) = 690
*February 24, 2015*

**probability**

for both red, we need 2/(n-2) * 1/(n-1) = 1/28 n = 8 Now you can do the other part.
*February 24, 2015*

**math**

1.12j = 8400 j = 7500 j/a = 5/3 a = 3/5 j = 4500 (14100-8400)/4500 = 1.267 so, Akinyi's money grew by 26.7%
*February 24, 2015*

**math**

2θ = 60,330,420,690,... θ = 30,115,210,345,...
*February 24, 2015*

**math**

r = 23.4/1.8 = 13 a = 1/2 * 13^2 * 1.8
*February 24, 2015*

**math**

∛(4/3 π) * 4.2 = 6.77
*February 24, 2015*

**Vectors Math**

so, have you picked out three vectors?
*February 23, 2015*

**MATH**

x y 1 4 5 1 8 6 10 5 not linear, since y changes direction not quadratic, since y changes direction twice. So, guessing cubic, I find -109/1260 x^3 + 1961/1260 x^2 - 2333x + 3130 I doubt that is what you had in mind, but anything else would be even more complicated.
*February 23, 2015*

**Math**

age now is x. x = 3(x+3)-3(x-3)
*February 23, 2015*

**pre-cal**

sounds like a plan.
*February 23, 2015*

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