Wednesday

April 1, 2015

April 1, 2015

Total # Posts: 30,251

**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*

**Physics**

the height in meters is h(t) = (Vo sinθ)t - 4.9t^2 Now just find t when h=0. You sure overload the symbol 0.
*February 23, 2015*

**Geometry**

where is R?
*February 23, 2015*

**algebra**

4/(x+4) + 4/5 = 5/2 4/(x+4) = 5/2 - 4/5 4/(x+4) = 17/10 4 = (17/10)(x+4) 4(10/17) = x+4 40/17 - 68/17 = x -28/17 = x
*February 23, 2015*

**Math**

looks like it. Are any of the numbers used more than once?
*February 23, 2015*

**Math**

no. 2*0^2 = 0 2*1^2 = 2*1 = 2 2*2^2 = 2*4 = 8 2*3^2 = 2*9 = 18 what you have is (2x)^2, not 2x^2
*February 23, 2015*

**geometry**

or, equivalently, if the angle with the ground is greater than the angle with the wall. Because then the side opposite the greater angle is longer.
*February 23, 2015*

**geometry**

if the angle the ladder makes with the ground is greater than 45 degrees.
*February 23, 2015*

**Math**

There are 9 values, so the median is the 5th value, or 4. No idea when you say that there is no data in 5. If these numbers are like bins containing data, then you have to count the values in each bin, and then find the middle value.
*February 23, 2015*

**Vectors**

a+(b•c) is undefined. You cannot add a vector and a scalar.
*February 23, 2015*

**[Calculus] U-substitution for Integrals**

u = 3x-5, so du = 3 dx, not just 3 ∫[1,5] (3x-5)^5 dx = 55552 Now, subbing in for u, the limits of integration are not 0-12. Rather, we have 1/3 ∫[-2,10] u^5 du = 55552 Where did you get 165888?
*February 23, 2015*

**Math**

count the vehicles: b+s = 21 count the wheels: 2b+4s = 54 assuming scooters have 4 wheels. If they also have 2 wheels, then the problem is bogus.
*February 23, 2015*

**Calculus 2**

at height y from the bottom, the area of the surface of the water is (5/4)y*10 ft^2 So, the weight of a thin layer of water at that height is (5/4)y*10*dy *62.4 = 780y dy lbs So, the work to lift all the sheets of water at height y is ∫[1,3] 780y(4-y) dy = 5720 ft-lbs
*February 23, 2015*

**Math**

nope C is a right angle obtuse means greater than 90. So, (b) D is a straight angle.
*February 23, 2015*

**Algebra**

Take a look here: http://www.wolframalpha.com/input/?i=solve+2x-3y%3C6%2C+4x%2B4y%3C8+
*February 23, 2015*

**Algebra**

g(f) = 1/4 f - 3 = (3x+4)/4 - 3 = 3/4 x - 2 f(g(-2)) = f(-7/2) = -13/2 f(g) = 3g+4 = 3(x/4-3)+4 = 3/4 x - 5 f(g)=5 means 3/4 x - 5 = 5 3/4 x = 10 x = 40/3
*February 23, 2015*

**Math**

since the new prism has dimensions 1/2 the old one, its volume is 1/8 the old volume.
*February 23, 2015*

**Math**

the slope is (-2-5)/(9-5) = -7/4 y-5 = -7/4 (x-5) so, f(x) = -7/4 (x-5) + 5
*February 23, 2015*

**Math**

Not quite sure what's what here, but I'll take a stab at it. h(-6) = -1 so, if 3-x = -6: x = 9 j(9) = h(-6)+4 = 3 A' = (9,3)
*February 23, 2015*

**Math**

see related questions below
*February 23, 2015*

**algebra**

you want to solve for h where π/3 * 25^2 * 7.2 = π * 15^2 * h
*February 23, 2015*

**algebra**

just add the exponents and multiply the numbers. What do you get?
*February 23, 2015*

**alleged math**

I looked it up in the answer key!!
*February 23, 2015*

**math**

∛(250z^4x^7) = ∛(125z^3x^6 * 2zx) = 5zx^2 ∛(2zx)
*February 23, 2015*

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