Friday

April 18, 2014

April 18, 2014

Total # Posts: 21,739

**math**

15(3 1/5) = 15*3 + 15*(1/5) = 45 + 3 = 48

**math**

If I pay, I don't get paid nothin'.

**Math**

y = a/c because as x gets huge, only x^2 matters. y = (a + b/x^2)/(c + d/x^2) as x gets huge, the fractions go to zero.

**geometry**

10^6 * 4/3 pi (.005^3) * 2 = pi/3 = 1.047 m^3

**Math**

#1 - forget about the blue pencils. There are 8 yellows in each package. So, in 426 packages, there are 8*426=3408. Note that the title of the problem set was "Multiply a 3- digit number by a 1- digit number". That should have been a clue that your approach was wrong...

**math**

well, yes, as I said. So, just work it out: 600 mi * 5280ft/mi * 12in/ft = 38,016,000 in Note how the units cancel, leaving you with just inches.

**math**

you know the circumference of the tire is 24pi or 75.4 inches. So, figure out how many inches there are in 600 miles, and divide that by 75.4 That's how many times the tire went around.

**Algebra**

you can get excellent graphing help here: http://rechneronline.de/function-graphs/ but you need to enter functions of x. So, solving for y, you have y = (5x+36)/3 y = (26-3x)/5 Enter those two functions and set the range on the y-axis to -10:10 and you will see the solution.

**Calculus**

Good idea. However, in an attempt to use your math, and also to apply to a possibly more general problem later, try implicit differentiation: x^2 + y^2 = 18 2x + 2yy' = 0 y' = -x/y So, if -x/y = 1 and x^2+y^2 = 18, 2x^2 = 18 x = ±3 Now, figuring y should not be ...

**Calculus: Integral**

recall that sec^2 = 1+tan^2, so you have ∫sec^4(4x) dx = ∫sec^2(4x)(1 + tan^2(4x)) dx = ∫sec^2(4x) dx + ∫tan^2(4x) sec^2(4x) dx = 1/4 tan(4x) + (1/4)(1/3) tan^3(4x) and you can massage that in several ways.

**math**

Glad to help. Translating words into symbols can be tricky, but the key is just to examine the data carefully. Believe it or not, it gets easier with practice, just like everything else.

**math**

2w + b = 36 2w + 2b = 56 well, since she walks 2 hours in each case, it is clear that the extra hour of biking gave her an extra 20 km. b = 20 so, since 2w+20=36, w=8

**algebra**

the amount of alcohol in the parts must add up to the whole. So, .3 * (8-x) + 1.0 * x = .5*8 x = 16/7

**Calculus Help Please!!!**

looks good to me

**Calculus Help Please!!!**

v = 4/3 pi r^3 dv = 4 pi r^2 dr c = 2pi r dc = 2pi dr so, dr = dc/(2pi) meaning that dv = 4 pi r^2 dc/(2pi) = 2 r^2 dr so, using the given numbers, dv = 2*(80/2pi)^2 * 0.5 = 1600/pi^2

**Calculus Help**

v = 2/3 pi r^3 (half a sphere) dv = 2 pi r^2 dr now just plug in the given r and dr watch the units.

**physics**

F = GmM/r^2 2.5*10^-10 = 6.674*10^-11 * x(4-x) / .25^2 x = 0.059 4-x = 3.941 Must be a bowling ball and a ping pong ball

**Algebra**

5m+3v = 29 2m+12v = 71 or 20m+12v = 116 2m+12v = 71 18m = 45 m = 2.50 v = 5.50

**math**

9x/4 = 1/3 x = 4/27

**Calculus Help Please!!!**

looking at a diagram, if A is a away from Q and B is b away from Q, then √(a^2+144) + √(b^2+144) = 39 a/√(a^2+144) da/dt + b/√(b^2+144) db/dt = 0 Now just plug in da/dt = 3.5 a = 5 b = 23.065 (from 1st equation when a=5) and solve for db/dt

**Calculus Help Please!!!**

when the water has depth x, the cross-section is a trapezoid with bases 30 and 30+x. So the volume of water at depth x is v = (60+x)/2 * x * 500 cm^3 = 250x^2 + 15000x so, knowing that dv/dt = (500x + 15000) dx/dt just solve that for dx/dt when x=20

**algebra 2**

I think you meant -17 instead of -7. If that's true, then the difference between terms is 15. So, the 10th term is -62 + 9*15 = ?

**math**

perimeter is 2(12+16) = 56 side of square is thus 14. So, the area is ? Note that 12*16 = (14-2)(14+2)

**math**

After 9 cycles he has made 18 miles On day 10 he swims the last 7 miles.

**math**

√(1^2+3^2) = √10 = ?

**engineering**

986.8*10^-9 * (1 - .5^(50/3.69*10^3)) = 9.225*10^-9 kg

**THE PARTS OF SPEECH**

A is present progressive B is present perfect C is present D is present tense, but passive voice

**Math**

just subtract the smaller volume from the larger: 10*10*15 - 7*9*12 = ?

**math**

You're kidding, right? 11 capable graduates in the whole state? Anyway, just work it out: P(1) = .56 P(2) = .56*.56 = .56^2 ... P(11) = ?

**Math - incomplete**

depends on how many she already has made. If she has made n shots so far, then we need x such that .70n + 1.00x = .75(n+x)

**math**

Good logic, but I wonder about the .47 in light of P(college) = .56

**Trig**

y=0 is not the y-axis. And anyway, in the stated domain, x=pi is the axis of symmetry.

**Trig**

see the graphs here: http://www.wolframalpha.com/input/?i=plot+y%3Dcos2x+and+y%3D-2sinx However, it might be better to graph y = -cos2x, then the final graph would be easier to visualize: http://www.wolframalpha.com/input/?i=plot+y%3D+-cos2x+and+y%3D-2sinx%2C+y+%3D+-cos2x-2sin...

**Trig**

sinC/c = sinA/a

**Trig**

you have two solutions for tanA, one positive, one negative. So, there are two values for A, one in each QI, one in QII.

**Trig**

period = 1/freqeuncy

**math**

assuming y is in billions, c(y) = 200 + .6y c(8250) = 200 + .6(8250) = 5150 or, 5.2 trillion looks ok to me. (b) subtract their spending from their income.

**math**

oops a+3x = 50

**math**

If adam is a now, then a - 1/2 = 12 a = 12 1/2 a + x/2 = 2x 12 1/2 + x/2 = 2x 25 + x = 4x x = 25/3 check: 25/6 + 25/2 = 25(2/3) = 2 * 25/3 25/2 + 75/3 = 100

**math**

heck no. a pipe of half the diameter only has 1/4 the area. So, for a given speed of the water, only half the volume is obtained.

**To Steve**

You could have added a posting to the original question. Anyway, There are 3 red marbles out of 9 total, so on the first draw, P(red) = 3/9 = 1/3 Without replacement, there are now 2 of 8 reds, so P(red) = 2/8 = 1/4 P(red,red) = 1/3 * 1/4 = 1/12 With replacement, both draws ar...

**math - incomplete**

and the other method is ... ?

**Calculus**

when dy/du = u^½ what does y = ? Just a simple power rule substitution. dy/du = u^n y = u^(n+1) / (n+1) + C

**algebra**

I can't make out the inequality sign, but that really makes no difference to the solution y = (x-4)(x+2)/(x-3)(x+5) You know there are roots at -2 and 4, and asymptotes at -5 and 3. So, the graph will cross the x-axis at -2 and 4. When x < -5, all factors are negative, ...

**algebra**

ok. sounds good to me.

**algebra**

42

**Stats**

Well, it must be immediately obvious that .9715 is bogus. A score of 10 is more than a whole SD below the mean, so there is no way that P(x<10) > 1/2 When I plugged in the data, I got P(x<10) = 0.1357 Still not one of your choices. However, the closest is 0.3, since y...

**Physics**

Oops. I just gave the mass of the seesaw on each side. That must be multiplied by the distance of its center of mass: 20(d/4.5)(d/2)+14*1.4 = 20(1 - d/4.5)((4.5-d)/2) + 39L

**Physics**

so, the mass of the seesaw can be ignored, since it is uniformly distributed. Now for the loads and their moments. They must balance on both sides, so 14*1.4 = 39*L Simple, no? Actually, since I have no diagram, I can't say that the seesaw mass may be ignored. If the pivot...

**math**

MTθF: 6/5 hr Sat: 4/5 hr W 1/5 hr 4 * 6/5 + 4/5 + 1/5 = 29/5 = 5 4/5 hr

**math help**

#2 is ok, but not yet complete: x^2-1 = (x-1)(x+1)

**Math 20-3**

v = 1/3 pi (3.4^2) (2.8) m^3

**Algebra**

If Literature, history and French are your life loves, then algebra probably will not be a great part of your life. But, life consists of a wide variety of experiences, and I guarantee you that some basic math know-how (yes, even including algebra) will make things easier. So,...

**Algebra**

Only you will know that. It's actually surprising how often basic math skills come in handy. Other than that, the mental skills developed when solving problems are definitely useful later in life, no matter the field of endeavor. Now, if you plan to find a sugar daddy/mama...

**Algebra**

well, 2(v-6) = 2v-12 Guess that's not it, eh? How about (v+4)(v-3) You need to find two numbers of opposite sign, which multiply to -12 and add up to +1. On the last one, (m^2 + 2m + 1/ m^3 + 3m^2 + 3m +1 (m^2/m - 3m/3) unless you can factor the polynomials, there's no...

**Algebra**

Just as in the previous two posts, you need to factor the trinomials and then cancel factors which occur in the top and bottom. Have you no ideas at all on factoring these? Or multiplying by the GCD to clear fractions?

**Algebra**

the GCD is x(x+3)(x+1), so if you clear fractions you get 3(x+3)(x+1) + 6(x)(x+1) = 8(x)(x+3) 3x^2+12x+9 + 6x^2+6x = 8x^2+24x x^2 - 6x + 9 = 0 (x-3)(x-3) = 0 x = 3

**Algebra**

r^2 + 6r +8 = (r+2)(r+4) So, what you have is (r^2-7r+12) / (r+4) * (r+2)(r+4) / (r^2-7r+12) Now lots of stuff cancels out and you are left with just (r+2)

**Statistics**

you can do lots of Z table stuff here: http://davidmlane.com/hyperstat/z_table.html

**Algebra**

Without actually checking, I'd have to say it better be yes. (Ohm's Law). Having said that, since amps*ohms is constant, 4y = .1*124 y = 3.1

**algebra**

The other root must be 4-i, so one factor is (x-(4+i))(x-(4-i)) = ((x-4)^2 + 1^2) = x^2-8x+17 So, now we have (x+3)(5x-1)(x^2-8x+17) at x=0, this is (3)(-1)(17) = -51, so y = -5/51 (x+3)(5x-1)(x^2-8x+17) = -5/51 (5x^4 - 26x^3 - 30x^2 + 262x - 51)

**Maths**

Well, if the sides are 5,4,4,4,3,3,3 you have the perimeter of 26 cm. So, you can build a nearly regular heptagon and just squash it till it has area 24 cm^2. Or, a trapezoid on the side of a rectangle might also do the trick.

**Need 8 grade math help quick please!**

there is only one number less than 2, and six possible throws. So, P(n<2) = 1/6 Similarly, P(n>=3) = P(n∊{3,4,5,6}) = 4/6 = 2/3 P(n≠4) = 1-P(n=4) = 1 - 1/6 = 5/6

**Calculus**

As a first check, I went to http://www.wolframalpha.com/input/?i=2%E2%88%AB[3%2C4]+%282%2F3+%E2%88%9A%28x^2-9%29%29+dx and saw that they show the area as 2.28 So, I suspect there is an error in the problem or the answer. Your calculation appears to be correct, within roundoff ...

**Algebra word problem**

C(x) is just a parabola, with minimum value at x = 1.2/0.2 = 6 so, 600 bikes will minimize the average cost per bike

**math**

we need <x,y> such that <x,y> + <6/√2,6/√2> = <0,-20> Now it's easy to get x and y, and thence the speed and direction.

**Math**

6 bags uses 5 pounds 12 bags uses 10 pounds, or 5 large bags.

**3rd grade math**

if there were x pieces to start, x/5 = 2 x = 10

**Mathematics**

T1 T2 = T1+1 T3 = T2+2 = T1+3 ... Tn = T1 + n(n+1)/2

**math**

change per year is (8750-23500)/8 = -1843.75 so, what's -1843.75/23500?

**Math**

a = 2pi*r(r+h) = 2pi*6*26 = 979.68 Hmmm. That includes both ends. The lateral area is a = 2pi*r*h = 2pi*6*20 = 753.6 Is there a typo somewhere? 2pi*6*24 = 904.3, so (D)

**MATH HELP**

If both lines go through (0,0), then one line has slope -2/3, so the other line has slope 3/2. so, s = 2/3

**math**

j=g j-2055 = g/4 = j/4 3/4 j = 2055 j = 2740

**geometry**

sinG/g = sinJ/j sinJ = sin(125)/63*36

**geometry**

B If two sides are a and b, the third side c must obey a-b < c < a+b because any side must be less than the sum of the other two.

**math**

recall that x^2-y^2 = (x-y)(x+y) so, substitute in x = a-b and y=c.

**math**

y = 3x-1 is probably the simplest, but it could also be y = 2x+7 y = 23 y = (1+∛x)^3 - 4 or lots of other possibilities. There are infinitely many relations which will take 8 to 23.

**Geometry**

given the three sides, with c being the largest, you know that the triangle is a right triangle if c^2 = a^2+b^2 if c^2 < a^2+b^2, then it's acute, and obtuse otherwise. So, start checking.

**Math**

(5+4-1)*3 = 24 (4-3+1)*12 = 24 (9-7)*(10+2) = 24

**Math**

4*4

**Math**

nope. You had to add 37x to both sides, so it should be y = 37x + 82

**senior problem solving**

x days from today, the number of ants n, is n = 350 * 2^(x/20) Now just plug in your values for n.

**Calculus**

y" = 5cos(t) so, when is that zero?

**Calculus**

-9.8 m/s^2

**Stop the Madness!**

I think part of the problem is that when a new question is posted, there is no list of related questions available to check. That only appears later when people open up the posting. And the search engine does not appear to be very good at providing relevant matches. And, most ...

**Calculus**

same as y": -cosx

**trig**

I meant 336.7°

**trig**

Assuming the normal 0° at North, and ignoring the misuse of the word "bearing", I'd say that the resultant heading of the plane is 246.7°

**Geometry**

not sure about your notation, but such a reflection takes (x,y) -> (-x,y)

**Geometry**

B Just add the components for X and Y

**Calculus**

The hands of a clock in some tower are 4.5m and 2m in length. How fast is the distance between the tips of the hands changing at 9:00 at time t hours after 12:00, (at t=0) the minute hand is at 4.5 sin(2pi*t) the hour hand is at 2.0sin(2pi*t/12) at 9:00, the distance d is d^2 ...

**Calculus**

Ahh. I see that I was interpreting 243^3/5 as (243^3)/5

**Calculus**

No one is bothered by the fact that 5 does not divide powers of 2 and 3?

**Calculus**

good except for this step: ln (243^3/5 *32^4/5) should be 1/5 ln (243^3 * 32^4)

**math**

If the balloon is between the stations h/x = tan45° h/(50-x) = tan60° h/tan45° = 50 - h/tan60° h = 50 - h/√3 h = 50/(1+1/√3) = 31.7 There is another answer if both stations are on the same side of the balloon.

**pre calculus**

C(x) = 2.00 for 0 < x <= 1 2.00 + .20(10x) for 1 < x < 2 since there are 10 charging units per mile.

**physics**

First you need to figure how much helium is the balloons. Each balloon's lift is reduced by the weight of the helium inside. Once you know how much each balloon weighs, subtract that from the lifting power of the displaced air, which is 1.225kg * .01716 * 9.8N/kg, since ea...

**Math!!!!**

this a line with slope -1/2 which goes through (0,-3) I think that should help some, eh?

**Calculus**

Since velocity is the derivative of position, v(t) = -32t + 160 now just solve -32t+160 = 32 -32t+128 = 0 t = 4

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