Sunday

March 1, 2015

March 1, 2015

Total # Posts: 29,297

**science**

I think you can find the formula you want here http://www.livephysics.com/physical-constants/mechanics-pc/moment-inertia-uniform-objects/ so you can check your derivation.
*January 2, 2015*

**algebra**

general advice: study your text and google online for problems that trouble you. do you have a particular question? Any general explanation we could type here will look very much like your textbook.
*January 2, 2015*

**Physics**

As you know, the trajectory of a body kicked with velocity v at angle θ is y = x tanθ - g/(2 (v cosθ)^2) x^2 For your ball, that is y(x) = 1.11x - 9.81/(2(20*0.67)^2) x^2 = 1.11x - 0.0273x^2 So, now check to see whether y(33) >= 3.05
*January 2, 2015*

**^Maths^**

6/3(4-2) 2(4-2) 2(2) 4 6/(3(4-2)) 6/(3*2) 6/6 1 Most people are sloppy online with parentheses. You will have to decide what you really meant to say.
*January 2, 2015*

**math**

3000(1+.08/4)^(4*1) - 3000 = 247.30
*January 2, 2015*

**Algebra**

if the width is w, then w(3w-8) = 35 Hint: 35=5*7
*January 2, 2015*

**Math**

y-4 = -2(x-3) y-4 = -2x+6 Now just rearrange things into the standard form: Ax+By=C
*January 2, 2015*

**mrr high school**

well, 1W = 1J/s, so that'd be 1.5*72/60 W
*January 2, 2015*

**math lit, accounting, business studies, cat**

surely your school has advisors. Why not have a nice discussion with one of them, rather than look for a sound bite here?
*January 2, 2015*

**math fractions**

1/4 + 3/4x = (1+3x)/4
*January 2, 2015*

**algebra**

sum: 4x+3x
*January 1, 2015*

**Maths**

well, since T10 = ar^9 a(5/3)^9 = 5859375 = 3*5^9 So, a = 3^10 = 59049
*January 1, 2015*

**calculus**

I'd say that (X1,Y1) is the position of the base of the ladder, if the base of the wall is (0,0). Since it is moving at the rate of 2 ft/s, after t seconds it is 2t feet from the wall. Naturally, the Pythagorean theorem says that the height of the top of the ladder obeys h...
*January 1, 2015*

**Math**

since cos(kt) has a maximum at t=0, -3cos(kt) has a minimum of -3 at t=0. So, since the bottom of the tire touches the ground, the axle is 3 feet up, and if the point starts at the bottom of the circle, it is indeed given by 3-3cos(3pi/2 t). The height is from the ground, not ...
*January 1, 2015*

**Math**

Yes, if the seat starts at the top. Though you should write cos(6pi t). As it is, it looks like cos(6pi) is the function. The parentheses are used to enclose the argument of the function.
*January 1, 2015*

**math**

If x is at 2%, the rest (10000-x) is at 12%. So, add up the interest and you have .02x + .12(10000-x) = 600 Now just solve for x.
*January 1, 2015*

**Geometry**

Can't quite fathom your terminology. How do you take the measure of a triangle? I'll assume that WXZ is an angle, not a triangle. If XZ is a side of the triangle ZXY, how can it bisect ZXY? I'll assume your logic is sound, even if your exposition is not.
*January 1, 2015*

**Math**

Let's see what's what here... QR = √((-2+2)^2 + (1-5)^2) = √16 RS = √((-5+2)^2 + (1-1)^2) = √9 SQ = √((-2+5)^2 + (5-1)^2) = √25 DE = √((5-5)^2 + (-2+6)^2) = √16 EF = √((2-5)^2 + (-2+2)^2) = √9 FD = √((5-...
*January 1, 2015*

**physics**

avg speed = total distance ----------------- total time = (1+1)km/(1/60 + 1/40)hr = 48 km/hr
*January 1, 2015*

**earth science**

but you would be wrong. Try looking it up, and I'm sure any explanation you find will explain why this is so.
*January 1, 2015*

**earth science**

and you would be wrong. Think about it. To study the universe you need to use a telescope to view distant objects. On earth, the atmosphere makes such sightings faintly blurry. On the moon, with virtually no atmosphere, the images will be a lot clearer. That is why we have ...
*January 1, 2015*

**chemistry**

2CS2 + 6O2 -> 2CO2 + 4SO2 So, now you know that you will need 6 moles of O2. What is the volume of 6 moles of any gas at STP?
*December 31, 2014*

**chemistry**

You know that PV=kT Since P is held constant, that means that V/T = k/P, which is constant. The temperature is reduced by a factor of (273+3)/(273+27), so the volume must also decrease by the same factor for V/T to remain constant.
*December 31, 2014*

**Algebra**

Or, you can think of it as looking for the intersections of a circle and a line. The circle: (x-2)^2 + (y-1)^2 = 45 The line: y=7 They intersect where (x-2)^2 + (7-1)^2 = 45 now the math proceeds as above
*December 31, 2014*

**Calculus**

f(x)=(10-3x) / 2√(5-x) quotient rule: f'(x) = [(-3)(2√(5-x))-(10-3x)/-√(5-x)] / 4(5-x) = [-6(5-x)+(10-3x)] / 4(5-x)^(3/2) = (3x-20) / 4(5-x)^(3/2) Using product rule: f(x) = 1/2 (10-3x)(5-x)^(-1/2) f'(x) = 1/2[(-3)(5-x)^(-1/2) + (10-3x)(-1/2)(5-x)^(-3...
*December 31, 2014*

**Calculus**

you know that h = 2d = 4r. So, the volume v is v = 1/3 pi r^2 h = 4/3 pi r^3 to find the height after 5 minutes, just recall that the volume will be 5min * 20m^3/min = 100 m^3 So, just solve for h (4r) when v = 100 To find how fast the pile is increasing, recall that dv/dt = ...
*December 31, 2014*

**Math- Max and Minimum**

better review your max/min topic. f has a max/min where f'=0 and f"≠0. f' = x^3-3x^2-6x+6 This does not factor over the rationals, so a graphical solution is probably the best bet. The graph of f' is at http://www.wolframalpha.com/input/?i=x^3-3x^2-6x%...
*December 31, 2014*

**Calculus**

You went to too much work on f" f' = (1-x)/(1+x)^3 f" = [(-1)(x+1)^3 - 3(1-x)(x+1)^2]/(x+1)^6 = [-(x+1)+3(x-1)]/(x+1)^4 = (-x-1+3x-3)/(x+1)^4 = (2x-4)/(x+1)^4 Just because the quotient rule says you wind up with v^2 in the bottom, you don't have to leave it ...
*December 31, 2014*

**Calculus**

(a) and (b) are correct (c) what's the problem? f"(x) = 2(x-2)/(x+1)^4 the denominator is positive always. So, f" < 0 for x < 2 f" > 0 for x > 2 Now you can say where f is concave up or down, and where the inflection point is.
*December 31, 2014*

**MATHS**

volume changes as the cube of the scale. So, each piece as 1/6^3 the volume of the container.
*December 31, 2014*

**Math**

use the distance formula many times to find the lengths of the sides. The two triangles with all congruent sides are congruent (SSS).
*December 30, 2014*

**equation and inequalities**

b-15 = -12 b = 3 So, on the number line, put a dot at 3.
*December 30, 2014*

**math**

Solve for x in x(1.15)^5 = 1400
*December 30, 2014*

**Pre-Calc/Trigonometry**

As I recall, our model is T(t) = 205 + (72-205)e^-kt So, solve for k in T(10) = 195 Then use that k to solve for t in T(t) = 180 What do you get?
*December 30, 2014*

**Pre-Calc/Trig**

T(t) = 350 + (350-50)e^-kt So, just solve for k in T(2) = 60 Then use that k to solve for t in T(t) = 150
*December 30, 2014*

**Pre-Calc/Trig**

T(t) = To + (50-To)e^-.1947t I think we can assume that To = 98.6, so T(t) = 98.6 - 48.6 e^-.1947t So, just solve for t in T(t) = 66. That will give the number of hours since death (t=0).
*December 30, 2014*

**Algebra 1**

Both equations are for straight lines. Two lines can intersect in only one point, so I can't imagine how you got two solutions. Did you work the problem two different ways, and the answers disagreed? It would be really nice if you showed your work. Now for the bad news: ...
*December 30, 2014*

**Algebra 1**

my answer is 5/17 or 16/17
*December 30, 2014*

**Algebra 1**

What is the solution to the system y=10x-3 y=-7x+3 answers from 5/17, 16/17 4/17, 16/19 -5/17, 16/17 3/17, 5/17
*December 30, 2014*

**math**

or, you can divide first: -5(4x+1) =15 4x+1 = -3 4x = -4 x = -1 Damon's way is nice if you expect to run into awkward fractions by doing the division first.
*December 30, 2014*

**algebra**

clearly the baby has gained 22/16 oz/day. So, starting with the birth weight of 50 oz, add that much for each day. And what does the no. of days in a month have to do with anything?
*December 30, 2014*

**maths**

subtract the inside volume from the outside volume: (29)(21)(13) - (28)(20)(12) = ?
*December 30, 2014*

**Math**

such a reflection takes (x,y) -> (x,-y), so it is neither set of vertices. You want (–1,2), (–4,4), (–3,5)
*December 29, 2014*

**Algebra 1**

They are the same line.
*December 29, 2014*

**Algebra 1**

How many solutions does the system of equations have? 3x+12y=20 y=-1/4x+5/3
*December 29, 2014*

**Algebra 1**

I think the Answer is 3x+5y=-10
*December 29, 2014*

**Algebra 1**

Write y=-3/5x-2 in standard form using Integers.
*December 29, 2014*

**Algebra 1**

thank you
*December 29, 2014*

**Algebra 1**

Tell whether the lines for then pair of equations are parallel, perpendicular, or neither? y=-2/3x +1 2x-3y=-3
*December 29, 2014*

**Pre-Trig/Calc**

e^-kt -> 0 as x->∞ 1-e^-kt -> 1 So, T(t) -> a+b at t=0, T(t) = b So, b is the initial temperature of the potato, 50°F a is the initial temperature difference between the potato and the oven: 300 in this case. You know this because it is clear that as time...
*December 29, 2014*

**Algebra 1**

Write the equation of a line that is perpendicular to the given line and that passes through the given points. y-2 = - 1/4(x+3), (-3,5)
*December 29, 2014*

**Algebra 1**

Write an equation in point slope form for them line through the given point with the given slope. (-4,6) m= 3/4
*December 29, 2014*

**Algebra 1**

actually, the slope of the V would be 2, not 1. |x| has slope 1. Extra credit: what if you had meant y=|2x+1|?
*December 29, 2014*

**Algebra 1**

Assuming you meant |2x| + 1, just like |x|, but squeezed half as wide, and shifted up 1.
*December 29, 2014*

**Algebra 1**

What would the graph look like for y= the absolute value of 2x plus 1
*December 29, 2014*

**Pre-Calculus**

Wow - a lesson in precision, as well as catching my typo! Still, if you just typed in 800 tan 33 /(tan 35-tan 33) I feel sure that the internal 32-bit precision would avoid any serious rounding error in this case.
*December 29, 2014*

**Pre-Calculus**

(800+x)tan33=xtan35 800 tan33 + x tan33 = x tan35 x tan35 - x tan33 = 800 x(tan35 - tan33) = 800 x = 800/(tan35-tan33) aside from the trig values (which are just constants), that's just good old algebra I, right?
*December 29, 2014*

**Algebra 1**

good question
*December 29, 2014*

**Algebra 1**

sorry I meant c(n) =.50n-.80
*December 29, 2014*

**Algebra 1**

Suppose soda is on sale for .50 a can and you have a coupon for .80 off the total price. Write a function rule for the cost of n cans of soda. My answer is c(n() = .80n -.50
*December 29, 2014*

**geometry**

the locus is the perpendicular bisector of the line from (0,0) to (-2,5). It's easy to find the midpoint, and the slope of the line. Now just find the line with perpendicular slope, through the midpoint.
*December 29, 2014*

**Algebra 1**

thank you
*December 29, 2014*

**Algebra 1**

What is the simplified form of the expression? 7(63/(5 sq-2 sq)-1)
*December 29, 2014*

**algebra**

x/50 = 0.8
*December 29, 2014*

**calculus**

you want dv/dt, where v = x^2 h The good old product rule says that dv/dt = 2xh dx/dt + x^2 dh/dt You have all the stuff you need, so just plug it in and watch the fun!
*December 29, 2014*

**Math and Finance**

5000(1+.08)^8
*December 28, 2014*

**Math**

d = √((5-1)^2 + (3-0)^2)
*December 28, 2014*

**Math**

as you recall, the slope-intercept form y = mx+b describes a line with slope m and y-intercept b. Surely you see the connection with the question above...
*December 28, 2014*

**math**

the stopping distance for a vehicle moving with speed v and deceleration a is s(t) = vt - 1/2 at^2 Since that is a quadratic function, the QF gives a handy way to solve it. And that's "brakes." Give me a break!
*December 28, 2014*

**geometry**

just use the Pythagorean theorem to find the resultant speed (the hypotenuse of the right triangle) √(130^2+20^2)
*December 27, 2014*

**Math**

4 is half the max displacement period = 8/3 seconds, = 2π/k, so k = 3π/4
*December 27, 2014*

**Math**

No, it is negative because I assumed that at t=0 the point is at its lowest elevation. (As my first sentence stated.) If we had started when the point was level with the circle's center, we'd have sin(kt). If we started with the point at its maximum height, then we'...
*December 27, 2014*

**Math**

If the point in question is at the lowest height at t=0 minutes, then y = -cos(kt) diameter is 6, so the radius is 3 y = -3cos(kt) the lowest point is 6" = 0.5 ft, so the axis of the loop is 3.5 ft off the ground: y = 3.5 - 3cos(kt) Since cos(kt) has period 2π/k = 1/...
*December 27, 2014*

**^Maths^**

You could just do the subtraction in base 6 and then convert the answer to base 7: 451-301 = 150 1506 = 1237
*December 27, 2014*

**math**

or, 4 3/8 inches :-)
*December 27, 2014*

**English**

a bit less than average in height. If the average height for men is 5'8", then someone who is 5'4" could be said to be a little on the short side. Similarly, a man 6'0" might be called a little on the tall side. A guy 6'8" is a long tall ...
*December 27, 2014*

**math**

Damon is correct. Surely you can see that 48:80 is not correct. There are only 80 records in all. 48:80 is the correct ratio, but the wrong total. For every 3 new there are 5 hot. So, 3:5 requires 8 records. There are 80 records. 80/8 = 10, so there are 10 sets of 3:5, or 30:...
*December 27, 2014*

**algebra (motion)**

if B takes x minutes, then the two runners cover the entire track in 15 minutes. 1/40 + 1/x = 1/15
*December 27, 2014*

**maths**

-5+20-8 = ?
*December 27, 2014*

**algebra**

almost anything 3/5 4/6 123/125 932789238/932789240 In fact, anything except 1/3 2/4
*December 27, 2014*

**Math**

or, A :-)
*December 26, 2014*

**Math**

The initial position (at t=0) is not -1/2 Or is that initial - just a separator? Also, if you're going to use parentheses, it ought to be d = 1/2 sin(pi/2 t) - 1/2
*December 26, 2014*

**Math**

If the line does not pass through the origin, it it not proportional. you clearly have y = 25x + 50 That "+ 50" wrecks the proportionality.
*December 26, 2014*

**Algebra**

What you had was 10P4 = 5040 10P6 = 10*9*8*7*6*5 10P4 = 10*9*8*7 So, you can see why 10C6 = 10C4
*December 26, 2014*

**math**

10% ≈ 12, so 60% ≈ 72
*December 26, 2014*

**math**

if it's just one item, I wouldn't call it a total cost. If the customer orders some number of each item, just add up all the item costs to get the total cost. 2c,3s,5d,1j is just a list of costs, not a total. So, how's that English class coming along? ...
*December 26, 2014*

**math**

hint: total cost
*December 26, 2014*

**math**

time = distance/speed, so it takes 1.5x10^11 / 3x10^8 = 0.5x10^3 seconds We all know it takes light about 8 minutes (480 seconds) to reach the earth. .5x10^3 = 500 is pretty close.
*December 26, 2014*

**physics**

three light sources at different directions, maybe of different sizes.
*December 26, 2014*

**statistic**

exactly half. It doesn't matter what the std is. In a normal distribution, 1/2 of the population is above or below the mean.
*December 26, 2014*

**Math**

There are only 3 colors. So, after 6 draws, it is possible to have two of each color. The 7th ball must be the one which matches two of the others.
*December 26, 2014*

**math**

3+4+5+6+7 = 25 So, the milk in the vessels is 3x,4x,5x,6x,7x If the capacity of each is z, then 25x = 0.6*5z = 3z x = 0.12z So, in terms of a full vessel, the amounts of milk in each of the vessels is .36, .48, .60, .72, .84 Looks like 3 of them are over 50% full.
*December 26, 2014*

**Math**

hint: the equation you have shown is called the slope-intercept form of a line. Just read off the slope and the y-intercept...
*December 26, 2014*

**Business Math**

have you no calculator? Heck, just type it into google, if nothing else.
*December 26, 2014*

**science**

the circumference is C = 2pi*r. Make sure your answer is in meters, not cm. Opposite ends of the diameter are half that far apart. Now just multiply that by 12.
*December 26, 2014*

**maths**

you surely have the formula for the surface area of a sphere. Divide that by 4.
*December 26, 2014*

**s.p s**

TODAY (actually, since it's not yet midnight here where I am, there's no way!)
*December 26, 2014*

**Algebra**

F(x) = -3cos(4(x+n/4))+6 So, the amplitude is 3 the period is 2pi/4 = pi/2 the phase shift is -n/4
*December 25, 2014*

**Math**

the period and amplitude do not change, whatever the value of t. The question asks for three things: amplitude frequency distance when t=3 The distance when t=3 is y(3) = -15cos((pi*3)/3)+5 = -15cos(pi)+5 = 15+5 = 20
*December 25, 2014*

**Math**

(a) The period is (2pi) / (pi/3) = 6 so, the frequency is 1/6 your amplitude and distance are correct. Remember, the period for cos(kt) is 2pi/k So, (b) should be no trouble now.
*December 25, 2014*

Pages: <<Prev | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | **16** | 17 | 18 | 19 | 20 | 21 | 22 | Next>>