Sunday

March 29, 2015

March 29, 2015

Total # Posts: 30,148

**Math**

If 30 spoke E&F and 10 spoke E&F but not S, then that means 20 spoke E,F,S. So, in your Venn diagram, the small triple intersection in the center has a 20 in it. But, since only 20 spoke French and Spanish, and all of them also spoke English, there are 0 who spoke no English.
*February 2, 2015*

**Calculus**

y^2 = (x^2−1)^2 y = ±(x^2-1) That is, y = (x^2-1) or y = (1-x^2) The area between two functions f(x) and g(x) is ∫ f(x)-g(x) dx. hat is, it is the sum of all the thin strips whose height is the difference between the curves.
*February 2, 2015*

**Calculus**

I get for the area ∫[-1,1] (1-x^2)-(x^2-1) dx = ∫[-1,1] 2-2x^2 dx = 2(x - x^3/3)[-1,1] = 8/3
*February 2, 2015*

**Precal**

The names monomial, binomial, etc, are usually reserved for polynomial functions, where all the terms are just powers of x (or other variables). The above function, with its 5^x factor is not a polynomial, as generally used. The terms of a polynomial are separated by + and - ...
*February 1, 2015*

**Math**

h(t) is just a parabola. The max height is reached at the vertex. For the parabola ax^2 + bx + c the vertex is at x = -b/2a. So, for h(t), the vertex is at t = 3. h(3) = 152 It hits the ground when h=0. To find that, just use the quadratic formula.
*February 1, 2015*

**Math**

c(x) = 25 for 0 < x <= 6 25 + 4⌊x-6⌋ for x>6
*February 1, 2015*

**Algebra 2**

√(79-5x) = 6x+4 79-5x = (6x+4)^2 = 36x^2+48x+16 36x^2+53x-63 = 0 (4x+9)(9x-7) = 0 x = -9/4 or 7/9 but, you have to check for spurious roots: √(79-5(-9/4)) = √(79+45/4) = 9.5 but, 6(-9/4)+4 = -9.5 This is the spurious root. If you square both sides, it becomes...
*February 1, 2015*

**Math**

GCF(32,40) = 8 so there are 4 rows of irises and 5 rows of tulips, all rows with 8 bulbs.
*February 1, 2015*

**Physics**

The horizontal and vertical speed are both 19.8/√2 = 14 m/s At that speed, it takes 35/14 = 2.5 seconds to get to the goal. The height of the ball is y = 14t-4.9t^2 y(2.5) = 4.375 m Looks like it will clear the bar by 1.325 m I'll let you decide the vertical velocity...
*February 1, 2015*

**Physics**

the height of the ball as a function of time is y = 31.1*sin37.0°*t - 4.9t^2 = 18.7t - 4.9t^2 So, you want to find t when y = 5.5 (and is on its way down). t=3.5 s Since the horizontal speed is constant, just multiply it by 3.5 to find the distance. v = 18.7 - 9.8t and ...
*February 1, 2015*

**Geometry**

Look. You had x^2+6x If you consider that as x^2+2ax, a=3. to complete the square, you have to add a^2, which is 9. Just follow the steps I took in my solution. I assume you can take the square root of 25 when you get that far...
*February 1, 2015*

**Geometry - ahem**

(x+a)^2 = x^2+2ax+a^2
*February 1, 2015*

**Geometry**

you were supposed to complete the square. x^2 + 6x = 16 x^2 + 6x + 9 = 16+9 (x+3)^2 = 25 x+3 = ±5 x = -3±5 and you get your solution. To complete the square, you divide the coefficient of x by 2, since (x+a)^2 = a^2+2ax+a^2.
*February 1, 2015*

**Math**

apparently you didn't bother to follow the hint, so I'll give you a push... dx/dt = -600 mi/hr h = 2 tanθ = 2/x sec^2θ dθ/dt = -2/x^2 dx/dt Now plug in your numbers and find dθ/dt. watch the units.
*February 1, 2015*

**physics**

Assuming constant acceleration, s = 1/2 at^2, so 1/2 * a * 3^2 = 41 a = 9.111 And, we all know that F = ma
*February 1, 2015*

**precalculus**

sinC/c = sinA/a, so sinC/52.8 = sin27°30'/28.1 sinC = 0.8676 C = 60°11' or 119°49' Since A+C < 180 in either case, there are two possible triangles. Knowing A and C, you now know what B can be, and then use the law of sines again to find b.
*February 1, 2015*

**Math 20-1**

(0,0) is 0 < 0.6? Of course, you haven't said what the region is. Maybe it doesn't include (0,0) . . .
*February 1, 2015*

**Math 20-1**

you can start by noting that 7*12 = 84
*February 1, 2015*

**Algebra**

the slope of the line joining the points is 1, so . . .
*February 1, 2015*

**math**

x + 4x = 90, so x = 18 ...
*February 1, 2015*

**geometry**

7' = 84", so the ratio is 84:16 Now reduce that to see which choice it matches.
*February 1, 2015*

**math**

split the terms into two pairs, and you can see that you have b(r+v) + y(r+v) (b+y)(r+v)
*February 1, 2015*

**math**

sorry. My bad. I was thinking we wanted the series to converge, not the sequence.
*February 1, 2015*

**math**

the sequence is 3/2, 6/5, 15/14, ... I don't see how it can converge, since each term is greater than 1.
*February 1, 2015*

**math**

10/2 (2a+9d) = 145 a+3d + a+8d = 5(a+2d) a=1 d=3
*February 1, 2015*

**math**

The vector from the sun to the object is <5,4,22> <5,3,-2>+<5,4,22>/11 = <60/11,37/11,0>
*January 31, 2015*

**advanced functions**

and then gets hit by a train.
*January 31, 2015*

**geometry**

sorry - no triangle has sides of 2,5,7. That is just a straight line of length 7. Whoever devised this problem was sadly unaware of properties of a triangle. Anyway, supposing you have a real triangle, if the sides are doubled, so is the perimeter.
*January 31, 2015*

**Calculus**

starting at time 0, when the first dose is taken, the amount after t hours is 80*(1/2)^(t/22) the 2nd dose: 80*(1/2)^((t-24)/22) the nth dose: 80*(1/2)^((t-24n)/22) Play around with that as t gets large.
*January 31, 2015*

**Pre Calculus**

oops. Since 0-length numbers are not allowed, we really have only 340 possibilities.
*January 31, 2015*

**Pre Calculus**

since there are 4 choices for each digit, and the numbers may be 1,2,3 or 4 digits long, you have 4^1 + 4^2 + 4^3 + 4^4 = (4^5-1)/(4-1) = 1023/3 = 341
*January 31, 2015*

**Calculus**

if y = u^n, dy/dx = n u^(n-1) du/dx, so y' = 3(x^2 + 8x +3)^2 * (2x+8)
*January 31, 2015*

**math**

looks good to me
*January 31, 2015*

**elementary differential geometry**

if you google the question, you will be directed to several discussions on the matter, in varying degrees of complexity.
*January 31, 2015*

**Calculus**

make that sec^2 x and not 1/sec^2 x
*January 31, 2015*

**math help**

since a^2-b^2 = (a-b)(a+b) (x+5)^2 - (x-5)^2 = (x+5-(x-5))(x+5+x-5) = (10)(2x) = 20x So, A=20
*January 31, 2015*

**Calculus**

-x^2+x+6 = -(x-3)(x+2) As you know, this is a parabola, opening downward. So, it is positive between the roots: -2 and 3. So, f(x) is concave up in that interval. So, (a)
*January 30, 2015*

**pre calculus**

P(-2) is the remainder when P(x) is divided by (x+2) P(-2) = -5, so P(x) = (x+2)(x^2-1) - 5
*January 30, 2015*

**Calculus**

h(t) = 2 + 0.2t so, plug that in to express v(t) Then find dv/dt Or, since dv/dt = pi/4 h^2 dh/dt, plug in h(t) and dh/dt=0.2 to express dv/dt
*January 30, 2015*

**math**

pi*(4x)^2 - pi(4y)^2 = 16pi x^2 - 16pi y^2 = 16pi (x+y)(x-y) Your mistake was in saying that pi(4x)^2 = pi*4x^2 Bzzzzt Parentheses are yur friends.
*January 30, 2015*

**Math**

u(t) = √(g(t)) So, if f(t) = √t, u(t) = f(g(t)) = (fog)(t)
*January 30, 2015*

**absolute value**

|x+1| - |x-1| If x>1, |x+1| = x+1 and |x-1| = x-1, so |x+1| - |x-1| = (x+1)-(x-1) = 2 Similarly, if x < -1, |x+1| - |x-1| = -(x+1) - -(x-1) = -2 If -1 < x < 1, |x+1| > 0 |x-1| < 0, so |x+1| - |x-1| = (x+1)- -(x-1) = 2x
*January 30, 2015*

**HELP ME PLEASE**

Just work your way out from the inside -1(-a)^2 + (-(-a))^2 = -1(-a)^2 + (a)^2 = -1*a^2 + a^2 = -a^2+a^2 = 0 recall that (-a)^2 = (-a)(-a) = +a^2
*January 30, 2015*

**Algebra 1**

the surface area is the two circular bases, plus the curved sides. a = πr^2 + πr^2 + 2πrh = 2πr(r+h) = 2π(3x-2)(3x-2+x+3) = 2π(3x-2)(4x+1)
*January 30, 2015*

**Algebra 1**

The radius of a cylinder is 3x-2cm. The height of the cylinder is x+3cm. What is the surface area of the cylinder?
*January 30, 2015*

**algebra**

just divide each term, then add them up: 6s^7p^8/3s^3p = 6/3 s^7/s^3 p^8/p = 2s^4p^7 6s^5p^6/3s^3p = 2s^2p^5 27s^3p^4/2s^3p = 9p^3 So, the final sum is 2s^4p^7 - 2s^2p^5 + 9p^3
*January 30, 2015*

**Math**

What is the GCF of the terms 8c^3+12c^2+10c? my answer is 2c
*January 30, 2015*

**Algebra 1**

How can the polynomial 6d^4+9d^3-12d^2 be factored?
*January 30, 2015*

**Algebra 1**

There is a circular garden in the middle of a square yard. The radius is 4x. The side length of the yard is 20x. What is the area of the part of the yard that is not covered by the circle?
*January 30, 2015*

**algebra**

well, 36x^3-73x^2-45x-6 = (9x+2)(4x^2-9x-3) Or, if you want to do the division manually, enter the polynomials at http://calc101.com/webMathematica/long-divide.jsp
*January 30, 2015*

**algebra 2**

(f◦g)(4) = f(g(4)) = f(-14) = -49 since g(4) = -2(4)-6 = -14 and f(-14) = 3(-14)-7 = -49 or, you can just define (f◦g)(x) = f(g) = 3g-7 = 3(-2x-6)-7 = -6x+25 so, (f◦g)(4) = -24-25 = -49
*January 30, 2015*

**math**

1 3/4 + 1 1/3 + 1 3/4 = ?
*January 30, 2015*

**algebra**

-4 2/3
*January 29, 2015*

**math**

23x = 127
*January 29, 2015*

**calculus**

the weight of a thin sheet of water of thickness dy and area 9x is 9x*62 dy At depth y, the width across the water surface is 2∜y, so the weight of the sheet of water is 1116∜y dy Now, since work = force * distance, we have to add up all the work lifting thin ...
*January 29, 2015*

**algebra**

visit here to see the graph. To do it yourself, get a sheet of graph paper. Plot two points on the line and connect them. Extend the connecting line segment on both ends, and that is the graph of the equation. If you cannot plot a point, you have some studying to do. http://...
*January 29, 2015*

**Albebra**

(10-2y)(10-2y)
*January 29, 2015*

**Algebra 2**

not clear just what is happening. A hiatus is a rest period. I don't expect much traveling is done during the stops. Anyway, I'd say that if the distance traveled is x, then since time = distance/speed, x/40 = 2 + x/50 Now just solve for x.
*January 29, 2015*

**College Algebra**

If the distance is x miles, the time Allison drove is x/65. So, since distance = speed * time, x = (x/65)(41) + 144
*January 29, 2015*

**Math**

use your sum/difference formulas: cos(x+y) = cosx cosy - sinx siny now use the difference formula, and then add them up and see what drops out.
*January 29, 2015*

**Math 20-1**

2x^2 - 7x + 3 > 0 (2x-1)(x-3) > 0 the roots of the equation are 1/2 and 3 Since the parabola opens up, 2x^2-7x > -3 when x is outside the interval between the roots: x < 1/2 or x > 3 See http://www.wolframalpha.com/input/?i=2x^2-7x%3E-3
*January 29, 2015*

**Math 20-1**

< or >
*January 29, 2015*

**college algebra**

x/6 + 8
*January 29, 2015*

**Calculus**

when the balloon is at height h, tanθ = h/300 so, sec^2θ dθ/dt = 1/300 dh/dt you know that when h=300, θ=π/4. You know dh/dt = 80, so just solve for dθ/dt
*January 29, 2015*

**Math Help!!**

you have a line containing the points (2,20) and (5,14) where y is 1000's of gallons of gas at price x. So, just use your two-point form for the line, and then find y(2.50).
*January 29, 2015*

**Algebra 2**

switch x and y, then solve again for y: x = 3y+12 x-12 = 3y (x-12)/3 = y or, y = x/3 - 4
*January 29, 2015*

**physics**

at 1329 m/s^2 it will take 43/1329 = 0.032 seconds to achieve takeoff speed. Now that is some acceleration!! You sure there's not a typo somewhere in there? Anyway, once you have the correct takeoff time, just plug in the distance formula s = 1/2 at^2 to find the required ...
*January 29, 2015*

**Math 20-1**

Since 7*12 = 84, you can start with a 7x12 rectangle. After that, just start increasing the width and using the appropriate length. There's no end of possible sizes with area greater than 84.
*January 29, 2015*

**Math**

1D ok 2A You count leading zeros for the exponent, but you cannot ignore any zeroes after the first nonzero digit. They are part of the value. 3A ok 4B 1/8000 = 0.000125 your answer is for 8000, not 1/8000 5C doubling n doubles then exponent, roughly equivalent to squaring the...
*January 29, 2015*

**maths contant**

the slope is -9/2 tan 77° = 9/2 tanθ is negative in QII, so our desired angle is 180-77 = 103°
*January 29, 2015*

**calculus**

The left term can be seen as (sin3x/x)^3 - 3 = 27(sin3x/3x)^3 - 3 now, we all know that lim sinu/u = 1 as u->0, so the limit of all that is just 27-3 = 24 multiply that by x^4 and you just wind up with 0 as x->0 Still not sure what that x^4 is doing in there. It just ...
*January 29, 2015*

**Calc II**

since the slope anywere is just y'=2x, you want to find where 2x = 5/2 x = 5/4 So, the desired point is (5/4,25/16) For the other one, figure the slope of the tangent line, and then follow the same steps.
*January 29, 2015*

**Calculus**

#1 I only see one point and a number. #2 at the given point, the slope is -1.2 So, the line there is y-0.36 = -1.2(x+0.6)
*January 28, 2015*

**math**

19g/(5cm)^3 = 19/125 g/cm^3 water has density 1g/cm^3, so the wood will float. (duh!)
*January 28, 2015*

**Math**

since cot t appears several places, let g(t) = cot(t) Now we have u(t) = g/(9+g) So, let f(t) = t/(9+t) and you have what you want. f◦g = f(g) = g/(9+g) = cot t/(9+cot t)
*January 28, 2015*

**Math**

If you mean A = 2600e^(0.08t) that is 8% compounded continuously
*January 28, 2015*

**Math**

g(x) = x/(9+x) f(x) = ∛x
*January 28, 2015*

**Math**

f°g°h = f(g(h)) = f(h^2) = √(h^2-5) = √((x^3+6)^2-5) = √(x^6+12x^3+31)
*January 28, 2015*

**Algebra**

get all the T stuff together: 5.5*10^2(1/T) = 17.8 Now you can start working the numbers: 1/T = 17.8/550 T = 550/17.8
*January 28, 2015*

**Math**

If their scores are x and y, we have x/y = 1/5 (x+48)/y = 5/9 Now just solve for y (x+48)/y = x/y + 48/y, so 1/5 + 48/y = 5/9 48/y = 16/45 y = 135 check: x = 27 (27+48)/135 = 75/135 = 5/9
*January 28, 2015*

**Math**

since we have our minimum at t=0, the function will look like y = -cos(kt) The brightness varies by ±0.30, so that is the amplitude. y = -0.30 cos(kt) Since the low is 5.5, the axis of the curve is at 5.5+0.30, so y = 5.80 - 0.30cos(kt) The period of cos(kt) is 2π/...
*January 28, 2015*

**Algebra**

you are correct
*January 28, 2015*

**math**

when x=0, y can be anything, so the points lie on the y axis. It just happens that (0,0) lies on both axes. Now show a counterexample.
*January 28, 2015*

**algebra**

w-9 = m+9 (m-9)/(w+9) = 1/2 Now, solve for w: m=w-18, so (w-18-9)/(w+9) = 1/2 2w-54 = w+9 w = 63 check: m=63-18 = 45 (45-9)/(63+9) = 36/72 = 1/2
*January 28, 2015*

**Trigonometry**

If the point in question is D, then we have AD+DC = 5 BD/AD = tan 60° BD/DC = tan 45° so, we have BD/√3 + BD/1 = 5 Now just solve for BD, the desired distance.
*January 28, 2015*

**Math**

1/5 1/20
*January 28, 2015*

**Math**

1/2 + (2/3)(1/2) = 1/2 + 1/3 = 5/6 That's how much he painted on Mon and Tue. So, that left how much for Wed?
*January 28, 2015*

**Math**

A $1,600.00 principle earns 7% interest, compounded semiannually twice per year. After 33 years, what is the balance in the account. My answer is $112,992.00
*January 28, 2015*

**Algebra 1**

does the rule y=-3x^5 represent an exponential function
*January 28, 2015*

**Calculus**

what does "x^3 x^4" mean?
*January 28, 2015*

**Algebra 1**

Is the following written in scientific notation? 4.8 X 100^7 My answer is no
*January 28, 2015*

**algebra**

just use your formula: 1500(1+0.035)^18
*January 28, 2015*

**Pre-Calc**

using the difference formula, you have sin(x)cos(π/6)+cos(x)sin(π/6) - sin(x)sin(π/6)+cos(x)sin(π/6) = 1/2 2cos(x)sin(π/6) = 1/2 2cos(x)(1/2) = 1/2 cos(x) = 1/2 x = π/3 or 5π/3 http://www.wolframalpha.com/input/?i=sin%28x%2B%28pi%2F6%29%29-...
*January 28, 2015*

**mathematics - eh?**

3/4 - 32 ?? I suspect that your problem is poorly stated.
*January 28, 2015*

**statistics**

1/17259390 times the value of the prize. So, if the prize it at least that much, a $1 ticket expects to make more than a buck.
*January 28, 2015*

**math**

Can't do it. If you draw A,B,C then D can be anywhere on a circle of radius 2.82 with center at C. There are two places on that circle where AD = 7.
*January 28, 2015*

**grammar**

people are there is no preposition here four quarters fell... past tense makes no distinction in number. (except for "to be") it is confusing sometime, all right. A good example is "the number" is singular "a number" is plural
*January 28, 2015*

**mathematics**

since the ratio is constant, 3/4 = x/120
*January 28, 2015*

**Chemistry**

A nice discussion is here: http://www.quora.com/Why-can-chlorine-atoms-not-form-hydrogen-bonds-even-though-they-have-very-similar-electronegativity-to-nitrogen-which-can
*January 28, 2015*

Pages: <<Prev | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | **17** | 18 | 19 | 20 | 21 | 22 | 23 | Next>>