Saturday

February 28, 2015

February 28, 2015

Total # Posts: 29,289

**math**

.42x = 210
*February 3, 2015*

**Pre-Calculus**

x-axis: (x,y)->(x,-y) y-axis: (x,y)->(-x,y) but, since h(x) is an odd function h(-x) = -h(x) reflecting h(x) through either axis flips it to the same thing. (x,x^3)->(x,-x^3) (x,x^3)->(-x,x^3) = (-x,-(-x)^3)
*February 3, 2015*

**Pre-Calculus**

swap x and g and then solve for g: x=(g+7)^3+4 x-4 = (g+7)^3 ∛(x-4) = g+7 g = ∛(x-4) - 7 check. If h is g inverse, g(h(x)) = h(g(x)) = x g(h) = (h+7)^3+4 = ((∛(x-4) - 7)+7)^3+4 = (∛(x-4)-7+7)^3+4 = ∛(x-4)^3+4 = x-4+4 = x and similarly for h(g)
*February 3, 2015*

**math**

and you'd be correct. But why guess? A brief check on the definition of a function would make it certain.
*February 3, 2015*

**Pre-Calculus**

nothing. x^2 is an even function. Reflecting it about its axis changes nothing. (x,y)->(-x,y) since f(x) = f(-x), the graph is unchanged.
*February 3, 2015*

**Pre-Calculus**

you surely know that this is a parabola which opens up, so there is no local maximum. So, find the f values at the ends of the interval, and pick the larger for the global max. No interval? No maximum.
*February 3, 2015*

**Pre-Calculus**

x then y: (x,y)->(x,-y)->(-x,-y) y then x: (x,y)->(-x,y)->(-x,-y) f(x) -> -((-x)^3+2(-x)^2-(-x)+5)
*February 3, 2015*

**Pre-Calculus**

reflect: (x,y)->(-x,y) f(x) -> -x^3+2x^2+x+5 shift: (x,y)->(x+4,y) f(x) -> -(x+4)^3+2(x+4)^2+(x+4)+5 = -x^3-10x^2-31x-23 see the graphs here: http://www.wolframalpha.com/input/?i=plot+y%3Dx^3%2B2x^2-x%2B5%2C+y%3D-x^3-10x^2-31x-23 the shift isn't real clear, but...
*February 3, 2015*

**math**

you need to divide the interval into 1/8 pieces, since the loaf holds 8 slices. Then shade in 6 of those pieces for the sandwiches.
*February 3, 2015*

**mathematic**

same as prob that it's not green: 1-3/8
*February 2, 2015*

**science**

likely a noble gas, since the others form molecules of more than one atom.
*February 2, 2015*

**Math**

a 50% profit margin means that the price is 2 times the cost. So the cost:profit ratio is 1:1 If the cost of raw material is doubled, then the ratio becomes 1.25:1 = 5:4, with the profit being only 4/9 instead of 1/2.
*February 2, 2015*

**vectors**

recall that every 5 minutes covers 30° That will give you the angle traversed.
*February 2, 2015*

**calculus**

cos(3π) is also -1, so x = 3π/2
*February 2, 2015*

**math**

14=-16-10n n=-3 the others look good.
*February 2, 2015*

**Algebra**

when it hits the ground, the height is zero. So, just solve -16t^2 + 1000 = 0 16(62.5-t^2) = 0 t = √62.5 = 7.9
*February 2, 2015*

**math**

27(1/8)
*February 2, 2015*

**Math**

you are right on.
*February 2, 2015*

**Physics**

momentum is conserved. In this case, we are considering only the horizontal motion, since the wagon cannot move vertically. So, the basic problem is that of what happens when the wagon's mass is suddenly increased by the addition of the bird's mass. To conserve ...
*February 2, 2015*

**Math Help**

that is not an answer. It is a restatement of the problem, using only sines. 2sin^2x-sinx-1 == 0 (2sinx+1)(sinx-1) = 0 sinx = -1/2 or 1, so x = π/2, 4π/3, 5π/3 http://www.wolframalpha.com/input/?i=cos2x%2Bsinx%3D0+for+x%3D0..2pi
*February 2, 2015*

**math**

where f'=0 and f"<0. So, what do you think?
*February 2, 2015*

**math**

y" = 4/(5-x)^3 f is concave up when f" > 0 So, what do you think?
*February 2, 2015*

**Grammar and Context**

I fail to see how this has anything to do with physics or chemistry. Maybe you have some ideas in that regard. when you say "would have" done, do you mean you want to know things that it failed to do?
*February 2, 2015*

**math**

1500 <= i+200 <= 2000
*February 2, 2015*

**algebra**

yes it is Though, to me, saying 3(4w+5)^2 is just as good. It is clear what the factors are. I wrote 4w+5 instead of 5+4w just because the polynomial was given in descending powers of w, so I made my factors read the same way.
*February 2, 2015*

**mathematics**

fractions are ratios 2/3 = 2:3
*February 2, 2015*

**Algebra 2!!!!**

checking the graphs at wolframalpha.com, http://www.wolframalpha.com/input/?i=4x^3-8x^2-15x%2B9 http://www.wolframalpha.com/input/?i=2x^3-5x^2-4x%2B10 it is clear that #1 has a root at x = 3 #2 has a root at x = 5/2 So, divide through and you can use the quadratic formula to ...
*February 2, 2015*

**maths..g.p**

a=5 d=4 n=(131-5)/4 = 31 check: 5+31*4=129 S = 31/2 (5+129)
*February 2, 2015*

**maths..a.p**

there are 49 of them, so 49*50/2
*February 2, 2015*

**math**

not possible. 1-4 is negative.
*February 2, 2015*

**maths**

a = 2r, so r = a/2 a/(1-r) = 8 a/(1-a/2) = 8 a = 8-4a a = 8/5 So, the sequence is 8/5, 32/25, 128/125, ...
*February 2, 2015*

**acceleration**

amazing. a = (Vf-Vi)/t
*February 2, 2015*

**Math**

As usual, draw a figure. If we start at the right angle and label the vertices clockwise, then we see that BD is the hypotenuse of an isosceles right triangle, with length 12√2. If the 4th side is x, then we have from the law of cosines, (12√2)^2 = 8^2 + x^2 - 2(8...
*February 2, 2015*

**math**

Sorry. 24 does not divide into 5 equal parts. You probably mean that there are 24 boys, or that there are 24 girls. If boys, then 24/g = 2/3 g = 36 If there were 25 total students, then (25-g)/g = 2/3 75-3g = 2g g = 15
*February 2, 2015*

**maths**

start distributing: (-2)(2x)(3x+1) + (-2)(-5)(3x+1) = -4x(3x+1) + 10(3x+1) I think you can probably take it from here, eh?
*February 2, 2015*

**maths**

Use a little synthetic division to find that the remainder is 4m+6 So, m = 1/2 x^3 + (1/2)x^2-5x+4 = (x+2)(x^2-(3/2)x-2)+8
*February 2, 2015*

**algebra**

you are correct
*February 2, 2015*

**algebra**

they are exactly the same. The * is optional, as you prefer.
*February 2, 2015*

**algebra**

correct
*February 2, 2015*

**math**

AP: 10,10+2d,10+6d GP: (10+2d)/10 = (10+6d)/(10+2d) d = 5 check: AP: 10,15,20,25,30,35,40 GP: 10,20,40
*February 2, 2015*

**math**

cone: v = 1/3 pi r^2 h using that v, then for the sphere, find its radius and then area: v = 4/3 pi r^3 a = 4 pi r^2
*February 2, 2015*

**math**

4 minutes is 240 seconds. So, the water will rise (27*240)/(60*30) cm I expect you can use that to answer the question.
*February 2, 2015*

**math**

as always, distance = speed * time. (x-20)(3.25) = 2x+30 x = 76 So, 56 km/hr * 3.25 hr = 182 km
*February 2, 2015*

**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*

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