Wednesday

January 18, 2017
Total # Posts: 47,946

**precalculus**

come on. just substitute it into the other equation. t = 2y+2 x = t^2-6 = (2y+2)^2-6 = ... It is clearly a horizontal parabola

*December 13, 2016*

**precalculus**

well, t = 2y+2, so use that

*December 13, 2016*

**precalculus**

well, |cos3?| <= 1 cos3?=0 when 3? is an odd multiple of ?/2

*December 13, 2016*

**precalculus**

the ellipse looks good, but y=-t^2 x=-t^2+4 becomes x-y=4 not x-y^2 = 4 same for t^3

*December 13, 2016*

**Pre-algebra PLease help**

I get: 1C 2B 3C (2/3 is just 2 * 1/3) 4A 5A 12 is 50% of 24 That is a good first estimate. However, you are correct that 11 is closer (.48*23.95 = 11.49). I would have guessed 12, but 48% is a bit less than 50%, and 23.95 is a bit less than 24, so my estimate was a bit high. ...

*December 13, 2016*

**chemistry**

how many moles of SiO2? Each mole needs 3 moles of C. similarly for the other parts. work with moles, then back to grams.

*December 13, 2016*

**ALGEBRA 2**

g(x) = f(x-8)-7

*December 13, 2016*

**algebra**

ummm ... h?

*December 13, 2016*

**math**

cosx/(secx-tanx) = cosx / (1/cosx - sinx/cosx) = cosx / (1-sinx)/cosx = cos^2x / (1-sinx) = (1-sin^2x)/(1-sinx) = (1+sinx)(1-sinx)/(1-sinx) = 1+sinx why the word "plus"? You have a missing + key?

*December 13, 2016*

**precalculus**

Hmmm. The standard form for the polar conics is r = ep/(1 + e sin?) While your equation looks correct, I'd have written it as r = (5/2) / (1 + (5/4) sin?) to make it clear what the parameters are. Or, if you dislike fractions, r = 10/(4+5sin?)

*December 13, 2016*

**Alegrebra 2**

horizontal axis or vertical? Where is the vertex or directrix? You need to give more information. Depending on the vertex, many parabolas have that focus.

*December 13, 2016*

**Math**

construct AB with length 4.8 find the midpoint M of AB with M as center, draw a circle P with radius 3.6 with B as center, draw a circle Q with radius 5.2 Where P intersects Q is C.

*December 13, 2016*

**precalculus**

the rotation angle ? is such that tan2? = B/(A-C)

*December 13, 2016*

**Math grade 6, two-part problem**

If the four had amounts w,x,y,z then we know that w+x+y+z = 60 w = (x+y+z)/2 x = 2/3 (y+z) y = 3z w,x,y,z = 20,16,18,6 Amy = w = 20 Nick-Cal = x-z = 16-6 = 10

*December 13, 2016*

**precalculus**

after all the similar exercises we have done, you still have no work of your own to show?

*December 13, 2016*

**Math**

no, the empty box is there so you can subtract it and see how much was crackers, not "tare weight" as it is called.

*December 13, 2016*

**Math**

the mass of crackers in one box is p-0.2 kg so, multiply that by 5 and you get 5p-1 kg So if you got several results, why did you not bother to show how you arrived at each? Maybe one of them was right!

*December 13, 2016*

**mathes for management**

enter your data at this site for all the details: http://www.gregthatcher.com/Mathematics/GaussJordan.aspx

*December 13, 2016*

**Calculus AB**

what you want is the moment when the particle changes direction. That is, when v(t) = 0. t=2 is the first such time, so you just want s(2) = 55 a(2) = -18 Makes sense, since the velocity is changing from + to -, that the acceleration would be negative. If you want the values ...

*December 13, 2016*

**Algebra**

the slope of AC is -1/3 Thus the line AC is y-2 = -1/3 (x+3) The diagonals are perpendicular bisectors. The midpoint of AC is (3,0). The slope of BD is 3, so the line BD is y-0 = 3(x-3) Note that you don't need the other vertices B and D.

*December 13, 2016*

**physics**

<0,4> + <3.46,-2> = <3.46,2> = 4N 30 degrees above the x-axis. Makes sense. Since the two vectors are the same magnitude, and 120 degrees apart, their resultant will be 60 degrees from each.

*December 13, 2016*

**Math**

the outside diameter d = 2.5/3.14 = 0.796 So, the inside diameter is 0.796-0.0625 = 0.734

*December 13, 2016*

**math**

100gal * 8.34lb/gal = 834 lb 100gal * 0.71kg/L * 2.2lb/kg * 3.78L/gal = 590 lb or, note that 1kg/L = 8.34lb/gal

*December 13, 2016*

**Math**

-3(x-5y)+1÷2 (4x-10y) assuming you mean -3(x-5y)+1 --------------- 2(4x-10y) that is (-3x+15y+1)/(8x-20y) Your syntax is a bit ambiguous. And where do you get off using the multiply operator (×) for a variable (x)? Seems a lot of trouble.

*December 12, 2016*

**h.p.**

your sum is n ? (3k-2)^2 = ?9k^2 - 12?k + ?4 k=1 ?k^2 = n(n+1)(2n+1)/6 ?k = n(n+1)/2 ?4 = 4n Add them up and you have 9n(n+1)(2n+1)/6 - 12n(n+1)/2 + 4n = n(6n^2-3n-1)/2

*December 12, 2016*

**math**

if we let t = ?x, we have t^2-6 = t t^2-t-6 = 0 (t-3)(t+2) = 0 t = 3 or -2 so, x = 81 or 16 But, since ?x is positive, only x=81 satisfies the original equation. So, the extraneous solution can arise even when the original equation is not squared. Trying to use x=16 gives us ?...

*December 12, 2016*

**Math**

3/10 * 5/10

*December 12, 2016*

**reading and preschoolers**

I don't know what the prevailing educational theory is, but the best thing you can do to help your kids enjoy reading is to read to them. That way they learn to associate reading with enjoyable experiences, and they will want to keep on reading even when story time is over...

*December 12, 2016*

**Math**

{-4,-3,-2}

*December 12, 2016*

**math**

10 - 6 = 4 4 + 10 = 14

*December 12, 2016*

**math**

land distance: x water distance: ?(1^2+(2-x)^2) Now determine the cost c(x) and find its minimum.

*December 12, 2016*

**math**

what platform? volume = area * height = 40h the well's volume is pi r^2 h = 56pi so, 40h = 56pi

*December 12, 2016*

**precalculus**

x = 4cos(t) y = 2sin(t) x^2+4y^2 = 16cos^2(t)+16sin^2(t) = 16

*December 12, 2016*

**Math**

again? What do you know about converses? p?q but what does that say about q?p ?

*December 12, 2016*

**math**

If you want to solve it, then let u=2/(x-2). Then you have u^2-3u+2=0 (u-2)(u-1) = 0 so, u=1 or u=2 Substitute that back for x and then you can find the solutions.

*December 12, 2016*

**Pre-Calculus**

think about it. (x+y)^n has + signs for all the terms. (x-y)^n has minus signs for all the odd powers of n, since that is the same as (x + (-y))^n I'm sure you can provide your own examples. You might also try this web site: http://www.wolframalpha.com/input/?i=(x-y)%5E5 ...

*December 12, 2016*

**h.p.**

time to do some thinking, using what you already know. Your sum is ?k(k+1)(2k+1) = 2k^3 + 3k^2 + k = 2?k^3 + 3?k^2 + ?k These are just special sums whose formulas you already know.

*December 12, 2016*

**math**

what do you mean by "completes"? The sequence goes on forever, as indicated by the ellipsis. And, you don't provide any choices anyway...

*December 12, 2016*

**math**

y:g = 15:3 = 5:1 = 35:7

*December 12, 2016*

**h.p.**

Your sum is 1/2^2 (1/1.2 + 1/2.3 + 1/3.4 + ...) = 1/4 ? 1/(k(k+1)) Didn't we just do this?

*December 12, 2016*

**differentiation**

http://www.wolframalpha.com/input/?i=derivative+(x%5E3%2Byx%5E3)xy%3Dx%5E5-y%5E5

*December 12, 2016*

**differentiation**

(x^3+yx^3)xy=x^5-y^5 Hmmm. I'd pull out that x^3 first, giving (1+y)x^4y=x^5-y^5 (y')(x^4y)+(1+y)(4x^3y+x^4y') = 5x^4-5y^4y' Now expand some to get the y' terms: x^4yy' + (1+y)(4x^3y) + (1+y)(x^4y') = 5x^4-5y^4y' y'(x^4y + (1+y)x^4 + 5y^4...

*December 12, 2016*

**Geometry**

think of your standard 30-60-90 right triangle, and the ratios of its sides.

*December 12, 2016*

**h.p.**

what question? You have the sum correct.

*December 12, 2016*

**h.p.**

since 1+2+...+k = k(k+1)/2 Your sum is n ? 2/(k(k+1)) k=1 see what you can do with that. Note that 1/(k(k+1)) = 1/k - 1/k+1

*December 12, 2016*

**Pre-Calculus**

well, just look at the differences: 1st differences: -2 2 6 10 2nd differences: 4 4 4 Looks quadratic to me.

*December 12, 2016*

**precalculus**

well, you know that the parabola y^2 = 4px has vertex at (0,0) focus at x=p directrix at x = -p The vertex is halfway between the focus and the directrix, at (6,-2) So, shift it right 6, down 2. Then note that the distance from the focus to the directrix is 8-4=4=2p, and you ...

*December 12, 2016*

**precalculus**

nope. See your next post. Your parabola has focus (-5/2, -2) directrix x = 11/2

*December 12, 2016*

**math**

1 2/3 (4 5/6 + 7 8/9) = 21 11/54

*December 12, 2016*

**Calculus**

the radius r is given by r^2+h^2 = s^2 r^2+h^2 = 25 h = ?(25-r^2) v = ?/3 r^2 h = ?/3 r^2 ?(25-r^2) So, for maximum v, find r when dv/dr = 0

*December 12, 2016*

**Math**

no. The converse is unknown.

*December 12, 2016*

**Algebra**

rearrange the equations. You have x+7y=34 x+7y=32 Surely it is clear that there is no common solution.

*December 12, 2016*

**Pre-Calculus**

Just replace every k with k+1. No math involved.

*December 12, 2016*

**Calculus**

just because the graph is a horizontal line, it is not necessarily the line y = 0

*December 12, 2016*

**Math (please check my answer)**

wow. Your two answers are the givens. Way to go. vertical angles SAS

*December 12, 2016*

**precalculus**

actually, you are correct, since 5pi/3 is in QIV, so with r = -14, the point is in QII, with y positive.

*December 12, 2016*

**algebra**

diam=8, so r=4 area = 2?rh = 52 h = 52/(2?r) = 13/(2?) v = ?r^2 h = 16? * 13/(2?) = 104

*December 12, 2016*

**Physics**

This article should get you off the ground, as it were ... https://en.wikipedia.org/wiki/Trajectory

*December 12, 2016*

**math**

3+4+5 = 12 You want their sum to be 24, so double each value.

*December 12, 2016*

**HELP MATH**

14 4/10 - 10 5/10 = ?

*December 12, 2016*

**Pre-Calculus**

well, just add on a k+1 term. Looks like (A) to me

*December 11, 2016*

**math**

p = 3 s = 3(2 1/2) = 7 1/2 r = 3 - 3/4 = 2 1/4 t = (2 1/4)(1 1/3) = 3 now just add 'em up

*December 11, 2016*

**Algebra**

the new dimensions are 6w and 6h, so the area is 24 * 6^2

*December 11, 2016*

**calc**

lim (1 + 1/x)^x log lim = lim x log(1 + 1/x) = lim log(1 + 1/x) / (1/x) = lim (-1/(x^2+x)) / (-1/x^2) = lim x^2/(x^2+x) = 1 so, log lim = 1 lim = e^1 = e

*December 11, 2016*

**Algebra**

you are correct: 2(x+4)(x+3)= 3(x+2)(x-1) 2(x^2+7x+12) = 3(x^2+x-2) 2x^2+14x+24 = 3x^2+3x-6 x^2-11x-30 = 0 Now use the quadratic formula.

*December 11, 2016*

**Math! Help!**

the domain of all polynomials is (-?,?). Remember that.

*December 11, 2016*

**Math! Help!**

type in your polynomials here, and see the details. http://calc101.com/webMathematica/long-divide.jsp

*December 11, 2016*

**Algebra**

h(2h-3)/2 = 126 2h^2 - 3h - 252 = 0 you are correct

*December 11, 2016*

**precalculus**

nope. The standard form for a conic is r = ep/(1 ± e sin?) try again

*December 11, 2016*

**Algebra**

there is always a b. In this case, it is zero! 2x^2+16 = 0 2(x^2+8) = 0 2(x+?8 i)(x-?8 i) = 0 ...

*December 11, 2016*

**Algebra**

so, did you apply one of the methods? recall that (a+bi)(a-bi) = a^2+b^2

*December 11, 2016*

**Algebra**

complete the square quadratic formula

*December 11, 2016*

**English**

yes. watch the capitalization, though: I said, "Where do you live?"

*December 11, 2016*

**English**

1 and 2 are common, and equivalent. #3 means that it was necessary to stay -- not leave. #4 would get raised eyebrows and a "huh?"

*December 11, 2016*

**Geoemtry**

just subtract the picture area from the total, and that is the area of the frame: (x+2*2)(x+2*4)-x^2 = 65.25

*December 11, 2016*

**Geoemtry**

No idea. What is x supposed to represent?

*December 11, 2016*

**Algebra**

-12 >= x^2+8x x^2+8x+12 <= 0 (x+2)(x+6) <= 0 You know the roots are -2 and -6, and the parabola opens upward. So, y is negative between the roots: -6 <= x <= -2 Clearly Philip did not check his answer.

*December 11, 2016*

**Algebra 2**

m = ks^6 m/s^6 = k a constant If m is replaced with 15m, then you want a new s such that 15m/s^6 = m/2^6 s^6 = 15*2^6 s = 2*15^(1/6) = 3.14 mi/hr

*December 11, 2016*

**AP Calculus AB**

If x = side parallel to river y = other sides So, you want to maximize a = xy subject to 12x + 4*2y <= 3600 See what you can do with that.

*December 11, 2016*

**algebra 1**

Or, just note that a horizontal line is expressed by the equation y = k for some k. Since we have the point (-3,-7), the line is just y = -7

*December 11, 2016*

**algebra 1**

x = -5 is a vertical line. So, you want a horizontal line through (-3,-7) Now, what might that be?

*December 11, 2016*

**algebra**

you want 55x to be a perfect square 55 = 5*11 So, if x=55, 55x=55^2 = 3025 and the square root is 55 Any value of x that is 55 times a perfect square will also work.

*December 11, 2016*

**math check my answer please**

looks good to me.

*December 11, 2016*

**Algebra**

Nope. eq #2 is better to use to substitute. It says x = 6y+49 So, use that in eq #1, and you have 8(6y+49)-11y = 170 48y+392-11y = 170 37y = -222 y = -6 x = 6y+49 = -36+49 = 13 So, where did you go wrong? How did you make this step? -11y = 170-8x y = -15.45 + 72 what did you ...

*December 11, 2016*

**Math**

(a) No If it rains, I'll eat chicken. If it rains, I'll go outside. No way to conclude that eating chicken means going outside. (b) Come on. You know that the converse of p?q id q?p. So, ... (c) ~q ? ~r (d) Nope. I'm sure you can come up with an example. (Use what ...

*December 11, 2016*

**math**

the volume will be reduced by a factor of (1/3)(1/3)(1/3) = 1/27

*December 11, 2016*

**math**

LCM(12,18) = 36

*December 11, 2016*

**Math**

google can provide many discussions, all of which will be more complete than what we can provide here.

*December 11, 2016*

**Math**

(9/4)^2 * 48

*December 11, 2016*

**Calculus**

p = 2x+y a = 1/2 y ?(x^2 - (y/2)^2) = 1/2 (p-2x) ?(x^2 - ((p-2x)/2)^2) = 1/4 (p-2x)?(4px-p^2) da/dx = p(p-3x)/?(4px-p^2) da/dx=0 when p = 3x That is, when the triangle is equilateral. I expect you saw that coming, eh?

*December 11, 2016*

**Geoemtry**

if the two sides given are corresponding sides, then they have a common ratio: x/4 = 1/(2x+7) x(2x+7) = 4 2x^2+7x-4=0 (2x-1)(x+4) = 0 x = 1/2 so the triangles have sides of 4,8 and 1/2,1

*December 10, 2016*

**Algebra2**

right you are

*December 10, 2016*

**Algebra2**

Name the property of real numbers for -2(x+4)=-2x-8 associative property of multiplication communitive property of addition distributive property associate property of addition my answer is distributive property

*December 10, 2016*

**algebra word problem**

1100x + 1800y = 288 There is no unique solution. The rates can be anywhere from x=0 and y=16% to x=y=9.93%

*December 10, 2016*

**Algebra**

solve for t when h=0.

*December 10, 2016*

**Math (calculus) (mean values)**

I agree with the average rate of change. So, you want s'(c) = 1395/28 s'(t) = 1395/(t+2)^2 so, 1395/(c+2)^2 = 1395/28 (c+2)^2 = 28 c = ?28 - 2 ? 3.29 So, the desired month is later than 3, namely 4.

*December 10, 2016*

**Functions - math**

fg = (2x+3)(ax^2+b) = 6x^2-21 2ax^3+3ax^2+bx+3b = 6x^2-21 when x=q, we have 2aq^3+3aq^2+bq+3b=6q^2-21 2aq^3+(3a-6)q^2+bq+(3b-21) = 0 check out the solutions to that at this url. The only easy one is in fact when q = -3. http://www.wolframalpha.com/input/?i=2aq%5E3%2B(3a-6)q%...

*December 10, 2016*

**Science**

The axial tilt is what causes the seasons, but it does not change. The language of the question and the choices is poorly constructed.

*December 10, 2016*

**h.p.**

google is your friend. This should help. http://www.askiitians.com/forums/Algebra/22/42109/progressions.htm

*December 10, 2016*

**Algebra**

well, what is h when it hits the ground? ...

*December 10, 2016*