Wednesday

February 22, 2017
Total # Posts: 48,712

**Mathematics**

xUy = {a,b,c,d,f,h,l,o,u} that is, everything in either set x?y is what is in both sets x'?y' = (xUy)' everything except xUy

*December 29, 2016*

**Maths sequence and series**

well, log(xy) = logx + logy so, you have an AP with a = logx d = logy S30 = 2(3^30-1)/(3-1) ? 3^30 (10^.477)^30 ? 10^14.313 log2 = .3010, so 10^.313 * 10^14 > 2*10^14

*December 29, 2016*

**SAT Math**

4w(z) = 4(6+3z) = 96 24+12z = 96 12z = 72 z = 6 the x in w(x) is just a placeholder. w(x) = 6+3x w(z) = 6+3z w(12)= 6+3*12 and so on.

*December 29, 2016*

**science**

KE = 1/2 mv^2 watch the units.

*December 29, 2016*

**Maths**

the angles A,B,C are x, x+y, x+2y They sum to 180, so 3x+3y = 180 x+y = 60 angle B is x+y, so ...

*December 29, 2016*

**Business mathematics**

the number who passed something is 18+17-11 = 24 Now just draw the usual Venn diagram to answer the questions.

*December 29, 2016*

**Maths**

12 = 1/3 of 36

*December 29, 2016*

**Calculus**

well, for any value except x=0, the quotient is 1. So, the limit is clearly 1. x/x is 1 everywhere except x=0.

*December 28, 2016*

**Science**

F = Gmm/r^2 = G*10*10/50^2 If their centers of mass are 50m apart.

*December 28, 2016*

**Physics**

looks right to me.

*December 28, 2016*

**Math**

The base is x^2 The height is 30/x^2 So, the area is x^2 + 4(x)(30/x^2)

*December 28, 2016*

**Math**

1.12(475/.95) = 560

*December 28, 2016*

**Math(A.P)**

a+3d = 3a a+6d = 2(a+2d)+2 now just crank it out.

*December 28, 2016*

**Computer Science**

there are lots of snippets of C code online. google is your friend.

*December 28, 2016*

**maths go with @MathMate**

My bad - I missed the fact that it was open on the ends.

*December 28, 2016*

**maths**

just compute the volume inside the metal and subtract it from the outside volume. ?/4 * 24^2 * 55 - ?/4 * 20^2 * 51 = 2820? cm^3

*December 28, 2016*

**Math**

Manufacturer's Recommended Price also seen as MSRP Mfgr's Suggested Retail Price

*December 28, 2016*

**math**

The distance from (5,12) to the line is |2*5 - 1*12 + 3|/?(2^2+1^2) = 1/?5 So, the circle is (x-5)^2 + (y-12)^2 = 1/5 check: The line and the circle must intersect at a single point. (x-5)^2 + ((2x+3)-12)^2 = 1/5 This has a solution only at x = 23/5

*December 28, 2016*

**Maximum value**

if x^2-ax >= 0, |x^2-ax| = x^2-ax That is, if x(x-a) >= 0 x <= 0 or x >= a These intervals are outside the domain. If x^2-ax < 0, then |x^2-ax| = -(x^2-ax) That is, if x(x-a) < 0 So, for 0<=x<=a the maximum value of f(x) is at x = a/2 See a sample graph...

*December 28, 2016*

**Finding Range - Maths**

the domain is (-?,-100)U(100,?) Since f(x) = 1/x^2, the range is the same on both those intervals. So, R = (0,1/100^2) if the domain is (-0.1,0.1) then f(0) is undefined. The range is thus (1/0.1^2,?) since f(x) is symmetric about x=0.

*December 28, 2016*

**maths**

They sum to 360, so 3x+5x+7x+9x = 360 find x, then you can get the angles. (not the lengths)

*December 28, 2016*

**Math**

If the smallest (2nd) piece is x, then just add up the lengths: 2x + x + x+4 = 10 now find x, and you can figure the length needed.

*December 27, 2016*

**geometry**

so the arcs will overlap, giving two points on the bisector line.

*December 27, 2016*

**Math**

My first estimate of T'(10) would be the average change from t=8 to t=12: (T(12)-T(8))/(12-8) = (86-73)/4 = 3.25 Using a forward estimate from t=10, it might be (T(12)-T(10))/(12-10) = (86-80)/2 = 3.0

*December 27, 2016*

**Algebra2**

12+x = .85(15+x)

*December 27, 2016*

**trig prove plz anyone help**

recall that tan(A-B) = (tanA-tanB)/(1+tanAtanB)

*December 27, 2016*

**Math**

10000*0.10*n = 5000 or, each year's interest is $1000, so ...

*December 27, 2016*

**Math**

j = t-5 jt = 126 Hint: 9*14 = 126

*December 27, 2016*

**Math - typo?**

you sure that h(2) = 44? I suspect it is h(3) = 44.

*December 27, 2016*

**Maths - Cartesian Equations**

(y/2)^2 = (t + 1/t)^2 = t^2+2+1/t^2 x^2 = (t - 1/t)^2 = t^2-2+1/t^2 y^2/4 - x^2 = 4 or in standard form, y^2/16 - x^2/4 = 1

*December 27, 2016*

**Maths - Parametric Equations**

(a) surely you can just plug in a value for t! x = 2cos(8?/3) = 2(-1/2) = -1 y = 2sin(8?/3) = 2(?3/2) = ?3 (b) Now you have a point and a slope, so the tangent line is y-?3 = 1/?3 (x+1)

*December 27, 2016*

**Maths**

5x * 2x = 360 Find x, and then you know the base and height.

*December 27, 2016*

**Math**

If there are x tables, then we have 9(x-2) = 6x+3

*December 27, 2016*

**Math**

just consult a map. I mean, there aren't that many oceans, right?

*December 27, 2016*

**value**

misspelled

*December 27, 2016*

**social stadies**

misspelled

*December 27, 2016*

**economics**

Note that the marginal cost is the derivative of the total cost. Note that the average variable cost is the integral of the marginal cost over the given interval. AVC is a minimum when MC = AVC. A nice video is here: https://www.youtube.com/watch?v=TBCAHbhhvTg

*December 27, 2016*

**calculus help me plz too hard show step**

R dQ/dt + Q/C = 0 R dQ/dt = -Q/C dQ/dt = -Q/(RC) dQ/Q = -1/(RC) dt lnQ = -1/(RC) t + k at t=0, we know that Q=Qo, so lnQo = k so, lnQ = -1/(RC) + lnQo ln(Q/Qo) = -1/(RC) t Q/Qo = e^(-t/RC) Q = Qo e^(-t/RC)

*December 27, 2016*

**Math**

x^2 = ln(xy) x^2 = lnx + lny 2x = 1/x + 1/y y' y' = y(2x - 1/x) y'(1) = e(2-1) = e or, more explicitly, ln(xy) = x^2 xy = e^(x^2) y = e^(x^2)/x y' = e^(x^2) (2x-1)/x^2 y'(1) = e^1 (2-1)/1 = e

*December 27, 2016*

**math**

first step: STOP REPEATING THE POST!

*December 26, 2016*

**math**

surely you can do 3*5 now just move the decimal point as needed. .5% = 1/200 I'm sure you can do 3200/200 .5 is 1/2, but .5% is not. 10.00 - 8*0.82 = ?

*December 26, 2016*

**algebra**

http://www.wolframalpha.com/input/?i=plot+y%3C%3D+3%2F2+x%2B3,+2x+%2B+y%3C10,+y%3E-1

*December 26, 2016*

**Proofing Maths**

of course. |x-y| = |x + (-y)| <= |x| + |-y|

*December 26, 2016*

**Physics**

7@N40W = <-4.50,5.36> 10@E = <10.00,0.00> now just add them up and use the normal distance formula.

*December 26, 2016*

**algebra**

They are the same equation, so infinitely many solutions.

*December 26, 2016*

**Math**

h'(1) = f'(f(1)) f'(1) = f'(3)f'(1) = 7*2 = 14

*December 26, 2016*

**algebra**

256z^2 -4 -192z^2 +3 = (256z^2-4) - 3(64z^2-1) = (16z-2)(16z+2)-3(8z-1)(8z+1) = 4(8z-1)(8z+1)-3(8z-1)(8z+1) = (8z-1)(8z+1) cool, eh?

*December 26, 2016*

**Maths(Surface integral/flux)**

recall polar coordinates: x^2+y^2 = r^2 dx dy = r dr d?

*December 26, 2016*

**calculus too hard**

just keep on ... dx/dt = k(a-x) dx/(a-x) = kt ln(a-x) = kt^2/2 or, using a new k ln(a-x) = kt^2 a-x = e^(kt^2) Now use your initial conditions to find a and k.

*December 26, 2016*

**math**

or, see how much 1 set of 9 notes is worth: 2*3+5*2+10*4 = 56 728/56 = 13 So there are 13 sets of 9 notes...

*December 26, 2016*

**math**

Add up the value of each kind of note: 2(3x)+5(2x)+10(4x) = 728 now find x, and the total number of notes is 3x+2x+4x = 9x

*December 26, 2016*

**ACCOUNTING**

<font Courier/> hello

*December 26, 2016*

**ACCOUNTING**

p { font-family: "Times New Roman", Times, serif; } Times Roman

*December 26, 2016*

**ACCOUNTING**

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*December 26, 2016*

**ACCOUNTING**

???? ???? ???? ???? ? ?? ? ?? ? ?? ? ?? ???? ???? ???? ???? ???? ???? ???? ???? ????????????????????? ? ????? Some Text ?? ? ????? in the box ?? ?????????????????????? ? ??????? ?? ? ??????? ?? ?????????????????????? ?????????????????????

*December 26, 2016*

**MATH**

recall that log and ^ are inverse operations: 10^(log x) = x log(10^x) = x so, now you can do 10^(log_10(a)) = 10^4 a = 10^4

*December 26, 2016*

**Math**

y-x = 4 3x = 2y+4

*December 26, 2016*

**Proofing Maths**

google is your friend. Lots of help with the triangle inequality. Once you have convinced yourself that |x+y| <= |x| + |y| then it is a simple extension to get to your problem.

*December 26, 2016*

**Math**

since tests count twice, add them twice. The total points must be the same as 5 scores of 70: 88+82+84+2x = 70*5 x = 48

*December 26, 2016*

**MATHS**

minar?

*December 26, 2016*

**physic**

If you mean a point on the edge of the wheel, then recall the relationship between linear speed v and angular speed ?: v = r? since arc length s = r?

*December 26, 2016*

**science**

Since the weight is proportional to 1/r^2, 5 times the weight means 1/?5 times the distance.

*December 26, 2016*

**Math**

The trick here is to recall that 1 = cos^a+sin^2A cos2A/(1+sin2A) (cos^2A-sin^2A)/(1+2sinAcosA) (cos^2A-sin^2A)/(cos^2A+2sinAcosA+sin^2A) (cosA+sinA)(cosA-sinA)/(cosA+sinA)^2 (cosA-sinA)/(cosA+sinA) (cotA-1)/(cotA+1)

*December 26, 2016*

**Math**

when you post something, bring it up so you can see what actually made it.

*December 25, 2016*

**Math**

better try again.

*December 25, 2016*

**Math**

nope. -8-2 = -10

*December 25, 2016*

**Math**

Nope. -8(2) + 5 = -16 + 5 = -11

*December 25, 2016*

**Math**

not complicated at all. Any group of n repeating digits (.ddd) can be represented by ddd/999 where there are as many 9's as digits. So, 0.583 = 583/999 To see why this is so, consider what happens if you assign the value to a variable, say x: x = 0.583583583... Now, shift ...

*December 25, 2016*

**math-integrating factor**

well, did you try it? d/dx (y/x) = 1/x y' - y/x^2 so, now you have an exact differential, which is what you wanted.

*December 25, 2016*

**Functions**

when after x months, .15x = 9.25

*December 25, 2016*

**mathes**

since a=25 and b=20, the ellipse can be represented by x^2/25^2 + y^2/20^2 = 1 Now just find 2x when y=10.

*December 25, 2016*

**Math**

let u = cotx. Now the integrand is just -u^4 du

*December 25, 2016*

**Math**

RT = 2.5 tan40° RQ = 2.5 tan65° so, tan <RQT = RT/RQ

*December 25, 2016*

**mathematics**

hint: find the center, then a=2 b=3 c^2 = a^2-b^2

*December 24, 2016*

**A.p.**

come on, guy. Algebra I here.

*December 24, 2016*

**a.p.**

1st 100 ft = 250 then .25(x-100)

*December 24, 2016*

**ellipse - ahem**

b^2+c^2 = a^2

*December 24, 2016*

**algebra**

well, did you try your solution to see whether it works? 3(-5)-(-6) = -15+6 = -9 4(-5)-(-6) = -20+6 = -14 Clearly not. SO, what happened? Using the first equation, we see that y = 3x+9 Using that in the other equation, we get 4x-(3x+9) = -4 4x-3x-9 = -4 x = 5 So, y = 3x+9 = 15...

*December 24, 2016*

**mathematics**

what, you can't plug in the numbers? 1/f = 1/(7/2) - 1/(15/2) 1/f = 2/7 - 2/15 1/f = (30-14)/105 1/f = 16/105 f = 105/16

*December 23, 2016*

**maths 10-3**

(3 ft 2 1/2 in)/2 = (3/2 ft)+(1 1/4 in) = 1 ft 6 in + 1 1/4 in = 1 ft 7 1/4 in

*December 23, 2016*

**algebra**

(t+2v)(t+2v) (z-3)(z+14)

*December 23, 2016*

**math**

just plug into your formula for volume: v = 4/3 * 3.14 * 10^3 = ?

*December 23, 2016*

**Math**

51 = .85x

*December 23, 2016*

**Alegebra**

dilation relative to (0,0) just means you multiply all the coordinates by the dilation factor. You are increasing all the distance from the origin by the scale factor. For example, with a scale factor of 3, (2,-12)->(6,-36) With a scale factor of 1/4, (-12,8)->(-3,2)

*December 23, 2016*

**physics**

maximum range is when the angle is 45° so, minimum velocity need for a given range is at 45° Now just use your range formula using 45° to see what initial velocity has that range. That is your minimum required velocity. Technically, a 1000km range has to include ...

*December 23, 2016*

**Algebra**

y = x+13 y = 2x+5 Looks like x=8 and y=21

*December 23, 2016*

**value**

since distance = speed * time, we need at time t hours 60t = 5(t - 1/3)^2 note: 10 km/hr is speed, not acceleration. I used a = 10 km/hr^2

*December 23, 2016*

**maths**

C is 1/4 of the way from A to B. So, add 1/4 of the difference in coordinates (B-A) to A: C = (-18,8)+(2,-4) = (-16,4)

*December 23, 2016*

**Maths**

The line has slope 3/4 So, the radius has slope -4/3 Thus, the line with that slope through P is y-4 = -4/3 (x+3)

*December 23, 2016*

**Maths**

Suppose the slopes are equal: -k/(k+1) = -4/(3k) -3k^2 = -4k-4 3k^2-4k-4 = 0 (3k+2)(k-2) = 0 k = 2 or -2/3 as you calculated. Using you pairs of equations, we see that they can be written 4x+6y=16 4x+6y=4 2x-y = -24 2x-y = 2 In both cases the lines are distinct and parallel, ...

*December 23, 2016*

**Pre-calc Math**

Draw a diagram. T = top of tree S = tip of shadow B = base of tree Draw a line L horizontally from the tip of the shadow. Drop a vertical from the tree base to intersect line L at P. Let x = SP y = PB h = BT (height of tree) y = 210 sin21° x^2+y^2 = 210^2 (y+h)/x = tan51&...

*December 23, 2016*

**Pre-calc Math**

Draw a diagram. If the height is h, h cot84° - h cot87° = 95 (a) the distance d is found via h^2+(h cot87°)^2 = d^2

*December 23, 2016*

**Pre-calc Math**

sinB/109 = sin24°/78 That gives you B Then C = 180-(A+B) then c/sinC = a/sinA

*December 23, 2016*

**Algebra**

just use the quadratic formula: x = (2a±?(4a^2+16a))/(2a) = (2a±2?(a^2+4a))/(2a) = 1±?(a^2+4a)/a or = 1±?(1 + 4/a)

*December 23, 2016*

**Transformations**

stretch = dilation shift = translation there is no reflection involved. It amazes me that you could not figure that out. I assume you know that x is horizontal and y is vertical...

*December 23, 2016*

**Transformations**

y = 4(2(x-3/2))^2+1 shift right 3/2 horizontal shrink by 2 vertical stretch by 4 shift up 1 or y=16(x-3/2)^2+1 shift right 3/2 vertical stretch by 16 shift up 1

*December 23, 2016*

**SAT Math**

m% of 50 is m/100 * 50 = m/2 20% of n is 20/100 n = n/5 m/2 = n/5 m/n = 2/5 check: m=50, n=125 50% of 50 = 25 20% of 125 = 25 m/n = 2/5

*December 23, 2016*

**math**

3(2tanA)/(1-tan^2A) = 2tanA+1 Now just solve for tanA

*December 22, 2016*

**Grade 6 Maths**

Looks like 7n+1

*December 22, 2016*

**SAT Math Question?**

p-q is largest if p is big and q is small: p-q <= 30-7 = 23 similarly, p-q >= 15-19 = -4

*December 22, 2016*