Saturday

January 31, 2015

January 31, 2015

Total # Posts: 28,515

**AP Calc**

one way to tell whether you did it right, is to show your work! Then we can see how you arrived at your result.
*November 22, 2014*

**AP Calc - PS**

Ignore the comment about the length of the normal line. The question did not actually ask for the distance, just the point on the curve. We just wanted to find the point where the normal to the curve passes through (8,0)
*November 22, 2014*

**AP Calc**

One way is to find the length of the line normal to the curve which passes through the point. The tangent at any point has slope 1/√x, so the normal line has slope -√x. The line through (8,0) and (x,2√x) has slope 2√x/(x-8), which we want to be -√...
*November 22, 2014*

**algebra**

#16 and #18 are correct though I must say the typesetting needs work The others, no idea. No diagrams or graphs here.
*November 22, 2014*

**AP Calc**

yep. nice feeling when it clicks, no?
*November 22, 2014*

**AP Calc**

the max or min area occurs where dA/dx = 0 dA/dx = 2(5x^4-6x^2+1) = 2(5x^2-1)(x^2-1) So, dA/dx = 0 at x = ±1/√5, ±1 Naturally, A=0 at x=1, so the obvious choice is x = 1/√5. Thus, the maximum possible area is 2/√5(1/5-1)^2 = 32/(25√5)
*November 22, 2014*

**Physics**

the tension is just the weight of the iron, minus the weight of the water displaced.
*November 22, 2014*

**Chemistry**

there are .03716*0.510 = 0.01895 moles of HCl Each mole of HCl neutralizes 1 mole of NaOH So, there were .01895 moles of NaOH in 0.025L of solution. Thus, the solution was 0.01895mol/0.025L = 0.758M
*November 22, 2014*

**statistics**

I found that the inversenorm is 83.5922, what do i do after this?
*November 22, 2014*

**statistics**

Ok but before we start we usually write downwhat the mean equals, standard deviation and X. Im still confused on which one would be the mean.
*November 22, 2014*

**statistics**

Hello i am doing assignment in my math 20-2 class. The question is: data collected of cars passing on the road revealed that the average speed was 90 km/h with a standard deviation of 5 km/h and the data which is normally distributed a policeman is assigned to set photo radar ...
*November 22, 2014*

**maths**

2score=40 2dozen=24 so, the cost will be 24/40 * 280
*November 22, 2014*

**mechanics**

stone 1: height is 40 + 20t - 9.8t^2 stone 2, 3 seconds later: 30(t-3) - 9.8t^2 they meet (or pass closely) when the heights are equal: 40 + 20t - 9.8t^2 = 30(t-3) - 9.8(t-3)^2 So, just solve that for t, and then plug that value into either equation.
*November 22, 2014*

**Algebra 2**

OK. But this is Algebra I, which you have presumably already passed. But, moving right along, 5x^2 - 16x + 6 = 0 The quadratic formula says that the roots of ax^2+bx+c are x = (-b±√(b^2-4ac)/2a So, for this problem, that means x = (16±√(16^2-4*5*6))/...
*November 22, 2014*

**Algebra 2**

Hmmm. By algebra 2 you should surely know how to solve quadratic equations... x^2+(4-2x)^2=10 x^2+16-16x+4x^2-10 = 0 5x^2 - 16x + 6 = 0 ...
*November 22, 2014*

**Math**

Review your text on partial fractions. You know that the sum will be (Ax+B)/(x^2+1) + C/(x-2) + D/(x-2)^2 If you put all that over a common denominator, you get (Ax+B)(x-2)^2 + C(x-2)(x^2+1) + D(x^2+1) ------------------------------------------------ (x^2+1)(x-2)^2 Expand ...
*November 22, 2014*

**Math**

I get (27x+14)/(x^2+1) - (27/25)/(x-2) + (8/5)/(x-2)^2
*November 22, 2014*

**calc**

I will assume you meant f'(0) = 3 in that case, f"(x) = 10 f'(x) = 10x+c 3 = 10*0+c -----> c = 3 f'(x) = 10x+3 f(x) = 5x^2+3x+c 1/3 = 5*0+3*0+c c = 1/3 f(x) = 5x^2 + 3x + 1/3 I figure you can now find f(1).
*November 22, 2014*

**calc - typo**

Sorry. If f'(x) is a constant, f"(x) = 0 Fix the typo and try again.
*November 22, 2014*

**Fencing Applicatio**

Clearly, P(f) = 500+20f There are lots of good graphing web sites. Or, heck, you can get out the old graph paper and a pencil!!! Imagine!!
*November 22, 2014*

**Calc**

You can set it up here to find out: http://mathworld.wolfram.com/RiemannSum.html
*November 22, 2014*

**Math**

clearly d=6 since T6 = T4+12 So, T1 = T4-18 = 19 T7 = T4+18 = 55
*November 22, 2014*

**math**

since the height:shadow ratio is the same for both, h/12 = 8/3
*November 22, 2014*

**algebra**

(3x)(4x^4) - (3x)(5x) 12x^5 - 15x^2 or, 3x^2(4x^3-5) Not sure what you mean by simplify
*November 22, 2014*

**algebra**

you have the formula. Just use your calculator, or any online calculator to find the values. For #1, just plug in t=5 For #2, just solve 3000 e^(0.07t) = 17700 e^(0.07t) = 5.9 0.07t = ln 5.9 ... For #3, just solve e^(0.07t) = 2
*November 22, 2014*

**jm**

study, study, study!
*November 22, 2014*

**MATH**

Cut a slice through the pyramid, perpendicular to one of the faces. You have a right triangle, with the base 18.6/2 and the hypotenuse of 14.8 Now just use the Pythagorean Theorem to find the height.
*November 21, 2014*

**Math**

First, you have to note that since log(z) is defined only for z>0, we must have x < -4/5 or x > 4/7 In that domain, there is an inverse. You made an error in your last line, which is where the trouble arises. It should be x = log (6y+5)/(7y-4) e^x = (6y+5)/(7y-4) e^x(...
*November 21, 2014*

**Precalc**

Since I have no diagram, I make the following assumptions. We can view the scene from high above, so the elevations of A and B make no difference. C is between A and B, but not on the line AB. The fire at F is on the opposite side of AB from C. So, we can use the law of ...
*November 21, 2014*

**math help**

so, if you divide f(x) by (x-4i) you get f(x) = (x-4i)(x^2+12x+32) = (x-4i)(x+4)(x+8)
*November 21, 2014*

**Algebra 2**

since we know y = 4-2x, use that: x^2+(4-2x)^2 = 10 5x^2 - 16x + 6 = 0 Now it's just the normal routine for solving a quadratic equation.
*November 21, 2014*

**Math**

well, it's 4 hours, and a full rotation is 12 hours. Tat help?
*November 21, 2014*

**Physics**

if we ignore air resistance, the bale's horizontal speed stays the same as the plane's. So, it remains directly below the plane till it hits the ground.
*November 21, 2014*

**Algebra 2**

you can easily check by substituting in those values to see whether they work in both equations. FYI, they do.
*November 21, 2014*

**math**

Actually, she can make 28 1/3 tacos. Since a taco uses .3 lbs, there is just .1 lb left over.
*November 21, 2014*

**Math**

m(x) = 4x, so m(f) = 4f = 12x^2 g(x) = 3x-1, do g(m) = 3m-1 = 36x^2-1 Using your nested notation, g(m(f(x)) = 3(m(f(x))-1 = 3(4(f(x))-1 = 3(4(3x^2))-1 =36x^2-1
*November 21, 2014*

**mixture problems**

If x is 25%, then the rest (10-x) is 50%. So, .25x + .50(10-x) = .40(10)
*November 21, 2014*

**Math Help!!!**

review your notes. There are whole numbers natural numbers integers rational numbers irrational numbers
*November 21, 2014*

**math**

9i/(-3+8i) = (9i)(-3-8i)/(3^2+8^2) = (72-27i)/73
*November 21, 2014*

**Math**

10000((1+.07*5)-(1+.06)^5) = 117.74
*November 21, 2014*

**English PLEASE!!!**

you picked a thermometer, which just measures. A thermostat affects things. So, the church influences society (by transforming the mores).
*November 21, 2014*

**algebra**

s = kv^2, so s/v^2 is constant. That means you want s where s/55^2 = 81/30^2
*November 21, 2014*

**math**

4/(5x-3)
*November 21, 2014*

**Math Help!!!**

#5,#8 no idea - no diagram #7 (-8,-10) (-8,10) is reflected through the origin. the others look good.
*November 21, 2014*

**Math**

amplitude is (125-99)/2 = 13 The center line is 125-13 = 112, not 122 2pi/k = 2*14, so k = pi/14 Since f(15) = 125, the max, that means that f(1) = 99, the minimum cosx has a max at x=0, so -cos(x-1) has a minimum at x=1. f(t) = 13 cos(pi/14 (x-1)) + 112 The graph at http://...
*November 21, 2014*

**MAth Help**

the vertical shift is +80 The curve oscillates about f=80 with an amplitude of 22. How did you arrive at 6?
*November 21, 2014*

**Math**

the vertex of y=a(x-h)^2+k is at (h,k)
*November 21, 2014*

**algebra**

you need x^2+y^2 = 100 So, which point works?
*November 21, 2014*

**math**

3:00 pm is 15 hours after midnight 6sin(pi/12 (15-11))+19 6sin(pi/12 * 4)+19 6sin(pi/3)+19 6*0.866+19 24.196 6+19=25 is the maximum possible temperature
*November 21, 2014*

**algebra**

x(2x+3) = 119
*November 21, 2014*

**math**

since the seat started at 60, it takes 1/2 turn to get back around to that height on the other side. That's 3.5 minutes
*November 21, 2014*

**Math**

√112 = ?
*November 21, 2014*

**math**

A B is an equation, not an expression C is too, and not even algebraic D is just a constant expression
*November 21, 2014*

**math**

Yes. In 1/4 turn, the wheel will be at the top.
*November 21, 2014*

**math**

any value between 64.8 and 65.4. For example, 65.0 64.800000000000000000000000000001 65.399999999999999999999999999999
*November 21, 2014*

**plz help. math**

Beckey! steven is mooching off your postings! Or, why not just use a single name, steven old buddy?
*November 21, 2014*

**plz help. math**

165 = 3*5*11 So, the table could seat 3,5,11,15,33,or 55 people The restriction on table size means that the only choices are 3,5,11,15
*November 21, 2014*

**algerbra**

If you mean √2, then the possibilities for a right triangle are 1 and 11, since 1^2 + √2^2 = 3 √2^2 + 3^2 = 11 If you meant 1/2, then the choices are √11/2 and √37/2 (1/2)^2 + 3^2 = √37/2 (1/2)^2 + (√11/2)^2 = 3
*November 21, 2014*

**algerbra**

try again. what is /2 ?
*November 21, 2014*

**algerbra - incomplete**

First, provide the length of the other side.
*November 21, 2014*

**algebra**

one way is to complete the square: x^2 + 5x + 3 = 0 x^2 + 5x = -3 x^2 + 5/x + (5/2)^2 = -3 + (5/2)^2 (x + 5/2)^2 = 13/4 x + 5/2 = ±√13/2 x = (-5±√13)/2 You get the same answer using the quadratic formula.
*November 21, 2014*

**geometry**

There are lots of sample constructions here: http://jwilson.coe.uga.edu/EMAT6680Fa08/Ruff/TriangleConstructions/TriangleConstructions.html
*November 21, 2014*

**math**

play around some at http://www.miniwebtool.com/hex-calculator/
*November 21, 2014*

**MATH**

Assuming hexadecimal, D5C
*November 21, 2014*

**math**

better check your text some more. With initial height h and initial velocity v, the height y at time t is y(t) = h + vt - 16t^2 So, you have y(t) = 20 + 5t - 16t^2 When working problems, don't forget what you know about parabolas -- their roots, vertex, etc.
*November 21, 2014*

**college algebra**

solve for t in 5300 * 1.077^t = 8000 * 1.056^t t = 20.9 years or, 21, since the interest is credited at year's end.
*November 21, 2014*

**math (fractions)**

mixed number
*November 21, 2014*

**DON VASCO PATNA**

you want 3/4 the concentration, so you must have 4/3 the volume.
*November 21, 2014*

**algebra 2**

just factor out the perfect squares: √25 = 5 √a^2 = a √b^5 = b^2 √b √(a^2/b^2) = a/b
*November 21, 2014*

**Algebra II**

you do have to remember to insert zeros for missing powers. Play around at http://www.mathportal.org/calculators/polynomials-solvers/synthetic-division-calculator.php
*November 21, 2014*

**algebra**

861bagels/3days = 287 bagels/day
*November 20, 2014*

**math**

father = 4y = brother+28 brother = father-28, so 4y-28
*November 20, 2014*

**Algebra**

31500(1+.068/365)^(365*30)
*November 20, 2014*

**math**

I'm sure you have a formula. For further reading, see wikipedia's article on Richter magnitude scale. It also has a table of sample values, so you can check your answer.
*November 20, 2014*

**Math**

2000(1+.05*18)
*November 20, 2014*

**Math**

Hmmm. Better make that Tn = n+3 if n is odd
*November 20, 2014*

**Math**

Looks to me like Tn = 2n+2 if n is odd Tn = 4*2^(n/2) if n is even
*November 20, 2014*

**Physics**

Vy = 7.4*1.7 = 12.58 m/s So, the final speed is √(8^2+12.58^2) = 14.91 m/s
*November 20, 2014*

**math**

Ah. missed the 5. So, yes, 5 3/2 = 13/2
*November 20, 2014*

**math**

12/8 = (3*4)/(2*4)= 3/2
*November 20, 2014*

**MTH001**

So, what are your truth tables?
*November 20, 2014*

**math**

If the original price is p, then the discounted price is 0.85p After the fee, the final price is 0.85p + 12 = 100 p = 103.53 That is not the greatest price; it is the only price which will cost exactly $100. It is the greatest price if she does not want to spend more than $100.
*November 20, 2014*

**Math Help**

#1. The figures are similar. Since 84/28=3, c/22 will also be 3. SO, c=66 #2. I believe you are correct. The 2nd triangle is 1/3 as big as the first. #3. No. Just double each coordinate.
*November 20, 2014*

**math ?**

no, but r=0, as a little synthetic division will show.
*November 20, 2014*

**Math**

s = 1/2 at^2 v = at, so t = v/a s = 1/2 a(v^2/a^2) = 1/2 v^2/a = 112^2/64 = 196 ft
*November 20, 2014*

**Math**

v(t) = 3/2 t^(1/2) s(t) = t^(3/2) + c 33 = 9^(3/2)+c, so c=6 s(t) = t^(3/2) + 6
*November 20, 2014*

**Calculus**

The slope at any point is y' = -7cosx The line through (0,0) and (x,-7sinx) has slope -7sinx/x. That means we need -7sinx/x = -7cosx x = tanx Solutions are x = 4.493409, 7.725252, 10.904122, ... The slopes of the lines are 1.521, -0.899, 0.639 See the first couple of lines...
*November 20, 2014*

**math, please help**

Looks to me like f(x) = (x-(-2))(x^2+4x+2)+0
*November 20, 2014*

**Math Help**

Looks good to me. Except "¢10 cents" is bogus notation. Try just 10¢
*November 20, 2014*

**Math (IB Calc)**

since cos(x) has it max at x=0, and you have a min there, A(x) = 50(1-cos(x)) now just plug in x = 2pi/3
*November 20, 2014*

**Math Help**

The scale factor of 2 means that (x,y) -> (2x,2y) So, you need to double the coordinates. Not sure what the heck you devised.
*November 20, 2014*

**Math help!!!**

First step: list the ordered pairs. (Mona,Kellie) ...
*November 20, 2014*

**Math**

You don't say whether these are courses from due North, or from the positive x-axis. I'll do the latter. the x- and y-components of the vectors are (450 cos65°, 450 sin65°) = (190.18,407.84) (35 cos20°, 35 sin20°) = (32.89, 11.97) Add them up to get (...
*November 20, 2014*

**english**

The current exchange rate is 1 yen = 0.0085 dollar So, a little less than 1 cent per yen. Or, a little more than 100 yen/$ 1/.0085 = 111.65
*November 20, 2014*

**English**

and you are correct.
*November 20, 2014*

**Pre Calculus**

P(0) = 160/(1+9) = 16 Just plug in t=6 for the population after 6 years. Note that as t->∞, e^-0.165t -> 0 So, the maximum possible population is 160 See http://www.wolframalpha.com/input/?i=160%2F+%281+%2B+9e^%28-0.165t%29%29+for+t+%3D+0..50
*November 20, 2014*

**Pre Calculus**

25*e^-5k = 18.4 So, k = ln(18.4/25)/-5 = 0.0613 So, now just solve for t in 25*e^-.0613t = 20.5
*November 20, 2014*

**Pre Calculus**

solve for m months in (1+.065/12)^m = 2
*November 20, 2014*

**Pre Calculus**

after t years, the population p is p = 127000 * 0.976^t Now just plug and chug
*November 20, 2014*

**algebra**

k = (h+j)+14 = x + x+4 + 14 = 2x+18
*November 20, 2014*

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