# Posts by Steve

Total # Posts: 50,544

**Pre-Cal: Limacons**

you might want to play around on wolframalpha.com by varying the parameters. You have the equations for sample curves, as well as examples in your book and online. You know the general form is r = b + a cos?. And of course, since sin? = cos(?/2 - ?) the two are just rotations ...

**Eigenvalues**

using your standard methods, see whether you come up with this: http://www.wolframalpha.com/input/?i=eigenvalues+%7B%7B10,10%7D,%7B-4,-3%7D%7D

**Mathematical Induction**

prove that P(1) is true: 8 = 1/2 1(3*1+13) = 16/2 = 8 Assuming P(k), see what P(k+1) means: 8+11+...+(3k+5)+(3(k+1)+5) = k/2 (3k+13) + (3(k+1)+5) = k/2 (3k+13) + 3k+8 1/2 (3k^2+13k + 6k+16) = 1/2 (3k^2+19k+16) = 1/2 (k+1)(3k+16) = 1/2 (k+1)(3(k+1)+13) = P(k+1) So, P(1) and P(k...

**Calculus**

f"(x) = 2+cos(x) f'(x) = 2x + sin(x) + c1 f(x) = x^2 - cos(x) + c1*x + c2 f(0) = -1+c2 = 5 --> c1=6 f(?/2) = ?^2/4 + c1*?/2 + 6 = -3 f(x) = x^2 - cos(x) + (?/2 - 18/?)x + 6

**calculus**

To evaluate ?e^(-?x) dx let z^2 = x 2z dz = dx and now you have ?e^-z 2z dz Now let u = z dv = e^-z dz du = dz v = -e^-z dz ?u dv = uv - ?v du, so ?2z e^-z dz = -2ze^-z + 2?e^-z dz = -2ze^-z - 2e^-z Now evaluate that at the limits and you get ?[0,?] e^(-?x) dx = 2

**calculus**

To evaluate ?x/(x^4+25) dx let z^2 = x 2z dz = dx now you have 1/2 ?1/(z^2+25) dz That is just a standard integral, giving (1/2)(1/5)arctan(z/5) evaluate at the limits and you get -?/20

**calculus**

let u=x^2 and it's a cinch.

**Algebra**

3/5 + 2/x = 1 now just solve for x

**Math**

P(red) = (#red)/(#total) Now take 90*P(red)

**calculus**

let u^2 = x-1 x = 1+u^2 2u du = dx Now your integrand is 1/(1+u^2) * u * 2u du = 2u^2/(1+u^2) du = 2(1 - 1/(1+u^2)) du That integrates to 2(u - arctan(u))+C = 2(?(x-1) - arctan?(x-1))+C

**calculus**

using partial fractions, you have 11 * 1/[(x-5)(x^2+5x+25)] = 11/75 * [1/(x-5) - (x+10)/(x^2+5x+25)] Doesn't look much better, does it? Well, the 1/(x-5) integrates easily enough. The other term has to be worked into something more standard. x^2+5x+25 = (x + 5/2)^2 + 75/4 ...

**Calculus**

Using a scaled-up 3-4-5 triangle, the distance after 10 minutes is 1/6 (500) miles. If z is the distance between them after t hours, then z^2 = (300t)^2 + (400t)^2 z dz/dt = 2*300^2 t + 2*400^2 t Now just plug in your numbers with t = 1/6

**calculus**

Using partial fractions, (x^2-x+12)/(x^3+3x) = 4/x - (3x+1)/(x^2+3) The 4/x integrates easily enough. The rest has to be worked on... 3x/(x^2+3) is just another log. For the rest let x = ?3 tan? x^2+3 = 3 sec^2? dx = ?3 sec^2? d? 1/(x^2+3) dx = 1/?3 d? Now, putting all that ...

**Algebra**

I'll assume the usual carelessness with parentheses and spacing, and go with (a+5/(4a)) + 11/12 = 2/(3a) Using a common denominator of 12a, that gives a(12a) + 3*5 + 11/12 (12a) = 2*4 12a^2 + 15 + 11a = 8 12a^2+11a+7 = 0 That has no real solutions. So, if I interpreted you...

**math**

all multiples of 3 and 7, added together: 3,6,9,12,... 7,14,21,28,... and sums of those rows in all combinations.

**calculus**

just use the chain rule twice y = v^2(u(x)) y' = 2v v'(u) u'(x) y' = -2sin(1/x) cos(1/x) (-1/x^2)

**Math calculus**

v = ?r^2h = 5, so h = 5/(?r^2) a = 2?r^2 + 2?rh = 2?r^2 + 10/r da/dr = 4?r - 10/r^2 da/dr=0 when 4?r^3=10 r = ?(5/(2?)) h = 5/(?r^2) = 5/(?(5/(2?))^(2/3)) = ?(20/?) since 3 > 2r > h, it works.

**Math bet**

(1.2 * 10^-4 km)(6.25*10^-5 km) = 7.5*10^-9 km^2 that's 7.5*10^-3 m^2

**geometry/check my work**

D it is When you combine translations all bets are off.

**physics**

recall that the standard equation of motion means that you will solve for h in h(t) = h + vt - 4.9 t^2 = 0 Just plug in your initial speed v and time = 30, then solve for h.

**English**

They are all used. #1 is most common in my experience. Take that -- I've had it! Get it?

**Math**

divide total miles by total days

**Math**

I'd say 20/4 = 5 Of course, if he didn't give away all of the quarters, that could change things.

**further algebra**

The nth bounce will have height 30 * (4/5)^n No idea how many bounces it will take before stopping. But, to get the total travel after n bounces, just add 30 to twice the sum of n terms of the sequence (a round trip is up and back down)

**Precalculus**

Multiply top and bottom by 1+sin?: 1/(1-sin?) * (1+sin?)/(1+sin?) = (1+sin?)/(1-sin^2?) = (1+sin?)/cos^2?

**Math**

it has as much symmetry as a regular octagon. Rotate it 1/8 of a circle and it looks the same as before. (As long as the cars are all the same color!)

**maths**

a+2d = a + a+d 5/2 (2a+4d) = 30 Now just crank it out.

**Calculus**

Review the chain rule d/dx f(u(x)) = df/du * du/dx

**Math**

(x,y) -> (x+1,y-1) so, move right 1 and down 1 It usually helps to plot the points on some graph paper, eh?

**Physics**

You should subtract the 5 meters from the calculated distance, or the package will land 5 meters from shore in the water, rather than on land! 2772 - 5 = 2767 You added an extra 5 m/s to the speed!

**Physics**

how long does it take to fall 125 meters? 4.9t^2 = 125 Using that value of t, how far will the package move at 550 m/s ? You might want to consider an extra 5 meters to hit at the center of the island?

**Physics**

clearly he wants the vertex of the parabola to be just above the wall. Recall that the trajectory can be described as y = x tan? - g/[2 v^2 cos^2?] x^2 Plugging in your values, that is y = 0.707x - 0.196x^2 The vertex is at (1.8,0.6) So, a distance of 1.8 meters will provide ...

**Physics**

good work

**Physics**

surely by now you know that distance = speed * time

**Physics**

how long does it take to fall 10 meters? 4.9t^2 = 10 The horizontal speed stays constant, so using the value of t calculated above, how far does he go? If less than 5.5 meters, he's gonna break something...

**PreCalculus**

I'd say that it would be 0.28 * 70000/12

**Math**

zero

**Calculus**

I'll do one, and you can follow the method for the others. g"(x) = 12x^2 g'(x) = 4x^3+c g'(1)=7, so 4+c=7, so c=3 g'(x) = 4x^3+3 g(x) = x^4+3x+c g(1)=3, so 1+3+c=3, so c=-1 g(x) = x^4+3x-1

**math**

x+10 = 4(x-5)

**Calculus - Riemann Sum help**

You just have to figure out how many intervals you want to use (n). Then, as it says, the interval width is ?x=2?/n The interval is apparently [0,2?]

**Math**

5a/6 - a/3 - 2/3 = -1/6 a(5/6 - 1/3) = -1/6 + 2/3 a(1/2) = 1/2 a = 1

**algebra**

n/s = 5/3 n-25 = s+25 5s/3 - 25 = s+25 2s/3 = 50 s=75 so, n=125 n+s=200

**Math - Calculus**

81 ? (2i-1) i=1 = 2?i - ?1 = 2*81*82/2 - 81 = 6561 since you know that n ?i = n(n+1)/2 i=1 Or, working with the sum as an AP, you have a=1 d=2 S81 = 81/2 (2+80*2) = 6561 Or, note that you have the sum of the first 81 odd numbers. That sum is 81^2 = 6561

**Algebra 2**

8000(1+.03/365)^(365*18) = ?

**Caclulus**

copy/paste does not work well here. Your special symbols are all mangled. Try just typing, using f^-1(x) for f inverse

**Calculus**

4tan(x)+5=2 tanx = -3/4 x will be in QII or QIV

**math**

solving for a and b, we get a=5 b=2 Now just plug in the numbers

**Calculus**

cosx = 0.2 x ? 1.37 since cosx > 0 in QI and QIV, your answers will be x = 1.37 and 2?-1.37

**math**

(a+b)^2 = a^2+2ab+b^2 = 16+2*8 = 32

**math**

a^b = x a^(b+2) = a^2 * a^b = a^2 x 2a^3b + a^2b + a^b = 2x^3 + x^2 + x

**Calculus**

you should review your basic standard triangles: 45-45-90: sides 1:1:?2 30-60-90: sides 1:?3:2 Each problem here will have two solutions, in the quadrants where the named function is positive or negative.

**Trigonometry**

Recall that sin? = y/r cos? = x/r tan? = y/x you have a triangle in QI where y=8 x=6 r=10

**Math**

The slope of the line is (5553-3518)/(2000-1995) = 407 That leaves only D as the choice.

**Geography (w/ geometry)**

I expect this article will help: https://astronavigationdemystified.com/latitude-from-the-midday-sun/ Since your time of day is 12:30 rather than 12:00, you will need to factor that in.

**Geography (w/ geometry)**

Since the distances involved are minuscule compared to the size of the earth, all you really need to know is your height and shadow length. And you are correct that arctan(1.2) = 50.19° You have listed the time as 12:30, so the sun should have been almost directly overhead...

**Math (Calculus) Integrals**

Looks like you have the answers correct. For the last one, recall that ?[a,b] f(x) dx = F(b)-F(a) So, ?[a,a] f(x) dx = F(a)-F(a) = 0

**Math**

Assuming that by "n" you mean "and" then there are 6 angles, and their sum is 4*180 = 720 So, add 'em up and see what x has to be to make 720.

**physicss**

it is related by the square of the inverse, since the 3" pipe has a cross-section area 9 times the 1" pipe. So, (C)

**Business statistics**

(a+b)/2 = 15 ?(ab) = 9 so, ab = 81 The only factors of 81 are (1,81),(3,27),(9,9) Clearly the only possible candidates are 3 and 27, which you can easily verify. If you want to do the algebra, you have a+b = 30 a^2+2ab+b^2 = 900 a^2 + 2*81 + (81/a)^2 = 900 a^2 + 6561/a^2 = 738...

**Math**

put the words into math: 8+x = 3(x-2) Now just solve for x.

**Calculus II**

it is clearly convergent, because (n^4)/(n^10 + 1) < n^4/n^10 = 1/n^6 and 1/n^p is convergent for p>1

**ALGEBRA CHECK ANSWERS**

duh - all the ones I didn't comment on.

**ALGEBRA CHECK ANSWERS**

Some are ok. I have listed the errors below --> 4g^2=25 your answer solves 4g^2=1 --> 64b^2=16 your answer solves 64b^2=1 --> 5z^2-45=0 5z^2 = 45 z^2 = 9 z = ±3 --> h^2=-49 see your solution to x^2+7=0 --> s^2-35=-35 s^2 = 0 ... Are you trying to take ...

**Math, Algebra**

Hmm. When you say There is then a 22m pointing to the dotted line marking the middle up the triangle. it sounds like the triangle has a base of 8 and an altitude of 22. So, its area would be 8*22/2

**Pre calc**

you did not try reviewing your basic trig functions and drawing a diagram. Look at the diagram and you will see that cos? = 7/20 The height h can be found from ?, since h/20 = sin? or, you can use the Pythagorean Theorem (have you forgotten that already?): h^2+7^2 = 20^2

**2nd university of NamÃbia programming 2d**

42

**math**

8C2*6C2 + 8C3*6 + 8C4 = 826

**science (chemistry)**

you get 0.25 moles of AgBr, so what is its mass?

**Maths**

The lateral surface area is a = 2?rs = 2?r?(r^2+h^2) = 2?*30*?3400 = 600??34 Assuming that the join is 0.5 cm at the base, but decreases approaching the point of the cone, the 0.5cm overlap at the base means that you have a portion of the surface area added. That amount is 0.5...

**Math**

consider that the rider will be 60 feet away both approaching and receding. So, you want |105-35t| = 60 for positive distance (still approaching), 105-35t = 60 t = 9/7 = 1.28 Going away, you have 105-35t = -60 t = 33/7 = 4.71 http://www.wolframalpha.com/input/?i=%7C105-35t%7C...

**math**

(22*30*33)/(10*12*13) * 3 = 41.88 minutes

**Algebra**

x^-4 * x^5 = 1/x^4 * x^5 = x^5/x^4 = x

**Math**

add up the total votes. Then see whether any candidate gets more than half of them. Or, does any candidate get more votes than all the others combined? It looks pretty close, since 16+8+12 = 36 So, it'll be the odd votes that make the difference.

**maths, statistics**

This is just Z table stuff. You can play around with the values at http://davidmlane.com/hyperstat/z_table.html

**Math**

what do you mean by "initial values?" Just values for x? If so, then since g(x) = -x+2,plug in the value for every x: g(4) = -2(4)+2 = -8+1 = -7 and do the same for any other desired values. I still don't get what you mean by behavior after "iterations."...

**Math**

6% off what? If you mean that the tax is 6% of the price, then it is of course .06 * 350

**math**

OK. v = Vanessa's score

**Math**

print "number: "; read value smallest = value while value ?0 read value if value < 0 then {print "Error: negative value"; continue} if value < smallest then smallest = value loop if value > 0 then ...

**physics**

since energy is conserved, and the balls are identical, the 2nd ball will rise just as high as the first ball was. Be prepared for the next step, where the balls are not identical in mass!

**Math1**

clearly it is supposed to represent ? C = 2?r so, to get r, just divide both sides by 2?

**chemistry**

Look at the balanced equation and see the relative amounts of reagents (or reactants, not reactors) needed. Compare that against the amounts available (in moles) and then the limiting reagent is the one that will be used up first.

**Physics**

well, KE = 1/2 mv^2 now just plug and chug ... (watch the units!)

**Algebra**

so, the length is 2w w^2+(2w)^2 = 20^2 solve for w, then the area a = w(2w) = 2w^2

**Physics**

it is 1/2 mv^2, so ... watch the units.

**Physics**

Without loss of generality, we may assume that the 1st ball is moving in the +x direction. After the collision, since the two balls are identical, the resultant v1+v2 will still have the same direction, which means that the two velocities are both 30° from the original v, ...

**Physics**

if all the forces are constant, their sum will be constant. F = ma so if F is constant, a is constant (I think we can assume that M does not change...) The resultant force F is just F = F1+F2 use normal vector addition.

**Math**

14x+21y = 7*2x + 7*3y = 7(2x+3y)

**SCIENCES**

look up the atomic masses in your periodic table. Add up the appropriate multiples to get the molecular molar mass. The rest is just algebra, if you know the yield of the Hall process.

**Precalculus**

how about using sin(a-b) = sina cosb - cosa sinb

**Mathematical Proof**

n+6 is odd if and only if 5n+1 is even if 5n+1 is even, 5n is odd. So, n is odd (only odd*odd = odd) If n is odd, n+6 is odd, (only odd+even=odd) if n+6 is odd, n is odd, since odd numbers differ by 2. so, 5n is odd, making 5n+1 even. USING odd*odd = odd since (2m+1)(2n+1) = 2...

**physics**

Since F = kx, k = F/x so its units are N/m, not Nm. So, just figure the F=KE=1/2 mv^2 and divide it by .37m

**Precal**

I think you mean 2sin^2(A/2)+cosA = 1 this follow directly from cos(2x) = 1-2sin^2(x) with x = A/2

**Maths**

1 - 1/4 - 1/3 = 5/12 1/5 of that is 1/12 So, 1/12 of the original amount is 40,000

**Precal**

that's -1 remember sec^2 = 1+tan^2 csc^2 = 1+cot^2 these come from dividing 1 = sin^2+cos^2 by cos^2 or sin^2

**Math**

5/6 * 1/2 = 5/12

**Math**

see related questions below

**precal Please Help**

you need to review your double and half angle formulas: tan(A/2) = (1-cosA)/sinA = sinA/(1+cosA) cos(2A) = 2cos^2A-1 = 1-2sin^2A = cos^2A-sin^2A sin2A = 2sinAcosA A little playing around with those should help. Also, you need some parentheses to make it clear just what the ...

**Math**

1422(1+.03*5) = ?

**precal**

(sinx+cosx)^2 = sin^2x + 2sinx cosx + cos^2x = 1+2sinx cosx Now you just take the last step, recalling that tanx = sinx/cosx

**Math(PLEASE HELP ME ASAP)**

you haven't described very well just how b and h and w are related to the alleged "star"