# Posts by Steve

Total # Posts: 50,414

**geometry**

Huh? The angle bisector is the diagonal AC. The diagonals of a rhombus are perpendicular.

**Math**

The original dimensions are width: w length: 2w+3 (w+4)(2w+3+4) = 434+196 w=14 so, the new length 2w+3+4 = ?

**math**

AD is also 6, so the height is 6 sin60°

**math**

Every 1+1=2 minutes it gains 5-3=2 meters. So, after 30 minutes it has moved 30 meters. In minute 31 it climbs the last 5 meters and is done.

**math**

Since the diagonals intersect halfway between the sides, we have the width = 2x length = 2(x+4) Now use those to get the perimeter: 2(2x+2(x+4))=56 Solve for x, and then you are home free.

**maths**

(?2+1)(?2+?3)

**mathematics**

log(8*20/1.6) = log(100) = 2

**Economics**

n?xy - (?x)(?y) ?(x-x...

**Economics**

n?xy - (?x)(?y) ?(x-x?)(y-y...

**Economics**

n?xy - (?x)(?y); ?(x-x?)(y-y...

**Economics**

n?xy - (?x)(?y); ?(x-x?)(y-y...

**Economics**

n?xy - (?x)(?y) ?(x-x?)(y-y?) ---------------------------------------------- = ------------------------------- ?(n?x^2 - (?x)^2) ?(n?y^2-(?y)^2...

**Economics**

n?xy - (?x)(?y) ?(x-x?)(y-y?) ---------------------------------------------- = ------------------------------- ?(n?x^2 - (?x)^2) ?(n?y^2-(?y)^2...

**Economics**

?(x-x?)(y-y?) ------------------------------- ?(?(x-x?)^2) ?(?(y-y?)^2)

**Economics**

?(x-x?)(y-y?) ----------------------------------- ?(?(x-x?)^2) ?(?(y-y?)^2)

**Economics**

n?xy - (?x)(?y) ----------------------------------------------- ?(n?x^2 - (?x)^2) ?(n?y^2-(?y)^2)

**Economics**

n?xy - (?x)(?y) ------------------------------------------- ?(n?x^2 - (?x)^2) ?(n?y^2-(?y)^2)

**Economics**

n?xy - (?x)(?y) --------------------------------- ?(n?x^2 - (?x)^2) ?(n?y^2-(?y)^2)

**Economics**

?(x-x?)(y-y?)

**Calculus II: Shell Method for finding volumes**

I guess the decimal values are ok, but I tend to prefer the exact values of (-1+?17)/2 and (-1-?17)/2. They are a bit cumbersome to work with by hand, but with the wealth of online web sites to do the calculations, that should be no obstacle. Your limits of integration are ...

**calculus**

starting with the peak at t=0 suggests use of a cosine function. Since the period is 11 years, start with y = cos(2?/11 t) The amplitude is (120-0)/2 = 60, so y = 60 cos(2?/11 t) The center line is at y=60, so y = 60 cos(2?/11 t) + 60 where t is the number of years since 2000.

**Algebra**

a^x is growth if a > 1 a^x is decay if a < 1 a^x passes through (0,1) for any value of a.

**math**

x + x+1 + x+2 = 3x+3

**math**

If you say so.

**Math**

see related questions below

**Math**

1/6 * 3/2 = 1/4

**Math**

It might help to express them all to the same precision: 0.125 0.100 0.120 ok?

**Science**

Take a look at Snell's Law. Then you can make up measurements and figure the related values. Not sure how you'd lay a protractor against the water container... I'm sure you can find some online examples, though.

**Physics**

Using the usual trig angles (not compass headings), 15 cis45° + 18cis5° = 28.54+12.18i = 31.03 cis23.11°

**calculus**

the initial value of 11 means that 2sin(2pi/3(t+1/4)) = 1 sin(2pi/3(t+1/4)) = 1/2 2pi/3(t+1/4) = pi/6 t + 1/4 = 1/4 t=0 With no displacement, y(0) = 0 since sin(x) is a rising curve at x=0, the graph needs to be shifted left to bring the value of 1 into play. Try playing ...

**Math**

well, 3577.4 is less than 3887, so ... how many feet in how many years?

**Math (Help please!)**

you're kidding, right? It has to be right below A and to the left of C. Where do those lines cross? Or, see that point D has to have the same x-value as A, and the same y-value as C.

**Algebra**

x^2 + 8x + 16 = 10+16 (x+4)^2 = 26 x+4 = ±?26 x = -4 ±?26 Nope, not A

**Math**

Since you don't say that each 5m portion is a separate attempt (like walking up stairs), I'll assume that his slippage is continuous as he climbs. That is, it takes him 5*5=25 seconds to climb 5-1=4 meters. So, his rate of ascent is 4/25 m/s It will take him 53/(4/25...

**trigonometry**

In QI, if cosx = 4/5, sinx = 3/5 (1+tanx)/(1-tanx) = (cosx+sinx)/(cosx-sinx) = (4/5 + 3/5)/(4/5 - 3/5) = (7/5)/(1/5) = 7

**enough already!**

impatient much?

**Maths**

oops. CA has radius 3.5

**Maths**

Huh? You already said that |AB|=4cm To construct the figure, bisect a 60° angle to get angle A. Mark B where an arc of radius 4 intersects one side of angle A. C lies on a circle CB of radius 5 with center at B. D lies on a circle CA of radius 5 with center at A. So, CD is...

**Maths**

surely you can plug in your numbers s = r? 60° = pi/3 (not pie!!)

**basic mathematic**

x-y

**Deduction**

P => Q says nothing about what happens if ~P.

**Math**

a/m = 3/7 a+92 = m+40

**math**

statement: for all x, either p(x) or q(x) There is some x where p(x) and q(x) are both false. The mod works just like any other function x ?3 1 Since x is congruent to 3 (mod 1) makes no sense, you must mean x ? 1 (mod 3) We all know what that means, so just negating the ...

**Pre-Calc**

I assume you mean r = 21/(5-2cos?) When you consider the general equation for a conic: r = ep/(1 - e cos?) you see that you must set it up as r = 21/[5(1 - 2/5 cos?)) Now you can read off e and p. Since e<1, you have an ellipse. Now you need to use the distance to the ...

**Maths**

since cos2x = cos^2x - sin^2x, you have 1 - 2cos2x = sin2x 1 - 4cos2x + 4cos^2 2x = sin^2 2x 5cos^2 2x - 4cos2x = 0 cos(2x)(5cos(2x)-4) = 0 cos 2x = 0 cos 2x = 4/5

**substitution and elimination methods**

x-4y = 20 -3x+4y = -12 add them together and you get -2x = 8 x = -4 Now use that to get y.

**Math**

Just divide the figure up into rectangles. You can then see the width and height of each part, and add up all the smaller areas.

**Math**

yep

**Math**

then the base is 24 and the height is 8 Now you can figure the area.

**MATHS**

You want a power of two to be x^2+4x+4 = (x+2)^2 So, which powers of 2 are perfect squares? There are lots of answers: (6+2)^2 = 64 = 6^2 + 4*6 + 4 = 1446 = 2^6 (14+2)^2 = 256 = 14414 = 2^8 All even powers of 2 are perfect squares.

**Physics**

m = 691/9.8 (c) F = ma

**maths**

This is exponential decay. No is the starting value e^-k is the decay rate N(t) is the amount remaining after t time intervals. Surely your text laid all this out in the section on exponential growth/decay.

**Calculus**

Hmmm. I found a proof in about 15 seconds. First, you need to understand what it means to be integrable: If a function is continuous on a given interval, it’s integrable on that interval. Now, if you show that f(x) is discontinuous at any point in [0,1] then it is not ...

**math**

clearly d = -k/3 S30 = 30/2 (2k + 29(-k/2)) = -375k/2

**go with @Reiny**

I calculated the difference, then erroneously assigned that to x. My bad.

**Trigonometry**

(25 - 1/2)/2 = 49/4 so, x = 49/4 (32 1/2)/(50/4) = y/(32 1/2) y = 169/2

**maths**

(x+10)/x = 3/1 x = 5 So the numbers are 5:15:20

**Math 8**

What do you mean "solve it separately" ? It's a system. If you rearrange things to match, you have 2x-y = 4 2x-y = -4 Clearly there are no solutions; the lines are parallel, and do not intersect.

**math**

that would be 1.20^3 = 1.728 or 72.8% increase

**calculus**

they are the same curves, just shifted. It usually depends on whether the initial value is a maximum or a center-value.

**Maths**

Just plug and chug. You know that f = sin f' = cos f" = -sin f(3) = -cos So, f(x) = f(?/3) + f'(?/3)(x-?/3) + f"(?/3)/2! (x-?/3)^2 + f(3)(?/3)/3! (x-?/3)^3 + ... f(x) = ?3/2 + (1/2)(x-?/3) - (?3/2)/2! (x-?/3)^2 - (1/2)/3! (x-?/3)^3 + ... f(x) = ?3/2 + (1/2)(x...

**Math**

3:1:4 = 9:3:12

**calculus**

If you mean ?[-3,3] x^2+2x+1 dx then the integral is x^3/3 + x^2 + x [-3,3] = (9+9+3)-(-9+9-3) = 24 or, ?(x+1)^2 dx = 1/3 (x+1)^3 [-3,3] = 1/3 (4^3+2^3) = 24

**maths**

just pick any element, say x. The the subset with that one element is {x}

**Calculus**

a simple google on the question turns up many discussions of the problem and its proof.

**Math (Integrals)**

You have the right idea, but you are working a definite integral, so there is no constant C. Instead, evaluate A(t) at the interval endpoints: A(x) = (4t + 4/t)[1,x] = (4x + 4/x)-(4*1 + 4/1) = 4x + 4/x - 8

**Math - Calculus**

It's just the chain rule. If F(t) = ?f(t) dt then ?[v(x),u(x)] f(t) dt = F(u)-F(v) so, taking derivatives, that gives you f(u) u' - f(v) v'

**Math**

j/p = 4/7 (j - j/2)/(p-60) = 1/2 (j/2)/(7j/4 - 60) = 1/2 4j = 7j-240 j = 80

**Pre Algebra**

In 1 hour he can bake 4 times as many. So, what is 4 times 2 1/2?

**Math (Definite Integrals)**

Do not assume r=1. But, it is a constant. And you are right, the graph is a sem-circle, so the area is ?/2 r^2 Not sure where the a comes in, but to do the integral, now is the time to start getting used to trig substitutions. Let x = r sin? r^2-x^2 = r^2(1-sin^2?) = r^2 cos^2...

**C++**

I assume you have picked up some C++ by now, so you know how to build an object and define variables. The logic can be something like what follows, but expressed in C++, using cout and cin print "enter mass: " read mass weight = mass * 9.8 print "Weight: ",...

**Algebra 2**

100(1 + .03/4)^(4*20) = ?

**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