# Posts by steve

Total # Posts: 51,159

**Calculus Math**

Not sure what you want in the first problem. The height varies along the interval. For the second problem, your equation is wrong. Both terms on the right have x in the denominator. There is no way to get the x/4 term on the left. And, find the arc length of what?

**Calculus Help**

Let's assume that f(0) = 0. Then ?[0,b] f'(x) dx = b^2 f(b) = b^2 f(x) = x^2 f'(x) = 2x ?[0,b] 2x dx = x^2 [0,b] = b^2 There are lots of other possible functions. Such as ?[0,b] e^x dx = 1/k e^(kb) - 1 1/k e^(kb) - 1 = b^2 e^(kb) = k(1+b^2)

**Math - Calculus**

shells of thickness dx: v = ?[1,2] 2?rh dx where r=x and h=y=x^2 v = ?[1,2] 2?x*x^2 dx = 15?/2 washers of thickness dy -- the curved portion plus a cylinder 1 unit high with radii 1 and 2: v = 3? + ?[1,4] ?(R^2-r^2) dy = 15?/2 where R=2 and r=x=?y v = ?[1,4] ?(4-y) dy =

**calculus please help!!**

what you do is use the rest of the information they gave you: y(0)=4 So, you have to find C using 4 = sin(2*0+C) Now, when does sin(?) = 4? Never!

**calculus**

not sure what's the trouble. You have the base lengths and the heights. Too bad you didn't show your work. It's just (1+2)/2 * 0.1 + (2+4)/2 * 0.1 + ... for the whole list of six intervals

**Math**

scalene

**math**

since all the sides are equal, the altitude is 10 sin60° = 5?3 To see this, draw the rhombus ABCD and drop an altitude from D to AB. Now you have a right triangle with hypotenuse=10 and base angle=60°

**math**

Since the diagonals bisect the vertex angles, if ?K is obtuse, ?PKE cannot be just 16°. Anyway, to solve this just remember that two consecutive angles of a rhombus are supplementary, so if half of one of them is 16°, half of the next one is 90-16=74° The diagonals...

**Pre calc**

see your earlier post

**PreCalculus**

draw a diagram. It is clear that (a) tan? = 20t/50 (b) solve when t=3 Since tan2??2tan?, (c) is clearly false.

**Math**

If the center is at O, then the slope of OP = 3/4 So, the tangent line, which is perpendicular to the radius OP has slope -4/3, and its equation is thus y-3 = -4/3 (x-4) The line's x-intercept is at Q=(7,0) Due to symmetry, the other tangent line through Q touches the ...

**Math**

I'll do (c) and the others work the same way. You just complete the squares: x^2 +y^2 +2x?8y = ?8 x^2+2x + y^2-8y = -8 x^2+2x+1 + y^2-8y+16 = -8+1+16 (x+1)^2 + (y-4)^2 = 3^2 Now you can just read off the center and radius.

**Math**

as in your other post, complete the squares and all will be clear.

**limit maths sir steve or damon or reiny help**

(e2x)1/x < (x+ex+e2x)1/x < (2e2x)1/x e2 < (x+ex+e2x)1/x < 21/xe2 Now take the limits, and since 21/x-> 1, e2 < (x+ex+e2x)1/x < e2 so (x+ex+e2x)1/x = e2 Or, you can take the log of the limit, and then use l'Hospital's Rule

**Physical Science**

A current of 6.0 amps passes through a motor that has a resistance of 5.0 ohms. calculate the power. my answer is 180 watts

**Calc**

#1 The LS can be written as sinx(cosx-sinx) ---------------------------------- sinx(cosx+sinx)(cosx-sinx) I think it's clear what comes next ... For #2, on the LS, multiply top and bottom by (1+sinx) and I think it will come ...

**Pre calc**

remember your double-angle formulas: cos(2?) = 4 ? 3 cos(?) 2cos^2(?)-1 = 4 - 3cos? 2cos^2(?)+3cos? - 5 = 0 (2cos?-1)(cos?+5) = 0 cos? = -5 has no solution cos? = 1/2 has solutions at ? = ?/3 and 5?/3 and other basic trig identities: cos(2?)cos? = sin(2?)sin? (2cos^2(?)-1)cos...

**math**

find the volume of 2 pyramids. 2* 1/3 Bh multiply by gold cos/mm^3 find area of 8 triangles. 8* 1/2 Bh multiply by paint cost/mm^2

**futher maths**

z^2 = 10^2 + 8^2 - 2*10*8 cos120°

**Precalculus**

Take a look here for an example: y = (x+2)^2 * (x-1)^3 http://www.wolframalpha.com/input/?i=(x%2B2)%5E2+*+(x-1)%5E3

**Precalculus**

if a factor (x-h) occurs n times, it has a multiplicity of n. Looking at the graph, if a root has odd multiplicity, the graph crosses the x-axis there. Think of y=x^3 If the multiplicity is even, the graph just touches the x-axis and turns back. Think of y=x^2

**Algebra**

y=4x and y=6x go through the origin (0,0) 10+4x starts off 10 units higher. 6+6x starts off 6 units higher. The y-intercepts are 10 and 6. The x-intercepts are -5/2 and -1 (2,18) is where the two lines intersect. It is the solution to both equations.

**Geometry**

take a look at the Angle Bisector Theorem AX/BX = CA/CB

**maths sir steve help me reiny**

The curve is just a parabola: x = y^2/4a So, the area is just A = ?[1,2] y(t) dx(t) dx = 2at dt A = ?[1,2] 2at * 2at dt = ?[1,2] 4a^2t^2 dt = 4/3 a^2 t^3 [1,2] = 4/3 a^2 (8-1) = 28/3 a^2 To check, express y as a function of x: A = ?[a,4a] y dx = ?[a,4a] 2?(ax) dx = 2?a (2/3 x...

**mathematics**

x = ky+m 7 = 5k+m 8 = 7k+m so, k = 1/2 and m = 9/2 ...

**science**

less dense

**Calculus**

Well, there are two regions, on the intervals [-3,0] and [0,1] Finding the areas is pretty straightforward, right?

**physics**

v = ?(20/2)^2 * 0.05 = 5? cm^3

**Math**

I guess that'd depend on how many yellow balloons there were at the start.

**math**

slide depends on the flips

**math**

the altitude of the triangle is the distance between the lines. A = 1/2 bh b is constant, h is constant.

**math**

If the diagonals AC ? BD then the figure is a rhombus. Since the area of a rhombus is the one-half product of the diagonals, (29/2)*BD/2 = 58 Not so hard now, eh?

**MAth**

500x-100(28-x)=9100

**Trigonometry**

nope. Although you do know that the period is 2?, so it crosses the x-axis at intervals of ?, with cos(0) = 1 cos(?/2) = 0 cos(?) = -1 cos(3?/2) = 0 cos(2?) = 1 Then sketch a smooth curve using those points. If you want other points, you can use known values at ?/6, ?/4, ?/3 ...

**Pre calc**

2sin ?/3 + ?3 = 0 sin ?/3 = -?3/2 ?/3 = 4?/3 or 5?/3 + 2n? ...

**Math**

4x-3y = -4 3x-2y = -4 8x-6y = -8 9x-6y = -12 Now subtract. The y disappears and you get -x = 4 x = -4 now use that to get y.

**Math**

a triangle of side 24/3 = 8 a square of side 24/4 = 6 and so on. If you want to include odd shapes, than just use your imagination...

**Calculus**

It is wrong. Your equation uses a rate of 2.1%. a .21% growth rate means 6.8 * 1.0021^(t-1988) You can use a base of e, but that would be ln 1.21 = 0.0021 6.8 e^(0.0021(t-1988)) or, if t is defined as the number of years since 1988, 6.8*1.0021^t or 6.8 e^(0.0021t)

**Trigonometry**

the domain must be at most 1/2 period, or 2?/.1 = 20?. In addition, it must not contain an asymptote or a max/min, because there the curve doubles back, so it fails the horizontal-line test. All of those intervals are short enough, so we need to consider that there is an ...

**math**

draw a diagram of the corner of the field. Since the rope is shorter than either dimension, the grazed area is just a 1/4 circle. You know how to figure the area of a circle, right?

**math**

cot sec^2 - cot = cot(sec^2-1) = cot(-tan^2) = -tan sin tan + cos = sin*sin/cos + cos = (sin^2 + cos^2)/cos = 1/cos = sec so, doing the division, you have -tan/sec = -sin/cos * cos = -sin = -cos tan^3 ...

**Calculus**

as with the other problem, y = ?[?x,x^3] 10?t sin(t)dt y' = (10?(x^3) sin(x^3))(3x^2) - (10?(?x) sin(?x))(1/(2?x)) = 30x7/2 sin(x^3) - 5/?x sin(?x)

**Caclulus**

the 2nd FT of Calculus is just the chain rule in reverse. If F(x) = ?f(x) dx ?[u,v] f(t) dt = F(u)-F(v) so, taking derivatives, F'(x) = dF/dv * dv/dx - dF/du * du/dx = f(v) v' - f(u) u' So, for this problem, g(x) = ?[2x,3x] f(u) du g'(x) = f(3x)*3 - f(2x)*2 = 3...

**Intro to Trig**

you know that cos(x) = -1 at x = ?,3?,5?, ... So, we can check 7? = 21.99 9? = 28.27 Looks like C to me

**Math**

His gain on selling one bottle is -3, sohis gain on selling x bottles is just -3x Now plug in x=7 I get -21, not -12.

**Math**

not quite. Remember the chain rule dx/dy = 6y^3*(3y^4+2)^-1/2 Now you can see that (ii) is true Doing things implicitly, it's much clearer: x^2 = 3y^4+2 2x dx/dy = 12y^3 dx/dy = 6y^3/x

**Algebra Word Problem**

Europe: .36c Asia: 24 S America: c - .36c - 24 = .64c-24 Without some further information it is not possible to find a numeric value for c.

**physics**

1 km/hr = 20 m/s periods are not measured in m/s a = (-20m/s)/(4s) = -5 m/s^2 F = ma s = 1/2 at^2 ...

**Math**

correct.

**Math**

huh? It's just like the other one, assuming you know how to change mixed numbers to fractions. (5 7/8 km)(18/4 km) = (47/8 km)(9/2 km) = 423/16 km^2

**Math**

correct, but I'd reduce the fraction to 49/5 and the area is mm^2, not just mm

**Algebra I**

I used a suitable function for f(x) Since you did not specify one of your own, I felt free to pick a good one. I felt that since 42 is the answer to the Ultimate Question of Life, the Universe and Everything, it would be a good answer to your rather ill-posed question.

**Algebra I**

42

**math**

why not just use a regular pentagon, with apothem equal to the radius of the pizza?

**Maths**

To find the base area, 70.56N / (4900N/m^2) = 0.0144 m^2 = 144cm^2 v = 1/3 * 144 * 15 = 720 cm^3

**MAth**

2(x+y) = 240 2(1.1x+0.8y) = 240 now solve for x.

**Mathmatics**

AB has slope (1+7)/(-8+12) = 2 Now you have a point and a slope, so the line is y-5 = 2(x-2)

**Geometry**

?(100/81) = ?

**Maths**

SRS: 10*5*9 RSR: 5*10*4

**Maths**

b^0 = 1, so a=3 Now you know that 3b^2 = 12 I expect you can find b now ...

**Physics**

what is the critical angle, using Snell's law?

**Math, Ratio and Proportion**

the whole lawn is 3/3, or (3/2)*(2/3) so, it will take 3/2 * 52 = 78 minutes As a proportion, x/1 = 52/(2/3)

**Math**

3/(3/5) * 85 = $425

**Math**

nope 5000/500 * 3 = 30 or, keeping track of the units, you want to convert $$ to weeks: $5000 * 3weeks/$500 = 30 weeks

**AP math**

Yes. logs and exponents are inverse operations; one undoes the other: e^(-0.244t) = 3 -0.244t = ln3 t = ln3/-0.244 = -4.5

**exponent**

sorry - can't be done v^-7 * v^7 = v^(-7+7) = v^0 = 1

**math**

every time 8 days pass, you multiply by 1/2. So, after t days, 8 days have passed t/8 times, and the amount left is 5*(1/2)^(t/8) After 7 days, then, you have 5*0.5^(7/8) = 2.726mg 1mg is 1/5 of the starting amount. 1/5 is a bit less than 1/4 = (1/2)^2, so it should take a ...

**maths**

If the side parallel to the wall has length x meters, and the other side is y, we have x+2y = 2000 The area a is just a = xy = (2000-2y)y = 2000y - 2y^2 This is just a parabola, with its vertex (maximum value in this case) at y=500. So, x=1000, and the maximum area is 1000*500...

**precalculus**

exponential growth is always of the form p = a*e^(kt) that's why it's called exponential... The starting amount (at t=0) is a, since e^0 = 1.

**Trig**

Surely you mean a depth of 45m, not 45km! Draw a diagram. The length x of fishing line can be found via (45+1)/x = sin35°

**precalculus**

or, you can get all the exponents over a single value: 23(.95^t) = 15(1.11)^t (.95^t)/(1.11)^t = 15/23 (.95/1.11)^t = 15/23 t log(0.856) = log(0.652) t = log(0.652)/log(0.856) = 2.751 You can see that Reiny's value will agree, since subtracting logs lets you divide numbers...

**Math**

.85 * 40 = ?

**Algebra**

The turns are in the ratios 2:9 and 3:5=9:15 So, they are in the ratios 2:9:15 Now things are clearer, right?

**Algebra**

(15÷(3×5)+4)×8=40

**word problem , MATHS**

I see you'd rather keep posting this than actually work on it. He has 7 sections of material to use. (a) the dog run could be 3x15 (1x5 panels) or 6x9 (2x3 panels) (b) 3x18 or 6x15 or 9x12 Now, was that so hard?

**math**

so, what are those dimensions?

**math**

I'd say no, in general.

**Calculus**

Since the region is symmetric, we can just double the volume from x=0 to x=1. using shells of thickness dy, v = 2?[0,1] 2?rh dy where r=7-y and h=x=?y v = 4??[0,1] (7-y)?y dy = 256?/15 using discs of thickness dx, v = 2?[0,1] ?(R^2-r^2) dx where R=7-y and r=6 v = 2??[0,1] ((7-...

**Math - Calculus**

the x=0 boundary is redundant, since the parabola intersects the x-axis at 0 and 2. Anyway, the area is just a bunch of vertical strips of width dx and height y, so a = ?y dx = ?[0,2] 2x-x^2 = 4/3 Now, you want the line to divide R into two equal areas. y=cx intersects the ...

**math**

if A = LW then the new area A' = (2L)(2W) = 4LW = 4A so, 4A = 160

**Chemistry**

We are asked to prepare a 100 ml solution with the following 3 concentrations: 53.00 mg/ml sodium salicylate, 0.020 M FeCl3, and 0.050 M HCl. Then, we are going to take the absorbance at 520 nm. What is the best way to prepare the 100 ml solution ?

**Math**

No way to tell. If C has the same base area as D, then since its height is 1/2 as big, so is its volume. However, if C is similar to D, then all the dimensions are cut in half, so the volume is (1/2)^3 = 1/8 as much.

**maths**

huh? 2 boxes is 2*12 = 24 apples Think that's enough?

**maths**

168 = 3*56 = 2*56 + 56 so, there are 112 girls and 56 boys 56 = 7*8 I think the rest is not so hard, right?

**Algebra**

recall the discriminant of a quadratic: b^2-4ac Here, that is 1^2-4(-3)(-4) = 1+48 = 49 since it is positive, there are 2 real roots.

**reiny steve reiny damon!!! Damon help maths**

huh? surely you recall the distributive property (x-h) divides P(x), so P(x) = (x-h)*p(x) similarly, Q(x) = (x-h)*q(x) So, P-Q = (x-h)*p(x) - (x-h)*q(x) = (x-h)(p(x)-q(x)) so, x-h divides P-Q And above, I showed you what that is, and even factored it for you! So, those factors...

**reiny steve reiny damon!!! Damon help maths**

This just the same as showing that property with numbers. If a|p and a|q then a|p-q p = a*m and q=a*n p-q = a(m-n) so a|p-q for your polynomials, P-Q = (ax^3+x^2-15x-18)-(ax^3-14x-12) = x^2-x-6 = (x-3)(x+2) that should get you going, right?

**Check My Math Answers Please!**

#12 recall that the vertex of a quadratic (max height in this case) occurs when t = -b/2a = -32/-32 = 1

**Math**

1/3 x + 6 = 5/6 x 6 = 1/2 x x = 12 Check: the 1/3-full tank has 4 gallons. Adding 6 makes 10 gallons, or 5/6 of 12

**Math**

Surely not 3, since it does not appear in both top and bottom 8/15 * 5/16 = 8*5 / 15*16 = 40/240 Now, 2 can be factored out to give 20/120, but that is not fully simplified Same for 5 and 8, done individually. However, if 5 and 8 are both factored out, the resulting 1/6 is as ...

**algebra**

for the parallel line, you have a slope and a point, so use the point-slope form to get y+1 = 3(x+1) The perp. line has slope -1/3, so it is y+1 = -1/3 (x+1) Now massage those into slope-intercept form.

**MATH HELP ASAP PLEASE!!!!!**

You and "Naomi" should get together. can't read you attempts at fractions. Try something like 11 1/4 or 10 1/2 Consider using a regular pentagon with an apothem of 7", or the radius of the pizza. Look online to find the area of such a figure. It will be the ...

**math,**

Just write algebra for the conditions. Let w = municipal bonds x = CD's y = T-Bills z = Stocks Then the conditions are w <= (w+x+y+z)/5 x <= w, x <= y, x <= z x+y >= 3/10 (w+x+y+z) x+y >= 1.2(w+z) Rearrange things a bit, and you want to maximize p = .085w...

**Calculus**

find v(t), and then just take its derivative to find its maximum: v(t) = 3t^2-6t-9 v'(t) = 6t-6 clearly, v'=0 at t=1 However, since v' is just a parabola, its vertex is a minimum, not a maximum. That means that its maximum is either at t=0 or t=2. So, just compare ...

**maths**

You have mangled the question, but if I parse the meaning correctly, you need to check the amount of alcohol before and after mixing in the water: .60(80) = .40(80+x)

**maths**

the fine is just an arithmetic progression where the amounts (in thousands) are a = 4 d = 1 You want n where 4+1(n-1) <= 60

**vector calculus**

Take the cross product and the normal to both lines is <2,-2,1> The plane is then just 2(x-1)-2(y-5)+1(z-12) = 0 2x-2y+z = 4

**Math (Integrals)**

since ((2x-1)^3)/6 = 4/3 x^3 - 2x^2 + x - 1/6 the two solutions differ by 1/6 Since C is an arbitrary constant, its value does not matter.

**Algebra**

just add up the interest amounts: .05x + .10(30000-x) = 2500

**advance maths**

to find n, just solve n/2 (9+105) = 741 Now find d, using 9+(n-1)d = 105