# Posts by Steve

Total # Posts: 51,170

**matha**

they are planed in a pentagram figure https://richardwiseman.wordpress.com/2013/07/29/answer-to-the-friday-puzzle-216/

**Trigonometry**

Draw a diagram. If the distance is x, then (h-85)/x = tan11°6' h/x = tan26°7' So, (h-85)cot 11°6' = h*cot 26°7' Use that to find h, and then you can get x.

**t**

If the sides are a and b, and the included angle is C, then c^2 = a^2+b^2-2ab cosC sinA/a = sinC/c A+B+C = 180

**Trigonometry**

These are all just about the basic trig functions. Draw a diagram and just decide which function to use.

**Calculus PLSSSSS HELP DUE SOON**

dp/dt=0.03p?0.00015p^2 This is a Bernoulli equation, with solution 200 e^0.03t / (e^c+e^0.03t) No idea what k is supposed to be, or the carrying capacity. You will need some more info to determine c. Since dp/dt = 0.00015p(200-p) its roots are at p=0 and p=200 So, p is growing...

**x**

Hint: an ellipse is the locus of points whose distances from the two foci have a constant sum. That sum is the length of the major axis. So, just write your equation in standard form for an ellipse, and you can read off the value of the semi-axes.

**Math**

there are 4 choices for each letter. So, that makes 4^3 = 64 codes in all.

**Maths**

j = 2d p = d+5 j+d+p = 95 Now just solve for the ages

**math**

take a look here, and you can play with your numbers: http://davidmlane.com/hyperstat/z_table.html

**Calculus 1 Help**

(a) using shells of thickness dx, v = ?[0,4] 2?rh dx where r=x and h=?x-x/2 v = ?[0,4] 2?x(?x-x/2) dx = 64?/15 using discs of thickness dy, v = ?[0,2] ?(R^2-r^2) dy where R=2y and r=y^2 v = ?[0,2] ?((2y)^2-y^4) dy = 64?/15 Now you try (b), recalling the formulas for volumes of...

**Math**

http://www.jiskha.com/display.cgi?id=1492718754

**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