# Posts by Steve

Total # Posts: 51,707

**trig**

(4cosx+1)(2cosx+1) = 0 now it's a cinch...

**maths**

There are 11P4 ways to pick the pained boats.

**Maths**

well, all the 400's and all the 900's and all the 40's & 90's of the other 8 hundreds and all the other 8 numbers ending in 9 for the other 8 hundreds

**maths**

a set of n elements has 2^n-1 non-empty subsets

**Algebra**

correct

**Math**

The way you have written it is incorrect. It should be 4tanx(1-tan^2(x)) ----------------------- (1+tan^2(x))^2 = 4tanx/(1+tanx^2) * (1-tanx^2)/(1+tanx^2) = [4sinx/cosx * secx^2][(1-tanx^2)/(1+tanx^2)] = (4sinx cosx)(1-tanx^2)/secx^2 = 2sin2x (cosx^2-sinx^2) = 2sin2x cos2x = ...

**Math**

take sin of both sides and use your sum and double-angle formulas to get a polynomial in x sin(2arcsin(x/?6)+arcsin(4x)) = 1 x = ?39 - 6 see http://www.wolframalpha.com/input/?i=sin(2arcsin(x%2F%E2%88%9A6)%2Barcsin(4x))+%3D+1

**Maths**

well, 22 tens is 220 ...

**algerbra**

looks ok to me, aside from using "square of" to mean "square root of"

**Math**

use a common denominator. Then you have 12/28 + 7/28 not so hard now, eh?

**algerbra**

right on.

**geometry**

their areas are in the ratio 2:3:4

**Math**

wrong in so many ways. First, 1.00 is 100% .609 = 60.9% so, 2.608 = 260.9% But you read the question wrong. You know that 2 is 50% of 4, right? So, you divide 2/4 = 0.5 = 50% But you divided 321 by 123. Surely you could see that 2.6% is a totally unreasonable answer. Actually...

**maths**

380 They have no common factors LCM(x,y) = xy/GCD(x,y)

**maths**

I guess he is then lost. Or, try the law of cosines.

**Maths**

A. usingsymmetry, the area is just ?[0,?/4] 2cosx - secx dx = ?2 - 2tanh-1 ?/8 Using discs of thickness dx for the volume, v = 2?[0,?/4] ?(R^2-r^2) dx where R=2cosx and r=secx v = 2?[0,?/4] ?((2cosx)^2-(secx)^2) dx = ?^2 Using shells of thickness dy, we have to split the ...

**Maths**

sides have shrunk by ?4 = 2

**Math**

well, just add up the numbers!

**math**

(8/6) * 48 = 64

**math**

well, the sum of all 9 angles is 7*180 = 1260 (1260-462)/6 = ?

**math**

x+x+9 < 24 2x < 15 x < 7.5 so, x <= 7 The sides of length x must each be > 9/2, so x >= 5 5 <= x <= 7

**Maths**

well, 05 is half of 10, so ...

**Math**

If R is the radius at the equator, then let r be the radius at 45°. That is Rcos45° = R/?2 So, the ratio is R/r = ?2

**Algebra**

6(j + j+10) = 780

**maths**

(A) subtract each number from 1000. Which difference is smallest in size, + or -? (B) 46+2x+96+x+62 = 306 (C) add up all the rentals and divide by the number of days

**geometry**

it can only be angle PKE=16° That means angle PKN=32° and KPM=148° MK and PN bisect the angles of the rhombus. Now you can get all the angles you need. Note that diagonals are perpendicular.

**Math**

do you not have a formula for the payments? Just plug your numbers into the formula.

**Math**

There are 10C3 ways to pick the females, and 10C2 ways to pick the males. So, that makes 10C3 * 10C2 ways to form the committee.

**Geometry**

If the area is 10, then 1/2(5+15)h = 10 1/2(20)h = 10 10h = 10 h = 1 If the radius is 7.5 (diameter=15) then 3.14*7.5^2/2 = 3.14*56.25/2 = 176.625/2 = 88.3125

**math**

6/(6+8+6) * 30 = 9

**math**

4 esses in 13 letters, so p = 4/13 * 3/12

**Math**

3/4 * 8 * 9? = 54? cm^3

**Math**

Not quite. There are 4 possible outcomes Coin1 Coin2 H H H T T H T T Two of those result in a tail and a head. So, the probability is 2/4 = 1/2

**algebra**

p = s + 12 + 2s+3 = 3s+15

**expression, not equation**

7(a-b)-8(a-2b) 7a-7b-8a+16b now you can finish it off

**Math**

(3/r)^2 + (4/r)^2 = 1 25/r^2 = 1 r = ±1/5 tanx = 4/3 x is in QI or QIII

**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