Posts by steve
Total # Posts: 50,530
well 45/3 = 15 = one-third of the bag. So, ...
well, each foot of rope has 3 pieces, right ? 1 foot = 3/3 So, how many in 8 feet?
If b is breaths, then 10^9 b * 1min/16b * 1day/1440min * 1yr/365day = 118.9 years This makes sense, since 1 billion seconds is about 30 years.
Draw the angle. It should be clear that y = -5 x = 2 r = ?29 Now just recall that sin? = y/r cos? = x/r tan? = y/x now let 'er rip
Note that ?STP is similar to ?RTS and angle SRT = angle TPQ (alternate interior angles)
dy/dt = k/y y dy = k dt 1/2 y^2 = kt + c y^2 = 2kt + c y = ?(2kt+c) E does not work, as you can easily see: y = ?(2kt)+4 dy/dt = 2k/(2?(2kt)) = k/?(2kt) but that is not k/y = k/(?(2kt)+4)
just plug and chug. The surface can be thought of as a stack of thin rings. A = ?2?r ds where r=y and ds=?(1+y'^2) dx so, y' = (9x^4-4)/(12x^2) A = 2??[1/2,1] (x^3/4+1/(3x))*((9x^4+4)/(12x^2)) dx = 1981?/3072
Calculus (trig derivatives)
Draw a diagram. It is clear that when the distance of the beam from the point P on shore nearest the lighthouse is x, tan? = x/2 sec^2? d?/dt = 1/2 dx/dt when x=4, we have (1+(4/2)^2)(2?/20) = 1/2 dx/dt dx/dt = ? mi/s
Since the bisector meets BC at a 45° angle, CB=PB = 15 so AP=25-15=10
on each trip, walk = 1/3 bus = 2 3/4 - 1/3 = 2 5/12
the hexagon has an apothem a=9. so, its sides are s=6?3 Now, you know how to find the area of a hexagon of side s. The area A of the pyramid is just A=6(sa/2) The volume is 1/3 (area of hexagon)*42
Let Ø = angle to bottom of picture ? = angle subtended by the picture x = distance from wall to observer tanØ = 4/x tan(?+Ø) = 9/x so (tan?+tanØ)/(1-tan?tanØ) = 9/x (tan?+(4/x))/(1-(4/x)tan?) = 9/x see what you can do with that. or, if y=...
g'(x) = ?((x^2)^3+2) * 2x now just plug in x=2
Alg II COT
cot ?/6 = ?3
Alg II COT
google radicals and examples
there are lots of sort routines you can find online. So, this will accept an arbitrary number of names: n=0 name="x" while name ? "" read name if name?"" then names[n++] = name end while snames = sort names,n while n>0 print names[n--]
(4cosx+1)(2cosx+1) = 0 now it's a cinch...
There are 11P4 ways to pick the pained boats.
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
a set of n elements has 2^n-1 non-empty subsets
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 = ...
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
well, 22 tens is 220 ...
looks ok to me, aside from using "square of" to mean "square root of"
use a common denominator. Then you have 12/28 + 7/28 not so hard now, eh?
their areas are in the ratio 2:3:4
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...
380 They have no common factors LCM(x,y) = xy/GCD(x,y)
I guess he is then lost. Or, try the law of cosines.
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 ...
sides have shrunk by ?4 = 2
well, just add up the numbers!
(8/6) * 48 = 64
well, the sum of all 9 angles is 7*180 = 1260 (1260-462)/6 = ?
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
well, 05 is half of 10, so ...
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
6(j + j+10) = 780
(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
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.
do you not have a formula for the payments? Just plug your numbers into the formula.
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.
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
6/(6+8+6) * 30 = 9
4 esses in 13 letters, so p = 4/13 * 3/12
3/4 * 8 * 9? = 54? cm^3
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
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
(3/r)^2 + (4/r)^2 = 1 25/r^2 = 1 r = ±1/5 tanx = 4/3 x is in QI or QIII
they are planed in a pentagram figure https://richardwiseman.wordpress.com/2013/07/29/answer-to-the-friday-puzzle-216/
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.
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
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...
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.
there are 4 choices for each letter. So, that makes 4^3 = 64 codes in all.
j = 2d p = d+5 j+d+p = 95 Now just solve for the ages
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...
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?
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!
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
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°
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...
see your earlier post
draw a diagram. It is clear that (a) tan? = 20t/50 (b) solve when t=3 Since tan2??2tan?, (c) is clearly false.
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 ...
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.
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
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
#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 ...
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...
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
z^2 = 10^2 + 8^2 - 2*10*8 cos120°
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
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
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.
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...
x = ky+m 7 = 5k+m 8 = 7k+m so, k = 1/2 and m = 9/2 ...
Well, there are two regions, on the intervals [-3,0] and [0,1] Finding the areas is pretty straightforward, right?
v = ?(20/2)^2 * 0.05 = 5? cm^3
I guess that'd depend on how many yellow balloons there were at the start.
slide depends on the flips
the altitude of the triangle is the distance between the lines. A = 1/2 bh b is constant, h is constant.
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?
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 ...