# Posts by steve

Total # Posts: 51,761

**math**

just divide and you get x+4 + 9/(x-2) So, as x gets huge, the 9/(x-2) vanishes, and the graph approaches the line y = x+4

**Math(Calculus)**

well, x^2-4 has a minimum at x=0, so e^(x^2-4) will as well. since x^2-4 has an absolute max at x=±2, so does e^(x^2-4) But, as long as this is calculus, dh/dx = 2x e^(x^2-4) dh/dx=0 at x=0 (minimum)

**Algebra**

Since F=ma, a = 8x/m. So, we have a(x) = -8/m x v(x) = -4/m x^2 + c1 s(x) = -4/3 x^3 + c1*x + c2 v(0) = 0 and s(0) = 10, so s(x) = -4/3 x^3 + 10 For the damping, just work with v(x) = -4/m x^2 + 8v and follow through to get s(x)

**Calculus**

Since the y-axis is also the axis of symmetry for the curves, we can just take one side and revolve it. This will be easiest using shells of thickness dx: v = ?[0,2] 2?rh dx where r=x and h=(8-x^2)-x^2=8-2x^2 v = ?[0,2] 2?x(8-2x^2) dx = 16?

**Math**

x = 3+4cos? y = 2+4sin? since cos^2? + sin^2? = 1, (x-3)^2 + (y-2)^2 = 4^2

**grades**

no way to tell all we know is that 15 students got above 74% and 15 got below 74%

**math**

x grows by 2 the slope is 2, so y grows by 4. so, y=2+4=6 or, more formally, (y-2)/(3-1) = 2

**Math**

each term is calculated using the previous term, not the next one!

**Calculus - Fundamental thm of Calc**

that would be ((1/4)(x^2)^2?1)^15 * 2x = 2x(x^4/4 - 1)^15

**Calculus - Related Rates**

Let the distance on the road be x. Then the distance z to the house is z^2 = x^2+25 z dz/dt = x dx/dt z(1) = ?26 ?26 dz/dt = 1 * 65 ...

**Calculus Help**

google hole in sphere. You will find many excellent discussions.

**Math**

well, you have a base perimeter of 30. So, the base of each triangular lateral face is 30/4 = 7.5 Now you can see what the side edges of the pyramid are, and there are four of them. let 'er rip...

**math**

reimburses? Did the doctor pay the patient something?

**maths**

L = km/h^2 So, Lh^2/m = k is constant You want L such that L*50^2 = 40*h^2/175 If you will supply the original height, then you can solve for L.

**math**

I don't see any graph ...

**algebra**

#1 The line of symmetry of y=x^2 is the y-axis (x=0). Similarly, for a translated parabola y=a(x-h)^2 + k the axis of symmetry is the line x=h. So, for y = -2x^2+3 the axis is x=0 #2 same for |x|. So, for y = 1/3 |x| - 2 the axis is x=0 #3 Huh? This stumps you? P(4,-4)-->(-...

**math**

I suspect there was meant to be a question in there somewhere, but I couldn't see it ...

**Math**

not mixed numbers. As a fraction, 87.3/8.73 = 10*8.73/8.73 = 10/1 = 10 In case you wanted individual fractions, 87.3 = 873/10 8.73 = 873/100

**Math**

I think Natavia is right. If P = 1/26 that is the theoretical probability. No mention is made of actually performing the experiment.

**Math**

recall that alternate interior angles between parallel lines are congruent. Draw a line through one vertex parallel to the opposite base. the rest is easy.

**Math**

4 * 1/4 = 1

**Math**

6 * 2/3 = 4

**Math**

4/12 = 2/6 = 1/3

**Intermediate Maths**

(a) 12(1/2)^(t/5700) = 10 t = 1499.3 years (b) 12(1/2)^(1599.3/5700) = 9.879 grams

**calculus please help!!**

The region is a semi-circle of radius 6, so the area ranges from 0 to 18? The integral is the area moving from the left side to some point x in [-6,6].

**Pre calculus**

set this up as a geometric series and then just apply the sum formula. a = 2 r = 0.9 allow for round trips up and down

**Probability**

to start out, there are 13 diamonds in 52 cards. As you draw the cards, the probabilities change, due to the makeup of the remaining cards. The result is 13/52 * 12/51 * 11/50 * 10/49 = (13*12*11*10)/(52*51*50*49) = 13P4/52P4

**Calculus**

A,C,D are true. The Intermediate Value theorem says that there is a zero in (-2,1) and (1,4) Rolle's Theorem (or MVT) says that f'(x)=0 somewhere in (-2,4) Mean Value Theorem says D is true

**Math 1**

I must say, these problems of yours are some cumbersome. I'm not sure they are teaching much about factoring, since most of them only have a single monomial common factor. Like this one: 48x^3 - 128x^2 -56x -16 8(6x^3-16x^2-7x-2)

**Math 1**

392n^4 + 168n^3 + 168n^2 +72n 8n(49n^3+21n^2 + 21n+9) grouping gives 8n(7n^2(7n+3)+3(7n+3)) 8n(7n+3)(7n^2+3)

**Math 1**

5x(16x^3-12x^2-12x-9) 5x(2x-3)(8x^2+6x+3) the discriminant of the quadratic is negative, so it has no real factors.

**Math1**

The x^2 is easy: 195x^5 + 168x^4 - 28x^3 - 24x^2 x^2(195x^3+168x^2-28x-24) Now things get tough. If there are any rational roots of the form p/q, then we know that p divides 24 q divides 195 Some time and effort should convince you that there are no other rational roots, so no...

**Maths ( Higher Level) Simultaneous Equations.**

the method is just to write the words as math: mother is x years old, her son is y years old and the sum of theirs ages together is 58 x+y = 58 Five years ago, the mother was five times as old as the son (x-5) = 5(y-5) Now it's easy! x+y=58 x-5y = -20 subtract and you get ...

**algebra 2, check my answer plzzzzzz**

x = 5y^3 x/5 = y^3 y = (x/5)^(1/3) Or, if you insist on rational denominators, y = (25x)^(1/3)/5 your notation would seem to indicate 3?(25x)/5 rather than cubrt(25x)/5 = ∛(25x)/5

**Check my math question please? Very urgent!**

correct

**Math**

5/8

**Math**

well 45/3 = 15 = one-third of the bag. So, ...

**Math**

well, each foot of rope has 3 pieces, right ? 1 foot = 3/3 So, how many in 8 feet?

**Math**

42

**Math**

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.

**Functions (Math)**

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

**MATHS**

Note that ?STP is similar to ?RTS and angle SRT = angle TPQ (alternate interior angles)

**Calculus**

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)

**calculus**

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

**MATHS**

Since the bisector meets BC at a 45° angle, CB=PB = 15 so AP=25-15=10

**Mathematics**

on each trip, walk = 1/3 bus = 2 3/4 - 1/3 = 2 5/12

**math**

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

**Math**

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=...

**Calculus**

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

**Math**

true

**programming**

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--]

**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...