# Posts by STEVE

Total # Posts: 50,503

**Math (Integrals) (Basic Integration)**

that is correct. There is no du to work with. I mean, I did all the algebra for you, and explained the 2nd example to show why it worked there.

**Math (Integrals) (Basic Integration)**

Nope. Not that easy. x^4+1 = x^4+2x^2+1 - 2x^2 = (x^2+1)^2 - (?2 x)^2 = (x^2+?2 x+1)(x^2-?2 x+1) 1/(x^4+1) = 1/[(x^2+?2 x+1)(x^2-?2 x+1)] = 1/(2?2) ((x+?2)/(x^2+?2 x+1) - (x-?2)/(x^2-?2 x+1)) = 1/(2?2) ((2x+?2)/(x^2+?2 x+1) - x/(x^2+?2 x+1) - (2x-?2)/(x^2-?2 x+1) + x/(x^2-?2 x...

**Math**

that depends on a number of values not given here, as I'm sure you can tell. Better research the topic a bit more.

**Math please help me**

just multiply all the numbers from 1 to 4: 4! = 1*2*3*4 = 24 5! = 1*2*3*4*5 = 120 and so on

**tangential velocity physics**

google is your friend. Or, consider that the moon is roughly 250,000 miles from earth It takes about 28 days to orbit once. So, its linear speed is 2pi*250000mi/28days Find out the actual values, in meters and seconds, and that will give you its tangential speed.

**Math (Integrals)**

Just expand the polynomial and integrate all the terms: 4x^2-4x+1 dx

**Math**

well, 363 = µ - 2? what does the rule have to say about that?

**Mathematics**

since cosine has a period of 2pi, either A=X or |A-X| = k*2pi Since both A and X are less than pi/2, A=X

**Maths**

(s/2)^3 / s^3 = (s^3/8) / s^3 = 1/8

**Math**

there are 26 choices for each letter. So, multiply.

**Math**

500 - 8*45 = ?

**algebra**

looks good to me

**math**

a = 3 Since the GP has a common ratio (a+3d)/(a) = (a+12d)/(a+3d) (3+3d)/3 = (3+12d)/(3+3d) 1+d = (1+4d)/(1+d) d = 2 a+9d = 3+9*2 = 21

**MAth**

950000 * 0.90^5

**Math - Calc**

see whether this helps http://www.wolframalpha.com/input/?i=plot+y%3Dx%2B6,+y%3Dx%5E3,+2y%2Bx%3D0

**calculus**

I get S-S6 = 5.93*10^-6 < 1.0*10^-5

**Algebra**

I disagree. Why are you subtracting the last terms?

**Math**

timothy's coins were 10 more than pearly's. So, if the new amounts were t' and p', 10 more than pearly = 10+p' t' = 10+p'

**Math**

t = 5/8 p t + 1/4 p = 10 + 3/4 p now just solve for t

**Calculus indefinite integrals**

(a) u = x^2-7 (b) u = 1/5 x^(5/3) + 2

**Pre calc**

looks like you need to review your basic identities and double-angle formulas: cos^4x-sin^4x = (cos^2x-sin^2x)(cos^2x+sin^2x) = cos^2x-sin^2x = cos2x (B) is clearly not true, since sin^2x/(1+cos^2x) has no asymptotes

**Calculus - Integrals**

The curves intersect at (-1,2) and (-1,-2) So, using horizontal strips of width dy, the area is a = ?[-2,2] (3-y^2)-(y^2-5) dy Using vertical strips, we have to split the region in two at the intersections, and then symmetry helps: a = ?[-5,-1] 2?(x+5)dx + ?[-1,3] 2?(3-x)dx

**Math**

add 30 ft to each dimension (15 ft on each side) then do perimeter as usual

**algebra**

vertical shrink by 4 shift down 12

**Algebra**

y = (3x^2+8x-10)/(x^2+7x+12) = (3x^2+8x-10) / (x+3)(x+4) no holes, since y is never 0/0 vertical asymptotes where the denominator is zero: x = -3,-4 as x gets huge, y -> 3x^2/x^2 = 3 so that is the horizontal asumptote http://www.wolframalpha.com/input/?i=(3x%5E2%2B8x-10)+%...

**Math**

The average lies somewhere between the lowest and highest values. There is no way that all the children in a group can be above the average for that group. However, it's possible that all the kids in Lake Wobegone are above the national average, or above the average of ...

**Math**

Of course it's possible. If 18 students get 100%, they all score above the average, which is less than 100.

**Math**

6C2 = (6P2)/2! = 6*5/2 = 15

**science**

C looks good to me.

**Math**

The wording is very strange. Maybe you mean there are thirteen 2/3-cup servings in a bag. In that case there are [13*(2/3)]/(1/2) = 52/3 = 17 1/3 half-cup servings

**Calc 1**

since e^(ln?) = ?, sin(?) is the first time the curve intersects the x-axis. So, the volume, using discs of thickness dx is v = ?[0,ln?] ?r^2 dx where r=y=sin(e^x) v = ??[0,ln?] sin^2(e^x) dx Now, sin(e^x) is not integrable using elementary functions. I guess you'll have ...

**Geometry**

Draw a horizontal line through M to intersect BT at P. Since MP is parallel to AC, angles TMP and TNC are congruent, making right triangles TMP and TNC similar. So, PT/PM = NT/NC since M is the midpoint of AB, P is the midpoint of BN. Thus, NC=2MP PT/MP = NT/2MP 2PT = NT Since...

**math**

10P4 = 10*9*8*7 = ? not sure what the 4- means

**math 105**

so, what is the rule? You have the mean and the std....

**physics**

correct

**physics**

T^2 is proportional to L/g. The mass has no effect. So, if A^2 = L/g B^2 = 2L/g B^2/A^2 = 2 B/A = ?2 ? 1.4 answer C

**physics**

Since the period T = 2??(m/k) y = 0.1 sin(?(k/m) t) = 0.1 sin(5t) the speed v = 0.5 cos(5t) which has a maximum value of 0.5

**Math**

4 6/12 - 3 8/12 = 3 18/12 - 3 8/12 = 10/12 so, 10" left or, 54"-44" = 10"

**Math help plz**

the area needed is ?r^2 + 2?rh = ?r(2h+r) = 6?(24+6) = 180? = 565

**Math**

#14 wants volume, not area the others look good.

**math**

For #1 and 2, type in your function at wolframalpha.com or any of many other fine graphing web sites. -4 = 4-8 7 = -4+11 so, (x,y) -> (x-8,y+11)

**Pre calc**

because I studied the double-angle formulas, as you need to do!

**Pre calc**

In QI, you have sinx = 5/13 cosx = 12/13 tanx = 5/12 sin2x = 2sinx cosx = 2(5/13)(12/13) = 120/169 similarly using the other double-angle formulas.

**Pre calc**

If this isn't even math to you, you clearly haven't been paying attention during the discussion of the double-angle formulas ... (1-cos2x)/sin2x = (1-(2cos^2x-1))/(2sinx cosx) = (2-2cos^2x)/(2sinx cosx) = 2sin^2x/(2sinx cosx) = sinx/cosx = tanx sin4x/sinx = (2sin2x ...

**Urgent math help!!**

Huh? You are in trig, and cannot tell what the x- and y- coordinates are? The point (-3,4) is at x = -3 y = 4 since r^2=x^2+y^2, r=5. sin? = y/r = 4/5 You got the right answer; was it just a guess?

**Urgent math help!!**

Draw your triangle in standard position. It is clearly a 3-4-5 right triangle. sin? = y/r

**Trig**

well, csc^2x-1 = cot^2x cot*cos/cot^2 = cos/sin * cos * sin^2/cos^2 = sin

**Math**

she kept 3/8 * 25 = ? kg

**Math**

you have to keep writing an equation; you just dropped off the right side. In any case, remember from your Algebra I that you usually need to set stuff to zero to solve. Now, using the fact that sin^2x+cos^2x=1, 2cos^2x+sinx-1 = 0 2-2sin^2x + sinx-1 = 0 2sin^2x-sinx-1 = 0 (...

**Trig**

oops, sorry Reiny...

**Trig**

Remember your double-angle formulas. You have tan(2*30) = ?3 Reiny dropped the 2 in the numerator...

**Trig**

(1+tan^2x)/tan^2x = 1/tan^2x + 1 = cot^2x + 1 = csc^2x

**laplace transformation:help damon or steve or rein**

In case you don't have an online calculator handy, L{1} = 1/s L{e^t} = 1/(s-1) = F(s) L{t^2 e^t} = (-1)^2 F"(s) = 2/(s-1)^3 L{y} = f(s) L{y'} = sf(s)-f(0) = sf(s)-1 L{y"} = s^2f(s)-sf(0)-f'(0) = s^2f(s)-s+2 L{y'"} = s^3f(s)-s^2f(0)-sf'(0)-f&...

**Bachelor of computer science**

I'm sure you can do the I/O. The calculation is if hours <= 0 print "error - no hours entered" else if hours <= 40 pay = rate * hours else pay = rate*40 + rate*1.5*(hours-40) endif

**math**

make up your mind -- BASIC or FORTRAN ? in any case, find the prime factors of each number, and select all those common to both. Their product is the HCF.

**algebra**

I don't see an inequality either. But, since the cost is not zero, you know that profit < earnings

**MAth**

If x is the middle integer, (x+4)/(x-4) = 2

**Calc**

yes. The f"' nonzero makes sure that it's not just a cup like x^4.

**Calc**

the slope has a minimum at x=c. f"(c)=0 means it's an inflection point. Think of the graph f(x) = x^3

**Math**

suppose k" = -5 on [-3,0]. You know that k'(-2) = -4. So, k'(0) would be -4+2(-5) = -14 Now, suppose that k" = -2 on [-3,0]. k' would be -4+2(-2) = -8 So, -14 <= k'(0) <= -8 remember that k" is just the slope of k'. k'(0) will lie ...

**Math**

you have ?y ? k'(-2)*?x You have k(-2), so use ?x=0.6 If the curve is concave down, it lies below the tangent line.

**Math**

6x10^-5 * 700 = (6*7)x10^(-5+2) = 42x10^-3 4x10^5 * 1.5x10^7 = (4*1.5)x10^(5+7) = 6x10^12 Now just divide as usual

**math**

just solve the system a(r^5-1)/(r-1) = 61 ar^4 = 81 Hint: 81 = 3^4

**math**

draw a diagram. Using similar triangles, h/(9+6) = 5/6

**MATH**

there are 20 students. So, the chance of getting the two freshmen is 2/20 * 1/19 = 1/190 This just selection without replacement.

**Precalculus**

the only constant I see is -1/5 I guess that make the power clear, eh?

**proofread**

I'll let bob handle the physics. You need to proofread your post this time -- get rid of the sloppy typos and finish the question.

**science URGENT!!**

hello. amplitude is wave height...

**science URGENT!!**

wave height is amplitude. and it still does not affect the frequency or the speed!

**Science**

F = ma, so a = -144/15 = -9.6 m/s^2 75 mi/hr = 33.53 m/s v = Vo + at = 33.53 - 9.6t v=0 at t=3.49 So, it will take 1.6 + 3.49 = 5.09 seconds to stop No deceleration occurs during the first 1.6 seconds, so the distance is s = 33.53*5.09 - 4.9*3.49^2 = 110.99 m

**Science. HELP!!**

not at all. The speed of light does not change because of the color. The speed of sound does not change because the pitch changes.

**Math**

The first step is to remove the parentheses. Then what do you have?

**SCIENCE**

The key to this is to come up with the equation of motion. If you review the relevant section, you will easily see that h(t) = 70+40t-4.9t^2 v(t) = 40-9.8t Now just apply your skills from Algebra I to find the vertex and find h(t)=70 and v(10)

**Math1**

just as the last one, 50=7*7+1

**Math1**

yep, it's a trinomial. If you want to factor it, recall that 15 = 2*7+1

**math**

without parentheses, exponents are done first. Remember PEMDAS? Parentheses, exponents, mult/div, add/sub So, -4^2 is evaluated as -(4^2) If you want +16, then that is (-4)^2

**math**

8×5-(6+12)÷6 = 37 as for simplifying, it doesn't get much simpler than 37 ...

**Science, Physics**

sinx(cosx-sinx) ---------------------------------- sinx(cosx+sinx)(cosx-sinx)

**Science, Physics**

sinx(cosx-sinx) ----------------------------- sinx(cosx+sinx)(cosx-sinx)

**Science, Physics**

(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

**Science, Physics**

(e2x)1/x < (x+ex+e2x)1/x < (2e2x)1/x e2 < (x+ex+e2x)1/x < 21/xe2

**Science, Physics**

(e2x)1/x < (x+ex+e2x)1/x < (2e2x)1/x e(2 < (x+ex+e2x)1/x < 21/xe21/x

**Science, Physics**

(e2x)1/x < (x+ex+e2x)1/x < (2e2x)1/x

**math**

no idea. how much area does a gallon cover?

**Math**

start by expressing each factor in scientific notation: 1.31x10^5 3x10^-4 Now just multiply as usual.

**math**

the whole pipe is the sum of its parts, right?

**Math**

Since you have two values for y, just set them equal to each other: x^2+2x = 3x+20 x^2-x-20 = 0 (x-5)(x+4) = 0 and you're home free...

**College Algebra**

Once you get to 5=5 you are done. There are infinitely many solutions. If you put the two equations to the same format, they are y = 5x-5 y = 5x-5 They are the same line. Any solution to one of them is also a solution to the other.

**Math**

a+12d = 27 a+6d = 3(a+d) Now just solve for a and d.

**Math**

I agree

**Maths**

you need to learn to use parentheses online to make it clear what you want, since otherwise, multiplication gets done before addition. You meant (cot^2(x)(sec(x)-1))/(1+sin(x))=(sec^2(1-sin(x))/(1+sec(x)) Apply some of your elementary identities: cot^2(sec-1) = (sec-1)/tan^2...

**Maths**

recall that (a^3-b^3) = (a-b)(a^2+ab+b^2) sin^2+cos^2 = 1 and it all drops right out.

**chemistry**

for the same reason that any other endothermic reaction is so considered...

**Calculus**

there are lots of online graphers. You already know what the graph looks like. The critical points are where secx=0 or tanx = 0 The concavity up or down is provided by the sign of y"...

**MAth**

so, just plug in the numbers, and you should get 129

**College Algebra**

no, it is actually .05x + .10y + .25z = 7.80 Now solve 'em!

**maths steve reiny bob reiny!!! Help**

One side is the distance from (3,4) to 2x+y+3=0 The other side is the distance from (3,4) to x-2y-1=0 The area of the rectangle is thus |2*3+1*4+3)/?(2^2+1^2) * |1*3-2*4-1|/?(1^2+2^2) = 13

**Calculus**

more like dy/(1+y) = x dx ln(1+y) = x^2/2 1+y = c e^(x^2/2) y = c e^(x^2/2) - 1

**Physics**

ever hear of google? http://www.physicsclassroom.com/class/1DKin/Lesson-1/Distance-and-Displacement

**Pre calc**

so, why don't you show us how you got it? I get in QIII tan? = sin?/cos? = 5/12, so sin? = -5/13 cos? = -12/13 in QIV sin? = -1/?10 cos? = 3/?10 You don't say what you want to do with that, but sin(???) = (-5/13)(3/?10)-(-12/13)(-1/?10) = 21/(13?10)