# Posts by steve

Total # Posts: 51,171

**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)

**Calculus**

See your other post.

**Math**

a nominal rate of r compounded n times per year is (1+ r/n)^(nt) So, since you have a 4t there, .01 = r/4, making r=4% (1 + .04/4)^(4t)

**Math**

3.3*0.006*10 = 0.198 12*0.005*200 = 12 .198/12 = 0.0165 = 1.65x10^-2

**Math**

ok, so we have a circle and a tangent. A tangent to the circle will always be perpendicular to a line through the center. Now what? Ever hear of a space bar?

**Math**

(3.25*8)x10^(-22+3) = 26x10^-19 = 2.6x10^-18

**pre calc**

what are all those sin() stuff still hanging around? The final expression should just be a bunch of fractions. tan(? + ?); cos(?) = ? 1/3 ? in Quadrant III, sin(?) = 1/4 ? in Quadrant II You really need to review the signs of the trig functions in the various quadrants. In ...

**pre calc**

me either. why don't you show your work? I've done several already; your turn. I'm sure we can pinpoint where you go astray.

**Pre calc**

Funny - you and Ruth need to compare notes ... in QI sin? = 5/13 cos? = 12/13 in QII sin? = 1/?5 cos? = -2/?5 sin(?+?) = (5/13)(-2/?5)+(12/13)(1/?5) = 2/(13?5)

**Pre calc**

Better check my math, then. Also, maybe they don't want the radical in the denominator.

**Pre calc**

in QIII sin? = -5/13 cos? = -12/13 in QIV sin? = -1/?10 cos? = 3/?10 sin(? ? ?) = (-5/13)(3/?10)-(-12/13)(3/?10) = 21/(13?10)

**Math - Calc**

You forgot the chain rule. The result is (x^2)^(x^2) * 2x if f(x)=?[1,u] t^t dt and u = u(x) then f'(x) = f(u) * du/dx

**Math**

well, 141 is 11/12 std above the mean...

**Calculus - Definite Integrals Please Help!**

If you want h'(x), then that would be [cos(sin^3(x))+sin(x)]*cos(x)

**Math - Calculus**

((3x+2)/(3x-5))*3 - ((6x+2)/(6x-5))*6

**statiictis**

So, you want 4 of each kind. Pick any set of 4 from each kind, and then there will be 12! ways to play them. 21C4 * 13C4 * 25C4 * 12! = 2.593*10^19

**Calculus**

((1/2)x^2-1)^8

**math**

It's easier if you rewrite it as u = 1/x^4 u^2-u+1 = u^2-u+1/4 + 3/4 = (u - 1/2)^2 + 3/4 = (1/x^4 - 1/2)^2 + 3/4

**Math**

If all the angles are in QI, then sinA = 4/5 cosA = 3/5 sinB = 12/13 cosB = 5/13 now just use the difference formula

**Science! NEED HELP URGENT**

none. wavelength and velocity are related, but not amplitude. Now, if you're talking about something like a water wave, where water molecules are moving up and down, then of course, amplitude is related to the velocity of the moving particles.

**Geometry**

102lb * 0.45kg/lb * 2.5mg/kg = 114.75 mg

**English**

I think google is very helpful for these idioms. You can get various definitions and explanations.

**Math**

Maybe if you explain what you mean by a road a pound we can work it out. However, the perimeter of the drawing is 16 inches, and 16*5 = 80 so ...

**Language Arts**

ever see a cold, dreary, misty winter day? (between dawn and dusk, both the boundaries of the dark of night)

**trigonomenty**

Try this. http://www.cut-the-knot.org/pythagoras/cos36.shtml

**math**

plug in 3 for p: 3 = 14-2q

**Math**

distance = speed * time so just crank it out

**Math**

well, what numbers multiply to 40? example: 8*5 = 40 so, -8*5 = -40 8 * -5 = -40 Now, pick any other factors of 40. As long as an odd number of them are negative, the result is -40 2(-2)(-2)(-5) = -40 -10 * 4 = -40 and so on. Not as hard as it looked, was it?