Friday

April 18, 2014

April 18, 2014

Total # Posts: 21,750

**Algebra 1**

any examples shown here will look just like the ones in your text. Why not post some specific type that is causing you difficulty? You can copy/paste the √ sign from here. √4 = 2 √8 = √(4*2) = √4√2 = 2√2

**algebra**

6n + 2b = 23.52 3n + 4b = 25.53 multiply the 1st equation by 2 and you have 12n + 4b = 47.04 3n + 4b = 25.53 now subtract to get 9n = 21.51 n = 2.39 Now you can get b

**Math HELP!!!!!!!!**

well, the max height is found at (6.25,625) So, as long as he's not too far away when he tries to leap over the building, yes, he'll make it (as long as the building is sufficiently narrow).

**Calculus**

∫2x e^-x dx = -2 e^-x (x+1) = -2(x+1)/e^x Now, e^x grows so much faster than x that the limit is zero as x -> ∞. That can be justified by using lHospital's Rule, to show that the limit is the same as -2/e^x -> 0 However, as x -> -∞ e^x grows very...

**MATH HELP**

to use factoring, get everything on one side of the equation. x^2 = -25 + 10x x^2 - 10x + 25 = 0 The only factors of 15 are 1,5,25 (x-5)(x-5) is what you want 2x^2 + 9x = 5 2x^2 + 9x - 5 = 0 you know that you will have (2x+?(x-?) or (2x-?)(x+?) since 9 = 10-1, I'd look at ...

**Math- please help!!!**

#1: 15p^7 - 14p^7 = ? #2: just collect like exponents (6-4)x^2 - (2+1)xy + (3+1)y^2

**pre-calc - NO**

log(a+b) is NOT log a + log b log(ab) = log a + log b 3-log(x+2)=5 log(x+2) = -2 assuming base 10, x+2 = 1/100 x = 1/100 - 2 = -1.99

**math**

75 or 175

**ROCKS :D just want to make sure.**

I'd say (A) The age difference is so small on a geological scale that the depth is not a good criterion.

**math word problem**

g = 2j-6 p = j+4 g+j+p = 54 22 gumdrops 14 jawbreakers 18 pops

**math**

just use the distributive law (m^2-2m+7)(3m+6) = m^2(3m+6) - 2m(3m+6) + 7(3m+6) = 3m^3 + 6m^2 - 6m^2 - 12m + 21m + 42 = 3m^3 + 9m + 42

**Maths**

if a and b are the times, since the speeds are in ratio 3/4, the times are related by a/b = 4/3 a = b+30 a = 120 b = 90

**Math**

$15 discount 15/40 = 37.5%

**Math**

B

**MATHS**

3/4 speed takes 4/3 the usual time. So, 4/3 x = x + 2.5 1/3 x = 2.5 x = 7.5

**Calculus Help STEVE**

oops. Well, if you got y', y" is just a bit down the road There are two ways to do it. First, use y' directly: y' = -2x^2 / 3y^2 by the quotient rule, y" = ((-4x)(3y^2) - (-2x^2)(6yy')] / 9y^4 = ((-12xy^2 + 12x^2y(-2x^2/3y^2)) / 9y^4 = -12xy(y - x(-2x...

**math (check please)**

1/3 * 9^2 * 4 = 108 good job

**Math - Hmmm**

taking a guess, we have 45/2 a^2 + 45b^2 = 90 a^2 + 2b^2 = 4 Can't think of any nonzero integers that will do the job.

**math**

a = pi d h = pi * 1.6 * 5.6

**Physics**

219 * (4.12/11.2)

**Math: Please help**

Well, if the 30th %ile is where 30% of the scores are less than that, then since there are 10 scores, and 3 are less than 113, I'd say you are correct. same for 200

**Math**

it's either gibberish or an equation.

**Calculus Help**

6x^2 + 9y^2 y' = 0 y' = -2x^2 / 3y^2

**Calc Help :(((((**

I figure it's y = ln (x+√(9+x^2)) y' = 1/(x+√(9+x^2)) (1 + x/√(9+x^2)) = 1/√(9+x^2) y" = -x/(9+x^2)^(3/2)

**math 1 question pls help!!!!!**

Ah - Damon is correct - I missed "similar"

**math 1 question pls help!!!!!**

since the length and height are not changed, the volume is 1/2 as much as before. 120 is correct.

**Calculus Help Please!!!**

f(θ) = arcsin(√sin9θ) since d/dx arcsin(x) = 1/√(1-x^2), f'(θ) = 1/√(1-sin9θ) * 9cos9θ/2√sin9θ = 9cos9θ / 2√(sin9θ)√(1-sin9θ) nasty, but that's how it is

**CALCULUS HELP**

r = f(g(h(x)) r'(x) = f'(g(h(x))) g'(h(x)) h'(x) r'(1) = f'(g(h(1))) g'(h(1)) h'(1) = f'(g(4)) g'(4) (3) = f'(5) (5)(3) = (7)(5)(3) = 105

**Calculus Help Please!!!**

y = secx y' = secx tanx at x = pi/6, y' = 2/√3 * 1/√3 = 2/3 so, now we have a point and a slope, so the line is y - 2/√3 = 2/3 (x-pi/6) http://www.wolframalpha.com/input/?i=plot+y%3Dsec%28x%29+and+y+%3D+2%2F3+%28x-pi%2F6%29+%2B+++2%2F%E2%88%9A3

**Calculus :(**

y' = -d sin(t) + 2t sin(t) + t^2 cos(t)

**Math - PreCalc (12th Grade)**

as with any arithmetic progression, Sk = k/2(a1 + ak) = k/2(1+5k-4) = k/2(5k-3) Sk+ak+1 = k(5k-3)/2 + 5k-4+1 (D) Don't forget your algebra I just because you're in pre-calc now.

**Math - PreCalc (12th Grade)**

ok so far, though they already told you that f(2)=3 because the point (2,3) is on the graph. As you know, the slope at any point (x,y) on the graph is 2x+2 So, the slope at x=2 is 6 Though, if this is pre-calc, how do you know the slope of the tangent to a curve? That's ac...

**Algebra**

visit calc101.com and click on the link for long division. You can see all the details and play around with other polynomials.

**Math - PreCalc (12th Grade)**

yes, that would be correct.

**Math - PreCalc (12th Grade)**

You are correct; you cannot divide by zero. So, f(x) is not defined for x=4. However, for any other value of x, f(x) = (x-4)(x+4)/(x-4) = (x+4) So, if you define f(4) = 8, then f(x) = x+4 for all values of x, and is now continuous.

**Math (check answers)**

A only has 4 faces, 4 vertices, 6 edges D has 5 faces, 5 vertices, 8 edges

**Math**

I think you mean ±3√3 The two solutions are √27 and -√27 But, √27 = √(9*3) = √9√3 = 3√3

**Math**

a quadratic always has two solutions.

**Math - PreCalc (12th Grade)**

I guess, though product is also involved lim(2x) = 2 * lim(x) But I guess C is closest.

**Math - PreCalc (12th Grade)**

D If for no other reason, the sum of nth powers will be a polynomial of degree n+1.

**Probability**

#3 ok #4 There are 3 odd numbers out of 6. P(odd) = 1/2 #5 ok #6 what happened to the coin? (A) #7 ok #8 ok

**Math (check answers)**

correct

**math**

nothing to solve. There is no equation. Now, if you just want to simplify the expressions, you wind up with -3+n and 23+N

**Math HELP!! Who can solve??**

If I tell you that 2x = 10 how do you find x? You just divide by 2 on both sides of the equation, so you wind up with x all alone. Similarly, if you have D = 2zv and you want to find v, just recognize that D = (2z)v and you want to divide by the 2z to get v by itself. Thus, D/...

**calculus**

I think you missed a t in the denominator ∫ 1/(t^2-5t) dt = 1/5 log((5-t)/t) the definite integral is log(6)/5

**Math help understand**

or, you can look at it like this. The bus makes up 195km in 3 hours, so it is going 195/3 = 65 km/hr faster than the bike. 15+65 = 80 km/hr

**geometry**

if two of the points are a diameter apart, the triangle is a right triangle. So, given that the first two points are on the same semi-circle, which they must be, then we want the probability that the third point is on the same semi-circle. If it is not, then the triangle will ...

**math question (CHECK)**

I get 1536

**Math**

a = j/2 a + j = 12 If all you want is the sum of their ages, it doesn't matter what a and j are.

**ama**

if the ladder reaches up h feet, sinθ = h/10 cosθ dθ/dt = 1/10 dh/dt You know θ and dh/dt, so solve for dθ/dt.

**math**

1/6, 1/4, 2/5, 6/11, 4/7, 5/8, 2/3, 15/22, 7/9, 9/10

**geometry - typo**

Read what you wrote. That's a might small fence.

**Math**

Hmmm. Once you have an equation for the line, which you have, there's no need to interpolate. You can just plug in 550 for x. Now, if you don't have a function, but just data points, then you need to interpolate. That is, use the slope between two points to estimate an...

**MATH (check please)**

Hmm. I got 285.7 typo?

**math algebra**

(w^2-x)(w^2-x)

**Math**

as it says, base 4. 16 = 4^2 the base is raised to the power.

**math help!**

well, since f(1) = f(5), the vertex is at x=3, midway between. So, y = a(x-3)^2 + k Now plug in your points to get a and k: 3-k = 4a k = 3-4a y = a(x-3)^2 + 3-4a Unfortunately, since all we have is two points, there's no way to determine exactly where the vertex is. It is ...

**trig**

tan(kx) has period pi/k, since x grows k times faster. So, I think you have a typo. It should be 3tan3x = 3 tan3x=1 now, you know that tan pi/4 = 1, so 3x = pi/4, and x = pi/12. Now, tan(3x) has period pi/3. 2tanx-2 = 0 tanx = 1 Since tan pi/4 = 1, tan (pi/4 + pi) = tan 5pi/4 ...

**trig**

as you said, sin and cos have period of 2pi and tan has period pi. That's how you know.

**math**

1-1/3 = 2/3 2/5 * 2/3 = 4/15 1 - 1/3 - 4/15 = 6/15 = 2/5 so, you want 2/5 x = 600

**math**

18 + (18+w) = 36+w

**algebra: functions**

f/52

**math**

cot -220 = 1.19 what, your calculator doesn't do tan x in degrees? cot x = 1/tan x

**Math (algebra)**

C(r) = .11(2pi r^2) + .06(2pi rh) since pi r^2 h = 500, h = 500/(pi r^2) C(r) = .11(2pi r^2) + .06(2pi r(500/pi r^2)) C(r) = .22 pi r^2 + 60/r For C to be a minimum, C'(r) = .44pi r - 60/r^2 = 0 .44pi r^3 - 60 = 0 r^3 = 60/.44pi r = 3.51

**Math**

vertical angles are always equal.

**Math**

the wall, ground, and ladder form a righ triangle, with the ladder as hypotenuse, so, x^2 + 4^2 = 12^2 supplementary angles add up to 180 degrees. So, x + 3x+2 = 180 one angle is x, the other is 3x+2. Solve for x, and you're almost there.

**math help**

Think of a regular hexagon. All the angles are 120°. Now, think of pressing down on the top, keeping the top and bottom sides parallel. The two side angles will decrease, and the other four angles will increase. Stop when the side angles have become acute.

**maths**

5^(logx) + x^(log5) = 50 Note that log(5^logx)) = logx * log5 log(x^log5)) = log5 * logx So, the two are equal. That means 2*5^logx = 50 5^logx = 25 = 5^2 logx = 2 x = b^2 where logs are base b. So, if natural logs, x = e^2 if common logs, x = 10^2 = 100 check (common logs): 5...

**College Math**

74 are taking something 36+52 = 88, so 14 are taking both. So, 22 are taking biology only 22/200 = .11 is P(biology only)

**MATH QUIZ PLS HELP!!!**

#6 baffles me. Pyramids don't have a radius. All the others look ok, as long as #5 wants the volume.

**MATH**

first step, as I have shown you on two of these, is do the substitution. Wherever there is an x, replace it with 789: -2.5(789^2) + 2825(789) - 115812.5 Now just pull out your handy calculator, or use an online calculator, and discover that the value is 556810

**Math - PreCalc (12th Grade)**

it asks for the first natural number ... the natural numbers are 1,2,3,... So, start trying them: n=1: (2n-1) = 1 1 <= 4*1-1 1 <= 3? yes n=2: (2n-1) = 3 1+3 <= 4*2-1 4 <= 7? yes n=3: (2n-1) = 5 1+3+5 <= 4*3-1 9 <= 11? yes n=4: (2n-1) = 7 1+3+5+7 <= 4*4-1 1...

**Math - PreCalc (12th Grade)**

start checking. How about showing some of your own ideas on further problems? I can do them - the goal is for you to work on them, eh?

**Math - PreCalc (12th Grade)**

Ummm. Do you not think it would be A? I explained how the discontinuity could be removed. Hence, it is removable.

**Math - PreCalc (12th Grade)**

at x=5, f(x) = 0/0, or undefined. At any other value, though, f(x) = (x-5)(x+5)/(x-5) = x+5 So, f(x) is undefined only at x=5. By defining f(5) = 10, the discontinuity is removed. f(x) = x+5 for all x.

**Math - PreCalc (12th Grade)**

just start checking: 5^1 <= 4^1 + 3^1 5^2 <= 4^2 + 3^2 5^3 > 4^3 + 3^3

**Math - PreCalc (12th Grade)**

Looks good to me. When finding a limit, first try just plugging in the given value. As long as you don't wind up with 0/0 or something else undefined, the resulting value is the limit, as long as the function is not defined differently on the two sides.

**Math - PreCalc (12th Grade)**

well, plug in x=2 and see what you get.

**math**

assuming the water heater actually sits in the pan, it occupies part of the area of the pan. So, the volume of the unused portion of the pan is π(11^2-9^2)(2)

**Grammar**

correct. The only mistake here is the word "grammaratically"! Maybe a bit too much capitalization, but I'm not sure that qualifies as grammar.

**Math - PreCalc (12th Grade)**

as x->3 from the left, (x-3) < 0, so |x-3| = -(x-3). -(x-3)/(x-3) = -1

**Math - Hmmm**

No, C is out because 8 ≠ 3.5+3.5+3 D is out because 10 ≠ 1+4+3

**Math - PreCalc (12th Grade)**

ok. so we have x + y + z = 15000 x = y + z + 3000 3x + 2y + 2z = 30000 Nope. That wasn't it. So, if #1 costs 3000 more than either #2 or #3, we have x + y + z = 15000 x = y + 3000 x = z + 3000 3x + 2y + 2z = 30000 No joy there, either. So, what about the choices? A: 3*9 + ...

**Math - clarify**

When you say Investing in the first venture will cost $3,000 more than investing in both the second and the third venture. do you mean #1 is 3000 more than both #2 and #3 combined, or 3000 more than either #2 or #3?

**Math - PreCalc (12th Grade)**

a = 5 b = 3 a^2 = b^2 + c^2 so, c = 4

**math**

p - .15p = 8000

**MATH**

Thanks for fixing the typo. What is the difficulty? What answer do you get?

**Physical Science**

well, if you buy into the fact that red+green+blue = white then the complement of any secondary color is the unused primary color.

**MATH**

and only two wrists!

**MATH**

and the rest are not

**math (check)**

3.14 * 8^2 * 5.7 = 1145.47

**math**

-2,5*780^2 + 2825*780 - 115,812.5 Is there a typo in -2,5? Evidently you use a period for the decimal point later on.

**derivatives - oops**

d/dx sinx = cosx

**math**

http://www.wolframalpha.com/input/?i=4x^2+cosx

**math - back to speed**

Reiny is, of course, correct, but given the choice of answers, I went with the assumption that they were just looking for knowledge of the area of the circle. And besides, who said that the flower areas were of any particular shape? Nothing says that the plants around the edge...

**math***1questions*****

pi * 7^2 = 154

**Calculus Help**

log y = log (e^-x) + log cos^2(x) - log(x^2+x+1) log y = -x + 2log cos x - log(x^2+x+1) 1/y y' = -1 - 2tanx - (2x+1)/(x^2+x+1) y' = -(1 + 2tanx + (2x+1)/(x^2+x+1)) * (e^-x cos^2x)/(x^2+x+1) Now, you can massage that for a few more steps, to get something that pleases you

**math**

(20)5/6-5A+11 50/3-5+11 17/3-5A

**math**

so, what number do you have to add to 7 to get 10?

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