Thursday

January 29, 2015

January 29, 2015

Total # Posts: 28,480

**Calculus**

Check out the graph, and scroll down to information on the extrema at http://www.wolframalpha.com/input/?i=%28x%2B1%29%28x-3%29^2 I think you are doing things right.
*January 11, 2015*

**Area and Perimeter Word Problems**

well, 15*15 = 225, so ...
*January 11, 2015*

**maths**

the cross-section area of the water is πd^2/8 = 49π/8 So, the water flowing through is (49π/8 cm^2)(2m/s)(10min) = 2309 liters
*January 11, 2015*

**Calculus**

Or, you can plot multiple curves to compare them: http://www.wolframalpha.com/input/?i=plot+y+%3D+%3Dx^2%2F%28x^2%2B5%29+%2C+y%3Dx^2%2F%28x^2%2B20%29
*January 11, 2015*

**Calculus**

I believe the answer is (A), given the interval [-3,-½]
*January 11, 2015*

**Calculus**

or, you can think of it as 5x^2-(2xy)+7y^2=0 and since (2xy)' = 2y+2xy', that gives 10x-(2y+2xy')+14yy' and go from there.
*January 11, 2015*

**Math**

(x+2)^2 = x^2+4x+4 So, the first is clearly not the same (x-2)(x+1) / (3x+1)/(x+1) = (x-2)/(3x+1) So, except for the missing division symbol, the 2nd is ok.
*January 10, 2015*

**Maths**

You can set up the vertices PQRS as points on the x,y,z axes such that P = (0,0,c) Q = (0,b,0) R = (0,0,0) S = (a,0,0) Then the midpoints are: PS: (a/2,0,c/2) QR: (0,b/2,0) PR: (0,0,c/2) QS: (a/2,b/2,0) You can easily show that the lengths of those segments are equal. There&#...
*January 10, 2015*

**PreCalc**

In general, there is no way to solve anything of 5th degree or higher. So, unless you can pick out some rational roots or easy quadratic factors, you are left with graphical or numeric methods.
*January 10, 2015*

**Integrals**

Assuming you meant 3x+5y < 15, just plug in allowable integral values for y: y 1: 3x+5 < 15 2: 3x+10 < 15 I think you can solve those, right?
*January 10, 2015*

**maths**

I get 8+7
*January 10, 2015*

**maths**

At a rate of 5 m/min, after 36 minutes he has gone 180m. On the 37th minute he swims 20m and is done. No need to let go and float backwards again.
*January 10, 2015*

**maths**

I have no idea what a sequence and series pattern involves, but surely you do. Try fitting Reiny's solution into your framework.
*January 10, 2015*

**maths**

37 minutes see previous solution for the explanation.
*January 10, 2015*

**Math**

What, still no work of your own to start with? Try factoring the quadratics, and see which factors cancel. And lose all those stupid parentheses around each term. Use one set of parens around the whole numerator or denominator.
*January 10, 2015*

**math**

No, because 4/9 = 16/36
*January 10, 2015*

**Math**

I don't see any way to simplify either one, except maybe to lose all the parentheses in the first one.
*January 10, 2015*

**english n engg.level to find d error related**

I hope it's not in English...
*January 10, 2015*

**Algebra 1a**

the lines meet where the x- and y-values are the same (9x+27)/3 = (72-8x/9) x = -153/19 so, y = 288/19
*January 10, 2015*

**Math**

since 12 = (3/2)*8
*January 10, 2015*

**Math**

helps to read: since 12 = (3/2)*8
*January 10, 2015*

**math**

just set x=0 and see what y is. That is the y-intercept -- where the graph crosses the y-axis (where x=0). Better review the examples and graphs in your text.
*January 9, 2015*

**algebra**

factor out the (x-y): (x-y)(4(x-y)^2-1) Now you have a difference of squares: (x-y)(2(x-y)+1)(2(x-y)-1) ...
*January 9, 2015*

**math**

because 15 minutes is 1/4 = .25 hour. So, 1 hr 15 min = 1.25 hours distance = time * speed, as calculated.
*January 9, 2015*

**math**

he ran 1.25 * 12 km Now convert that to meters.
*January 9, 2015*

**Intro to computers**

there are 4 memory sticks, each with 1GB of storage.
*January 9, 2015*

**Math**

well, what is a when t=0?
*January 9, 2015*

**Math 6 help**

By the way, the scale is 1:9, not 1:9cm Or, you could say it is 1cm:9cm It could have any units, for example, 1cm:100m 2mm:3km etc. If there are no units, you can make them both anything you want, as long as they are the same. If one unit is given, the other must also be ...
*January 9, 2015*

**math**

She's also good with spelling. Just sayin' ...
*January 9, 2015*

**algebra 2**

Apparently you want to simplify the expressions. I'll do #19. y^3/y^2 = y z^5/z^3 = z^2 So, you are left with -yz^2 Try the others. What do you get? If in doubt, show your work.
*January 9, 2015*

**Algebra 1**

Surely by this point you know how to graph a straight line. Graph both lines, and you will see that they intersect at (5,75). Just put a nice dot where the lines intersect. You might want to have a larger scale on the y-axis than on the x-axis. Just sayin'.
*January 9, 2015*

**math**

Check the differences: 9,16,25 Those are perfect squares. The next square is 36, so the next term in the sequence is probably 57+36 = 93
*January 9, 2015*

**Math**

So, can you not just plug in some values and check the results? I'll show you my work if you'll show me yours... Note: Usually it's easiest to use -1,0,1 for checking stuff.
*January 9, 2015*

**CAL**

Come on. There's no way to have two relative maxima without either a minimum or an inflection point in between them. y' = 15(x^4-5x^2+1) y" = 30x(2x^2-5) y" =0 when x = ±√(5/2) So, f is concave down in (-∞,-√(5/2)) up in (-√(5/2...
*January 9, 2015*

**Deductive reasoning help**

From your question, I deduce that you haven't studied hard enough. This seems to be a pattern among students.
*January 9, 2015*

**Calculus**

How about just writing down a few terms? 5^1 - 5^0 5^2 - 5^1 ... 5^100 - 5^99 --------------- 5^100 - 5^0 If you want to do it the hard way, you could note that ∑(a-b) = ∑a - ∑b
*January 9, 2015*

**math**

since circumference is just a constant multiple of radius, the circumferences are in the same ratio as the radii.
*January 9, 2015*

**Multiplying Polynomials**

The order of distribution does not matter, since the associative property guarantees that the order of grouping does not matter, and the commutative property guarantees that the order of multiplication does not matter. You can drag it out by brute force: (t-1)(t+1)(t^2+1) t(t+...
*January 9, 2015*

**Math**

well, did you check 24 to see whether it satisfies the conditions?
*January 9, 2015*

**math(5th grade)**

Since there are 1440 minutes in a day, that would be 1000000/(25*1440)
*January 9, 2015*

**MATH**

Try plugging x=0. Does 7(-2) = 0+14? Do the same for #2. Does -10(-2) = 10-10(-1)?
*January 8, 2015*

**trigonometry**

well, you know that sinθ = 3/10
*January 8, 2015*

**Algebra 1**

The amount of oil should be twice the amount of lemon juice. y = 2x How much of each ingredient is needed to make 45 ounces of floor wax? x+y = 45
*January 8, 2015*

**Algebra**

http://www.wolframalpha.com/input/?i=solve+2x+-+y+%3D+1%2C+x+%2B+2y+%3D+3+
*January 8, 2015*

**6th grade math**

I parsed the words as 7 times as long as (x minutes more than a platypus)
*January 8, 2015*

**6th grade math**

7(x+10) = 112
*January 8, 2015*

**Algebra 1**

12 x^-2 y^4 = 12y^4/x^2 -3xy^-7 = -3x/y^7
*January 8, 2015*

**Algebra 1**

for any nonzero number, x^0 = 1 by the way, your brackets are not balanced. Try using [ ] instead of those block graphics symbols. They're right there on your keyboard.
*January 8, 2015*

**MATH**

If there are x married men (or women) y single men z single women, we want 2x/(2x+y+z) (x+z)/2 = z (x+y)/6 = y x = z x = 5y so, z = 5y 2x/(2x+y+z) = 2*5y/(2*5y+y+5y) = 10/16 = 62.5% Check: If 1 single man 5 single women 5 married couples there were 16 people in the room. 10 ...
*January 8, 2015*

**Ap calc**

You should also know that y = tanh(x) y' = sech^2(x) = [2/(e^x+e^-x)]^2
*January 8, 2015*

**Advanced Algebra**

Is the ball punted straight up? If so, then h = 1+22t-4.9t^2 So, just solve for t when h<=5. If the ball was kicked at an angle, things get a bit harder, as the equation is then h = 1 + 22sinθ t - 4.9t^2
*January 8, 2015*

**Math**

since the fill rate is volume/time, and is constant, then just solve (40^2 * 5)/50 = (36^2 * 4)/x
*January 8, 2015*

**math**

check your other post for the answer.
*January 8, 2015*

**MATH**

15-.75x = 13-.50x Now just find x 15 - 3/4 x = 13 - 1/2 x 60-3x = 52-2x 8 = x After 8 games, Julia has 15-8*.75 = 15-6 = $9.00 Kevin has 13-8*.50 = 13-4 = $9.00
*January 8, 2015*

**math**

k = 2a k-7 + a-7 = 31 so, 2a-7 + a-7 = 31 3a-14 = 31 3a = 45 a = 15 So, Amy is 15 and Ke is 30 7 years ago, 8+23 = 31
*January 8, 2015*

**algebra**

google is your friend. So is any used book store. This site is not for long instruction, but for homework help.
*January 8, 2015*

**Algebra 1a**

First off, there are no "these two points." Second, the solution to the system of inequalities is a whole region of the plane, not just a point. If we are dealing with equations, then the point where the lines meet is (-126/11, 21/11) The shaded region in the plot ...
*January 8, 2015*

**Geometry**

since 24/16 = 3/2, the perimeter will be 3/2 as big. All the sides are 3/2 as big.
*January 8, 2015*

**Math**

S = 9/(1-r) If r>=0, you can see that 9/1=9 would be the smallest possible sum, if r=0. That is, if the sequence is 9 0 0 0 0 0 ...... Any other value for r will produce nonzero terms, and therefore a larger sum. You can investigate what happens when r < 0.
*January 8, 2015*

**Math**

These are just your basic trig functions. Review them and the first three are trivial. I'll do #4. sin^2x / 1+cosx (1-cos^2x)/(1+cosx) (1-cosx)(1+cosx)/(1+cosx) 1-cosx If indeed your teacher assigned these problems with no prior introduction to the trig functions, then don...
*January 8, 2015*

**math**

x+19 >= 8.2
*January 8, 2015*

**Useless Science Question**

the question is "what is missing?" It doesn't want an answer to what it should be.
*January 8, 2015*

**Useless Science Question**

the units km/hr, m/s, cm/ms ?what?
*January 8, 2015*

**Pre-Calc/Trig**

eq #1 says that 7y^2=5x^2-20x+3, so plug that into eq #2 and you have 5x^2+3(5x^2-20x+3) = 209 20x^2-60x-200 = 0 x^2-3x-10 = 0 (x-5)(x+2) = 0 And I think you can take it from there, no?
*January 8, 2015*

**science**

I'd get the answer by looking it up somewhere. Unless you have some data about amounts and times, there's no way to figure it out yourself. Can you see what a useless question you have posted? Sheesh! And, I think you have missed your deadline...
*January 8, 2015*

**Math**

x^2-3x-4 = (x-4)(x+1) So, you have x/[(x-4)(x+1)] - 4/(x+1) (x - 4(x-4))/[(x-4)(x+1)] (16-3x)/[(x-4)(x+1)] As usual, you cannot divide by zero, so any values of x that make a denominator zero are excluded. Now show us what you got on the others you posted. As it says, show ...
*January 8, 2015*

**Math**

what the heck are tries toom points ? If this is the old trisection problem, then you know that AQ = 2 PQ, so 7x+6 = 2(3x+5) 7x+6 = 6x+10 x = 4 PQ = 17 Now, AB = 3 PQ = 51
*January 8, 2015*

**math**

-1 < 2x/3 + 1 <= 4 Add 1 to get 0 < 2x/3 <= 5 times 3/5 to get 0 < x <= 3 Looks like x is any of 1,2,3
*January 8, 2015*

**calculus - ps**

Sorry - I forgot to include the factor of a on the right side. It will just multiply the final answer, since it's a constant.
*January 8, 2015*

**calculus**

(x-y)(x^2+y^2) = 2axy In polar coordinates, that is r(cosθ-sinθ)(r^2) = 2ar^2 sinθ cosθ r(cosθ-sinθ) = asin2θ r = asin2θ/(cosθ-sinθ) The loop is the area enclosed as θ goes from π/2 to π, so the area is a = &#...
*January 8, 2015*

**MNHS_HELP**

63^3+63^2+63+1 = (63^4-1)/(63-1) = 63^4-1 So, your number is 63^4-1+1 = 63^4 = 3^8 * 7^4 So, the prime factors are 3 and 7 I'll leave it to you to determine how many factors there are in all
*January 8, 2015*

**Math**

Remember that r is the ratio between successive terms. As for the infinite sum, remember that S = a/(1-r), so 7/(1-r) = 4
*January 8, 2015*

**maths**

gotta know radius or angle, or something else. s = rθ a = π/2 r^2 θ = π/2 rs
*January 8, 2015*

**maths**

h/60 = tan 30°
*January 8, 2015*

**math**

T5 = T2 + 3d, so d = (22-7)/3 = 5 The sequence is 2 7 12 27 22 Technically, the above is an arithmetic sequence. If you truly wanted an arithmetic series, then you'd have to use the sums S2 = 2/2(2a+d) = 7 S5 = 5/2(2a+4d) = 22 a = 16/5 and d = 3/5 The sequence is 16/5 19/5...
*January 8, 2015*

**College Algebra**

well, just answer the question. Find x when y=9. -0.146x+11.074 > 9 -0.146x > -2.074 x < 14.2 So, 1980-1994 Do the other in like wise.
*January 8, 2015*

**Algebra**

1/(x-5) = 7/2x 1(2x) = 7(x-5) 2x = 7x-35 5x = 35 x = 7 Whenever you have fractions, make sure that your answer does not make a denominator zero. In this case, things were ok. x=5 and x=0 would be excluded as solutions, since you cannot divide by zero.
*January 8, 2015*

**Algebra**

correct.
*January 8, 2015*

**Algebra**

t/8t + 16/8t = 17/8t t+16 = 17 t=1 That's the only solution I can see.
*January 8, 2015*

**Math**

well, it will triple 4 times. So, what is 3^4 * 1mm ?
*January 7, 2015*

**math**

I believe the angles should sum to 90°
*January 7, 2015*

**Math**

#2. I believe we are discussing a square, not any rectangle. x = -6 y = 6
*January 7, 2015*

**AP Calc. AB**

so, now you just want to find where -2*3x^2 e^x^3 = -6 x^2 e^x^3 = 1 I think a graphical solution will be easiest.
*January 7, 2015*

**Math**

k^-3 = 1/k^3 So, you want 1/k^3 = 1/256 k^3 = 256 Now, 256 = 2^8, so k = 2^(8/3) = 2^2 * 2^(2/3) = 4∛4 Hmmm. we got trouble. (a) and (b) are out. 4^-3 = 1/4^3 = 1/64 64^-3 = 1/64^3 = 1/4096 I think that 1/256 is a mistake. Or, the -3 is a mistake. 16^-2 = 1/16^2 = 1/256 ...
*January 7, 2015*

**Math**

wow - use run-on sentences much? If the base is x by y and the height is z, just break the area into rectangles lateral area: 2(xz+yz) total area: 2(xz+yz+xy) Now just plug in your numbers. You didn't say which was which.
*January 7, 2015*

**math**

2/3 = 16/24 3/4 = 18/24 So, what do you think is in between them?
*January 7, 2015*

**Math**

5^13 = 5^6 * 5^7 The powers on the right can be any two numbers which add up to 13.
*January 7, 2015*

**physics**

how long does it take the dart to fall 0.32 m? 4.9t^2 = 0.32 with that t, find the distance is 13.5t
*January 7, 2015*

**Algebra 1**

2(x+y) = 100 But, y = 3x+2, so 2(x+3x+2) = 100 ...
*January 7, 2015*

**math**

20*2πr = 45*60 must be a wind turbine.
*January 7, 2015*

**math**

6C2 = (6*5)/(1*2) = 15
*January 7, 2015*

**Algebra**

remember that log x^n = n*log x So, log(4)(3x + 4)^5 = 15 5log(4)(3x + 4) = 15 log(4)(3x + 4) = 3 3x+4 = 4^3 = 64 3x = 60 x = 20 Expanding (3x+4)^5 would not help, because then you just have a polynomial, but the log of a sum cannot be simplified. log(x^2+3) is not log x^2 + ...
*January 7, 2015*

**Math**

Assuming you mean (1+2x)/(x+4) = 3/(x-1) (1+2x)(x-1) = 3(x+4) 2x^2 - x - 1 = 3x + 12 2x^2 - 4x - 13 = 0 Now it's just a regular old quadratic equation to solve. Note that right from the beginning you have to exclude -4 and 1 from the domain.
*January 7, 2015*

**Math**

2x^2-2 = 2(x-1)(x+1) 2x^2-8x+6 = 2(x-1)(x-3) Now rewrite your fractions with the factors, put all over a common denominator of 2(x-1)(x+1)(x-3) and simplify the numerator. Probably something will cancel out. Non-permissible values are those that make any denominator zero.
*January 7, 2015*

**Math**

#1 I haven't followed the argument through yet, but I think if you play around with 0 < a < 1 (x+1)^a <= (x+1)^1 x^a + ax^(a-1) + ... + ax + 1 <= x+1 things will work out.
*January 7, 2015*

**MATHS**

well, (42)^2 - (38)^2 - (17)^2 = 31 so that should get you started...
*January 7, 2015*

**maths**

well, you have a = π/2 r^2 θ So, plug in a and r to find θ, and then you know that the arc length s = rθ Or, do it at one fell swoop: s = r(2a/(πr^2)) = 2a/πr
*January 7, 2015*

**Maths trigonometry**

secAcotB-secA-2cotB+2=0 secA(cotB-1) + 2(cotB-1) = 0 (secA+2)(cotB-1) = 0 Now it should be clear what the angles are.
*January 7, 2015*

**Math**

yes, as long as x≠1 (1-x) = -(x-1), so you have -x^2(x-1) ------------- (2x-3)(x-1) and the x-1 factors cancel.
*January 7, 2015*

**Algebra**

well, does it work in both equations x-3y = -5-3*2 = -11 ≠ 1 So, I'd say a resounding NO.
*January 7, 2015*

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