Tuesday

January 17, 2017
Total # Posts: 47,908

**Algebra**

1,2 ok 50 is 62.5% of 80

*December 1, 2016*

**Math check my answers**

all correct.

*December 1, 2016*

**Algebra**

product, not sum better review what that means.

*December 1, 2016*

**Algebra 2**

Having n in the index and in the expression is bogus. If you mean n-1 ? (3k-2) k=1 Then that can be split up into ?3k - ?2 Now, ?2 is 2(n-1) ?3k = 3?k You know that n ?k = n(n+1)/2 k=1 So for you example, 3?k = 3(n-1)(n)/2 The final sum is thus 2(n-1) + 3(n-1)(n)/2 If I ...

*December 1, 2016*

**Math**

x values change sign. y stays the same try experimenting with this -- it should be quite clear.

*December 1, 2016*

**Calculus**

(a) is correct (b) solve for t when h=0 (c) use the above t to get v

*December 1, 2016*

**Gk**

sounds like a cheap pair of glasses...

*December 1, 2016*

**Math**

524 - 520

*December 1, 2016*

**calculus**

g has a critical value where g'=0: x = ±4 max/concave down if g" < 0 g" = (x^2-4x+16)/(x-2)^2 For the tangent line, you have a point (3,4) and a slope (-7), so the line is y-4 = -7(x-3)

*December 1, 2016*

**Math**

23 years before Columbus set sail. or, maybe 11 14 16 19 ...

*December 1, 2016*

**Math**

You now have expressions for AN and NC Using the law of cosines, AN^2 = CN^2 + 2^2 - 2*CN*2 cos?ACN

*December 1, 2016*

**Math**

an angle is in degrees, not feet. Your height h is found via h/80 = tan70°

*December 1, 2016*

**math**

surely you can tell which is the greatest odd digit. Just take it from there.

*December 1, 2016*

**Algebra**

Fix the typo Then just expand and subtract, knowing that (x-4)^2 = x^2-8x+16

*November 30, 2016*

**Algebra**

w(3w-2) = 60 solve for w, then get length

*November 30, 2016*

**Calculus**

just find c such that 3c^2 = (65-9)/(4-2)

*November 30, 2016*

**Algebra 1**

double the 1st equation and you have 6x + 2y = -22 6x - 2y = -2 Now you can add them to eliminate y, or subtract to eliminate x.

*November 30, 2016*

**Math**

g = b+12 b/(b+g) = 35/100 ...

*November 30, 2016*

**calc**

since C = 2?r, dC/dt = 2? dr/dt Since dC/dt = ?/4, dr/dt = (?/4)/(2?) = 1/8 cm/min v = ?r^2h dv/dt = 2?rh dr/dt + ?r^2 dh/dt = 2?*4*12*(1/8) + ?*16*(1/2) = 20? cm^3/min

*November 30, 2016*

**Pre-Calculus**

r can be anything. So, the equation is just a straight line at that angle. tan? = -?3 y/x = -?3 ...

*November 30, 2016*

**Math**

add up the value of all the coins: 5n + 10(n+15) = 255

*November 30, 2016*

**Math -Calculus**

you need to do a definite integral. So, what are the limits? The curve intersects the x-axis at x = ±?(3/2) so now all you have to do is evaluate 6x - 4/3 x^3 at those points. Due to symmetry, you can just take twice the value from 0 to ?(3/2) f(?(3/2)) = 6?(3/2) - 4/3...

*November 30, 2016*

**Algebra**

If he makes x more cheese sandwiches, then (12+20+4+x) = 5(4+x) x = 4 check: 4+4=8 cheese out of 40 total. You can try other settings as well.

*November 30, 2016*

**Math (Geometry)**

well, if x is the number of hours worked, and y is the pay, y = 320 if x <= 40 8x if x > 40 It's a piecewise function, since it changes at x=40 hours

*November 30, 2016*

**Algebra**

no question. But there will be a guaranteed pay of 8*40 = $320

*November 30, 2016*

**Math**

If x>0, 4x > 5*2 x > 5/2 If x<0, 4x < 5*2 x < 5/2 But x<0, so x < 0 < 5/2 is a solution. Naturally, since if x<0, 2/x < 0, so 4/5 > 2/x for any negative x.

*November 30, 2016*

**6th grade math**

well, 30 = 6*5, so five times as far.

*November 30, 2016*

**Algebra**

from the axis and vertex, we know y = a(x+2) - 6 Now use the point to solve for a. You already know that a is positive (why?)

*November 30, 2016*

**Pre-Calculus**

C depending on how the equations are set up. Position involves x and y, and both depend on the time t.

*November 30, 2016*

**Math - Aglebra One (Ratios)**

If there are d dollars and h halves, then what do we know? d = h+133 d/h = 5/2 Now you can easily find h and d.

*November 30, 2016*

**precalculus**

man, this has been an agonizingly slow homework dump. Absolutely no work shown on your part, and general questions that indicate a total lack of preparation or study.

*November 30, 2016*

**College alegbra**

If the field has length x and width y, then xy = 1500 y = 1500/x p = 2x+3y = 2x + 4500/x See what you can do with that.

*November 30, 2016*

**Super Hard Math**

2*9 + 5*9 = (2+5)*9 = 7*9 did you actually try using some arrays?

*November 30, 2016*

**Math**

the height difference is 15 ft. So, now just use the Pythagorean Theorem for a triangle with legs of 15 and 36 Hint: Recall that one of your standard right triangles is 5-12-13

*November 30, 2016*

**Calculus**

exponentials grow and decay faster than polynomials. So, b,d -> 0 logs grow more slowly than polynomials, so c -> 0 a does not go to zero, since f(x)g(x) = 2x^2 + x/5 e^-x -> 2x^2

*November 30, 2016*

**Math**

so, if x is the number, what is one half of the number ?

*November 30, 2016*

**Calculus**

if you google MVT you will see diagrams of what it means. All it amounts to is that if the conditions are met, there will be one or more places in the interval where the tangent line is parallel to the line joining the points at the ends of the interval. In this case, the line...

*November 30, 2016*

**math**

with a little practice, which you evidently have, you can just read it from the equation, since b^0 = 1 for any nonzero b. But, when the function is a little more complicated, you may have to plug in t=0 and evaluate a more involved expression. Read about Newton's law of ...

*November 30, 2016*

**Math**

no. A is -6-24

*November 30, 2016*

**geometry**

6 * 8 * sin45°

*November 30, 2016*

**Math**

y=kx so, y/x = k is constant. You want y such that y/9 = 15/5 You don't really need to know what k is.

*November 30, 2016*

**math**

well, 2+3+4 = 9

*November 30, 2016*

**Math**

difference, not product

*November 30, 2016*

**math**

T-C = ((40-4t)-(-10+45t),(20+75t)-(40+70t)) = (50-49t,-20+5t) at t=0, then d=|T-C| = ?(50^2+20^2) = ?2900 = 53.9 d^2 = (50-49t)^2 + (-20+5t)^2 = 2426t^2 - 5100t + 2900 Not sure how you got your expression

*November 30, 2016*

**precalculus**

Your answer gives x = -1 y = -3/2 z = 1 but these do not satisfy the original equations. I already gave you a url which shows the row-echelon details.

*November 30, 2016*

**pre calculus**

the equation fits. Could you not do as I did, and just plug in t=1,2,3 to see that the result is as desired?

*November 30, 2016*

**Maths**

AC = AB+BC = 3+4 = 7 AC/CD = 2/1, so CD = 7/2 AD = AC+CD, so BC/AD = 4/(7 + 7/2) = 8/21

*November 30, 2016*

**Algebra**

The nth row, Rn = 41-4(n-1) = 45-4n So, there are 11 rows, since 45-48 < 0 You have an AP with a = 41 d = -4 S11 = 11/2 (2*41 + 10(-4)) = 231

*November 30, 2016*

**Algebra**

This is easy if you just break it up: 8 ?(-1)^k + 1.5k 1 = ?(-1)^k + ?1.5k The first sum is -1+1-1+1... = 0, so = ?1.5k = 1.5 ?k Now you know that n ?k = n(n+1)/2 1 so, your sum is just 1.5* 8*9/2 = 54

*November 30, 2016*

**Calculus 1**

the area of the water's surface is 4? m^2 since volume = area * thickness, the thickness is changing by 2/(4?) = 1/(2?) m/min

*November 30, 2016*

**college**

it must show all debits and credits, and they must all add up to the same value.

*November 30, 2016*

**Calculus MATH**

it's just a step function, so just add up the areas of the rectangles with heights 0,1,2,...9 and width 0.1

*November 30, 2016*

**math**

If you mean ?54 * 3i then that is 3?6 * 3i = 9?6 i ?(54*3)i = 9?2 i

*November 30, 2016*

**college algebra**

1/x = 1/60 + 1/55 x = 660/223 So, it will take them about 28.7 minutes. Sounds right, since their times are so close, it would just about cut the time in half, working together.

*November 30, 2016*

**Calculus**

A is true, since (0,1) is an open interval. Pick any nonzero x, and the limit from both sides is just 1-x. B and D are also true. C is false, since f(0) = 0 and f(x)->1 as x->0+ f is continuous on (0,1] There is a problem with the definition. It should be f(x) = 0 if x=0...

*November 30, 2016*

**precalculus**

for the row echelon stuff, enter your data here to check your results: http://www.math.odu.edu/~bogacki/cgi-bin/lat.cgi 2-7 look ok #8 is missing data, but the equation looks typical. Must be for a flight from ground level at 64 ft/s

*November 30, 2016*

**Calculus**

a is ok, since it is differentiable on (-1,1). It has a vertical tangent, bot only at the endpoint(s). b is ok for the same reason c is ok d is real on the interval, since it's ?(x+1), so it's ok too Looks to me like they all work, since the MVT requires 1. f(x) is ...

*November 30, 2016*

**pre calculus**

enough of these. Lots have been done, and a url provided.

*November 30, 2016*

**additional mathematics**

just use the formulas you know: a+d = 15 a+3d = 23 Note that T2 and T4 differ by 2d=8.

*November 30, 2016*

**math**

If you draw the diagram, you can see that you have a right triangle with 45° angles. So, the two legs are the same length: 60km.

*November 30, 2016*

**Math**

I think they want you to approximate it with the Taylor series, which involves y", y"', etc.

*November 30, 2016*

**Math - good catch**

thanks. leftover from a previous attempt.

*November 30, 2016*

**Math**

The volume of a spherical cap of radius r and height h (the water depth) is v = ?/3 h^2(3r-h) = ?/3 h^2(30-h) = ?/3 (30h^2 - h^3) dv/dt = ?(20h - h^2) dh/dt Now, when is h increasing most slowly? When the water surface area is greatest (since dv/dt is constant, and dv = a*dh) ...

*November 30, 2016*

**cal**

v = 10xy dv/dt = 10(y dx/dt + x dy/dt) So, using your numbers, dv/dt = 10(8(-4) + 6(2)) = -200 looks good to me.

*November 30, 2016*

**cal**

good call. So, rather than being where f' goes negative, it is where f' starts decreasing: where f" goes negative. That is, when the graph changes from concave up to concave down. Unfortunately, this graph starts out concave down, so the growth rate is always ...

*November 30, 2016*

**cal**

well, what does diminishing returns mean? Sales decrease while numbers increase. So, you want to find t for max N.

*November 30, 2016*

**cal**

f = 4x^3-21x^2+36x-4 f' = 12x^2 - 42x + 36 = 6(2x^2-7x+6) = 6(2x-3)(x-2) f" = 24x-42 = 6(4x-7) f is concave up for x > 7/4 f is decreasing for x in (3/2,2) So, both for x in (7/4,2) f has a max at x=3/2, so it cannot be concave up there.

*November 30, 2016*

**math**

what do you mean by one of the line of ax2+2hxy+by2=0 ?? That is a conic section, unless it's degenerate. In that case, you have (ax+c)(x+d)=0 ad+c = 2h cd = b Now work with the slopes of the lines, using the condition that one is twice the other.

*November 30, 2016*

**Math**

You sure there are no typos? I suspect you want (sinx+siny)/(cosx+cosy) Anyway, you can get things a bit simpler by taking the tan of both sides, since tan(arctan(x)) = x also, tan(x/2) = (1-cosx)/sinx

*November 30, 2016*

**Math Logs**

raise both sides as powers of 4, and you have, since 4^log_4(u) = u x+5 = 64(x-2) 63x = 133 x = 19/9

*November 29, 2016*

**math**

you want P where Pe^(.065*4) = 3000

*November 29, 2016*

**precalculus**

since (0,3) and (2,3) have the same y-value, the vertex is at x=1. So, (1,4) is the vertex, and we have y = a(x-1)^2 + 4 At x=0, a(1)+4 = 3 a = -1 y = -(x-1)^2 + 4 = -x^2+2x+3

*November 29, 2016*

**precalculus**

If you have to ask such a general question, you must not have read the relevant section in your text. So, you can start here: https://en.wikipedia.org/wiki/Row_echelon_form

*November 29, 2016*

**precalculus**

I get 2/x - 1/(x-1) + 2/(x-1)^2 what do you get?

*November 29, 2016*

**algebra**

m = (-5-(-10))/(-2-(-9)) = 5/7

*November 29, 2016*

**Calc**

what's the comma for? anyway, d/du arcsec(u) = 1/(x?(1-x^2)) the ?(x^2-4) I'm sure you can handle.

*November 29, 2016*

**Finite Math**

If there are x and y of A and B, then the constraints are 3x+4y <= 1000 6a+3b <= 22*60 The profit function is p(x,y) = 2.00x + 1.50y When you figure out the question you want to answer, this should help.

*November 29, 2016*

**MATH PLZ HELP**

After all the time you've spent posting this problem, don't you have any ideas? There just aren't that many ways to add at most three numbers to get 4: 1+1+2 1+2+1 2+1+1 ... Consider that she might have missed one or two throws.

*November 29, 2016*

**Math**

If the number is ab, then a.b = (a+b)/2 That is, a + b/10 = (a+b)/2 10a+b = 5a+5b 5a=4b Given the one-digit values of a and b, the number must be 45

*November 29, 2016*

**math**

so, for the 1st problem, x-2 = -2 8 = 2^3, so 2=8^(1/3) 4 = 2^2 so, 2^2 = (8^(1/3))^2 = 8^(2/3) That means 2/x = 2/3 and x=3

*November 29, 2016*

**math**

Font garbling, but (1/4)^(-2) = 16

*November 29, 2016*

**math**

27 = 3^3, so 3^(2x?1)= 1/(27^x) 3^(2x?1)= 1/3^(3x) 3^(2x-1) = 3^(-3x) 2x-1 = -3x 5x = 1 x = 1/5 25=5^2, so 5^(3x?8)·5^(4x) = 25^(2x) 5^(3x?8)·5^(4x) = 5^(4x) 3x-8+4x = 4x x = 8/3

*November 29, 2016*

**6th grade math**

(6 3/4)/6 = 1 3/24 = 1 1/8

*November 29, 2016*

**Math**

If you can do 5/6 and 3/4, then you can do 5/8. Just use a common denominator of 24 instead of 12.

*November 29, 2016*

**math**

true, 60+60=120, but there were 150 oranges.

*November 29, 2016*

**Math**

c + 5c = 24 solve for c, then get r, and then r-c

*November 29, 2016*

**maths**

clearly, 150/6 heaps so, how much profit per heap?

*November 29, 2016*

**Pre-calc**

as you know, sin(kx) and cos(kx) have period 2pi/k tan(kx) has period pi/k sin^2(x) = (1-cos(2x))/2 use a similar trick to find the period of tan^2(x) Looks like you need to review the basic info on trig functions.

*November 29, 2016*

**Math**

2*9 + 4*9 = (2+4)*9

*November 29, 2016*

**Math**

since the denominator is never zero, there is no vertical asymptote. as x gets huge, the fraction becomes tiny, so y=0 is the horizontal asymptote.

*November 29, 2016*

**Math**

well how many divisors of 24 are at least 4?

*November 29, 2016*

**Correcting Math**

65*1.07 + 65*.20 = ? Better try again. How did you arrive at your answer?

*November 29, 2016*

**pre calc**

see the post just below yours.

*November 29, 2016*

**pre calc**

don't forget your Algebra I! (2sin?+1)(sin?-1) = 0 now you can use your trig...

*November 29, 2016*

**Linear Algebra**

This is easiest with vectors, as discussed here: http://math.stackexchange.com/questions/371649/distance-of-a-3d-point-from-the-parametric-form-of-a-line Using the distance formula and calculus is a bit more involved.

*November 29, 2016*

**Math**

y = -3 x = 7 r = ?(9+49) = ?58 Now recall that sin? = y/r cos? = x/r tan? = y/x

*November 29, 2016*

**Algebra2**

8^3x=60 rewrite using the rules for exponents

*November 29, 2016*

**math**

3 * 1/6 = 3/6 = 1/2

*November 29, 2016*

**pre calculus**

see all the details here: http://www.gregthatcher.com/Mathematics/GaussJordan.aspx

*November 29, 2016*

**Pre-Calculus**

the larger of a and b is the major axis. In this case, the ellipse is vertically oriented. Think about it. When (x-2)=0, (y-4)=4 When (y-4)=0, (x-2)=3

*November 29, 2016*