Saturday

February 25, 2017
Total # Posts: 48,786

**algebra word problem**

the nice wikipedia article on trajectory gives the equation, including the angle involved.

*January 6, 2017*

**Algebra**

how can y be negative? x+y+z = 54000 .08x + .06y + .09z = 4260 .08x = 6*.06y x,y,z = 36000, 8000, 10000 ----------------------------------- So, where did you go wrong? 8x+6(9y)=0 Why do you think the interest sums to zero?

*January 6, 2017*

**algebra**

G <= 2

*January 6, 2017*

**Math**

come on, man! It's still ?2u^(2/3) du = 2 (3/5) u^(5/3) + C = 6/5 (x^2+2x-3)^(5/3) + C

*January 6, 2017*

**Math**

I suspect a typo. You probably meant ?(4x-4)(x^2-2x-3)^(2/3) dx or ?(4x+4)(x^2+2x-3)^(2/3) dx Assuming the first, Let u = x^2-2x-3 du = 2x-2 dx Then you have ? 2u^(2/3) du which is easy. ... right? If there is no typo, you are in deep trouble.

*January 6, 2017*

**Math**

1+i = ?2 cis ?/4 (1+i)^101 = (?2)^101 cis (?/4 * 101) = 2^50 ?2 cis (25? + ?/4) = 2^50 ?2 cis 5?/4 = 2^50 (-1-i) or -2^50 (1+i) There appears to be a typo.

*January 6, 2017*

**math**

what do you mean 4y=6y-2x+1 ?? Why not just say 2y = 2x-1 If that is the case, then we have 2(2x-1) + x+2x-1 = 28 Use that to find x, and then y. If there's a typo in the problem, fix it and use the same logic.

*January 6, 2017*

**Math**

8000 * 0.095 * 120/365

*January 6, 2017*

**math - eh?**

by "cover the entire room" do you mean all the walls, floor, and ceiling? Strange use for border.

*January 6, 2017*

**maths**

Not sure what you mean. The angle of reflection on the surface is always equal to the angle of incidence. The critical angle for total internal reflection decreases with increasing index of refraction.

*January 6, 2017*

**More calculus**

the fence length f = 2r + r? since the area is constant, 1/2 r^2 ? = A ? = 2A/r^2 so, f(r) = 2r + 2A/r Now just find r where df/dr = 0 Use that to get ?

*January 6, 2017*

**Calculus**

You have the right idea. But a must be chosen so that lim (ax-10) = 1 as x->3+ f(x) is defined every where except at x=3, so define it to be 1 there, since the limit from both sides is 1. That is, f(3) = B a = 11/3 B = 1

*January 6, 2017*

**Math**

true for any number except zero.

*January 6, 2017*

**calculus**

the points are (4,1) and (4,-1). -2x + 4yy'+3 = 0 y' = (2x-3)/(4y) So, now you have a point and a slope, and the lines are y-1 = (x-4) y+1 = -(x-4) See http://www.wolframalpha.com/input/?i=plot+-x%5E2%2B2y%5E2%2B3x%3D-2,+y-1%3Dx-4,+y%2B1%3D-(x-4)

*January 5, 2017*

**Algebra2**

rearrange things to 3x^2+6x-5 = 0 Now use the quadratic formula.

*January 5, 2017*

**Algebra2**

Which expression gives the solution -5+2x^2=-6x?

*January 5, 2017*

**calculus**

5y + 5xy' + 2xy + x^2 y' = 0 Now collect terms and solve for y'

*January 5, 2017*

**Calculus**

f' = 3h^2 h'

*January 5, 2017*

**Math**

.450km / 10km/hr = .045 hr

*January 5, 2017*

**pre calc**

do this just like your previous exercise.

*January 5, 2017*

**math**

They are separating at a rate equal to the slower truck's speed. Since It took 6 hours to separate by 204 miles, that is 204mi/6hr = 34 mi/hr So, the slower truck is going 34 mi/hr You found the faster truck's speed, but then botched it again since 2*8 ? 146 !!

*January 5, 2017*

**pre calc**

If x is at 6%, then the rest y=(3000-x) is at 7%. So, x+y = 3000 .06x + .07y = 194.00

*January 5, 2017*

**Calculus**

look up the arc length. The particle must move along a curve. At each small interval, the curve can be considered a straight line, with "arc length" ds^2 = dx^2 + dy^2 Thus, adding up all the tiny hypotenuses gives. s = ?ds = ??(dx^2+dy^2) = ??(1+(y')^2) dx

*January 5, 2017*

**Math Calc**

I assume you mean y = 3x^(1/3) + 6x^(4/3) = 3(2x+1) x^(1/3) y' = x^(-2/3) + 8x^(1/3) = (8x+1) x^(-2/3) y" = -2/3 x^(-5/3) + 8/3 x^(-2/3) = 2/3 (4x-1) x^(-5/3) y'=0 at x = -1/8 y"(-1/8) < 0 so y(-1/8) is a maximum y"=0 at x = 1/4 so that is an ...

*January 5, 2017*

**math**

double the 1st equation and you have 6x-2y = -6 It is clear to see that since -6 ? -8 the two equations cannot have a single solution. The lines are parallel, and do not intersect.

*January 5, 2017*

**Pre calc**

adjust things a bit to get 6x+12y = 57 -12x+12y = 12 subtract and you get 18x = 45 x = 5/2 Now use that to find y. Or, note from the 2nd equation divided by 3 that -x+y=3 y=x+1 Use that in the 1st equation to get 2x+4(x+1) = 19 2x+4x+4 = 19 6x = 15 x = 5/2

*January 5, 2017*

**pre calc**

add the two equations to get 0 = x-1 x = 1 now use that to get y.

*January 5, 2017*

**Algebra**

f+s = 106 f+3 = 2(s-4) now crank it out.

*January 5, 2017*

**math**

PV=kT So, PV/T = k a constant. So, you want T such that 48*1/T = 1*16/305

*January 5, 2017*

**Math**

Any line not in the plane of GH is skew to GH. Skew lines necessarily cannot intersect. Not having any details of your diagram, it's hard to say much more. Maybe google can provide you with further relevant examples.

*January 5, 2017*

**math**

Try this. I just googled ABCD is aparallelogram in which BCis produced to E such thatCE = BC and got this hit, among others: http://www.meritnation.com/ask-answer/question/abcd-is-a-parallelogram-in-which-bc-is-produced-to-e-such-th/areas-of-parallelograms-and-triangles/...

*January 5, 2017*

**math**

F=ma v = at = Ft/m so, t = vm/F = 2.8*10^3*7.9 /6.7*10^5 = 0.033 seconds

*January 5, 2017*

**math**

a = v^2/(2s) = 0.614/0.744 = 0.8263 m/s^2 F = ma = 61.0*0.8263 = 50.34N

*January 5, 2017*

**math check my answer ?**

as written, f(x) has a range of all real numbers. Pick any value for f(x), and you can easily find an x to produce it. For example, suppose you want f(x) = 11 -3x-1 = 11 -3x = 12 x = -4 So, you must have a typo. Your range of f < -1 could be produced with f(x) = -3x^2 - 1 f...

*January 5, 2017*

**geometry**

a = 1/2 r^2 ? Use that once you supply the missing data.

*January 5, 2017*

**Science**

google is your friend. Try it.

*January 5, 2017*

**math**

g = 12.50 r = 0.75g y = 2.5r = 2.5 * 0.75 * 12.50 = 23.4375 or, using mixed numbers, y = 2 1/2 * 3/4 * 12 1/2 = 23 7/16

*January 5, 2017*

**computer applications (commerce)**

knowing that they differ by 2, start at 19 1: n=19 2: print n 3: n=n-2 4: n>0? 5a: yes: go to step 2 5b: no: stop Or, you could test every number against some criterion. In this case, divisibility by 2. 1: n=20 2: divide n by 2 3: remainder=0? 4: yes: print n 5: n=n-1 6: n&...

*January 5, 2017*

**further maths**

m•p = 0 n = h(i+j-2k) a+b-2c = 0 a+h = 2 b+h = -1 c-2h = 2 This gives the result m = 5/2 i - 1/2 j + k n = -1/2 i - 1/2 j + k I don't see a second solution, but maybe you can find one.

*January 5, 2017*

**Maths**

5/7 as many men, so 7/5 as much time.

*January 5, 2017*

**MATH**

well, p^2+pq-6q^2 = (p+3q)(p-2q) p^2-3pq+2q^2 = (p-q)(p-2q) So, what cancels?

*January 5, 2017*

**maths**

a = 2 pi r l

*January 5, 2017*

**Math**

look at the differences: 2,3,_,_,6

*January 5, 2017*

**Math Calc**

1. Think of f(x) = x^3 2. If f'=0, f does not change. 3. think of f(0). It clearly does not satisfy the MVT on [-1,1] The typos do not help...

*January 5, 2017*

**Math Calc**

clearly it is either 3 or 4. Since 4 is not a choice, 3 seems logical. Now, knowing what you do about cubic curves, and that this one comes down from the left, its first turnaround will be a minimum. So, it will be at x=3. Or, note that f" = -4x+14. f"(3) = 2 > 0...

*January 5, 2017*

**Math Calc**

This is a typical exercise using the chain rule. 2x^2 + y^3 =10 4x dx/dt + 3y^2 dy/dt = 0 Now, using your values, when x=1, y=2. So 4(1) dx/dt + 3(2^2)(3) = 0 4 dx/dt + 36 = 0 dx/dt = -9

*January 5, 2017*

**Math!!!**

recall that if z = x+iy zz* = (x+iy)(x-iy) = x^2+y^2

*January 4, 2017*

**Math**

well, 2mn-16m = 2m(n-8) so what can you do to get from -7n+? to ?(n-8) ?

*January 4, 2017*

**algebra**

p +4- 2 (p-10)> 0 p + 4 - 2p + 20 > 0 -p + 24 > 0 24 > p or p < 24

*January 4, 2017*

**math**

check out your sum-to-product formulas. sin(2A+C)+sin(A+2B) = 2sin((2A+C)/2)cos((A-2B)/2) C = ?-A-B 2A+C = 2A+?-A-B = A-B+? sin (2A+C)/2 = sin(?/2 + (A-B)/2) = cos(A-B)/2 and similarly for the others. Much stuff will cancel out.

*January 4, 2017*

**Math**

h(5)-h(2) -------------- = ? 5-2

*January 4, 2017*

**Mathematical reasoning**

p = rain does not come q = crops are ruined and the people will starve p -> q You have ~q, so ~p (contrapositive) since q = r and s, ~r or ~s = ~q

*January 4, 2017*

**Math**

recall that (x+a)^2 = a^2+2ax+a^2 You have 2a = 16

*January 4, 2017*

**Math**

C and D divide BE into three equal parts. 1.64-(-2.83) = 4.47 4.47/3 = 1.49 So, now you know how much goes in each blank space.

*January 4, 2017*

**Calculus**

x23-5

*January 4, 2017*

**Calculus**

x&sup4;-5

*January 4, 2017*

**Calculus**

x&sup0;¹²³-5

*January 4, 2017*

**Calculus**

x&sup5;-5

*January 4, 2017*

**Calculus**

x^²-5

*January 4, 2017*

**Calculus - incomplete**

Fill in the missing text that did not get pasted

*January 4, 2017*

**Math**

well, sums are not directly related products and quotients, so I think more information is needed. I will note, however, that a product is just a short way to write a sum. 4*3 = 4+4+4 or 3+3+3+3

*January 4, 2017*

**Math**

Note that the two points lie on the same horizontal line: y = -6 Graph that line segment, and it will be clear where the other two points lie, since width+height=13

*January 4, 2017*

**pre Calc**

come on. A vertical line has the same x-coordinate everywhere. Look at the point it passes through. Vertical lines are of the form x=h Horizontal lines are y=k

*January 4, 2017*

**pre calc**

Looks like you need to review your Algebra I... 8x ? 3y + 6 = 0 3y = 8x+6 y = 8/3 x + 2

*January 4, 2017*

**pre Calc**

clearly the slope is -3.5 Now think f=mt+b

*January 4, 2017*

**Number Theory**

4x+7 = 2(mod 15) 4x = -5(mod 15) 4x = 10(mod 15) so, what can x be so that 4x = 5k+10 for some k?

*January 4, 2017*

**precalc**

you know that cos(u) might either decrease or increase, depending on u. surely t/52 increases when t does. !!

*January 4, 2017*

**Math**

x = -2 is a vertical line. So, what is the horizontal line through the point? x = -5 is parallel to x = -2, not perpendicular.

*January 4, 2017*

**Math**

F = ma just add up the vectors, after dividing by the mass.

*January 4, 2017*

**Math**

surely you can graph the lines y = 2/3 x y = -2 y = -3x+1 then just take the pieces you need.

*January 4, 2017*

**Pre Calc**

the slopes must be the same, so (8-1)/(2-a) = (1-6)/(a+2-12) a = 30

*January 4, 2017*

**profit&loss mathematics**

come on. B paid 20% of A's cost: 1.20A C paid what A did: B/1.20 = 08.3333B So, B lost 16.666 % maybe my math looks as funny to you as your texting does to me...

*January 4, 2017*

**profit&loss mathematics**

B = 1.20A C = A = B/1.20 = (5/6)B so, that is ...

*January 4, 2017*

**Maths**

?x^2 h = 800 h = 800/(?x^2) The area is two circles plus a curved cylinder, so a = 2?x^2 + 2?xh the result follows, so just find x such that da/dx = 0

*January 4, 2017*

**mathematics**

sorry - a=20 ! :-(

*January 4, 2017*

**mathematics**

you must be studying geometric sequences, so just note that this is the sum of the sequence with a = 12 r = 3/5 Now just use your formula for the infinite sum: s = a/(1-r)

*January 4, 2017*

**am I right ? or no?**

you are correct that b^x is growth if b>1 and decay if b<1

*January 4, 2017*

**Maths**

convert the area to m^2 area = length*width, so divide by the width to get the length

*January 4, 2017*

**Maths**

0.5ha * (10000m^2/ha) / 12.5m = 400m

*January 4, 2017*

**algebra - duh @Reiny**

It usually helps to read the problem carefully. Nice work, Reiny

*January 4, 2017*

**algebra**

x+y <= 9x-5y 6y <= 8x 3y <= 4x You might also want to consider x+y <= |9x-5y|

*January 4, 2017*

**Maths**

n = d-7 (n+2)/(d+9) = n/d (d-7+2)/(d+9) = (d-7)/d d(d-5) = (d-7)(d+9) d^2-5d = d^2+2d-63 7d = 63 d=9 check: original fraction: 2/9 modified fraction: 4/18 = 2/9

*January 4, 2017*

**Maths**

if you have a unix system, you can use the bc calculator for arbitrary precision, as well.

*January 4, 2017*

**Math**

1/4 of 2 cups is ??

*January 4, 2017*

**math**

((3200/1.60)/4)^2 * 1.20/100 = 3000

*January 4, 2017*

**Math**

there are 4 choices for each digit, so 4^4 total ways. 3/4 of them start with a number less than 7, right?

*January 4, 2017*

**Math;);)**

the two points lie on the unit circle, clearly 54-19 degrees apart. The angle is 35°. or, use the distance formula to find the vector u-v. Then use the law of cosines to find the included angle.

*January 4, 2017*

**math**

Just plug in n=5: a5 = 6*5-4 = ?

*January 3, 2017*

**Math**

well, the equations is (x-(2-3i))(x-(2+3i)) = 0 ((x-2)+3i)((x-2)-3i) = 0 (x-2)^2 + 3^2 = 0 x^2-4x+4 + 9 = 0 x^2-4x+13 = 0 a+b = -4+13 = 9

*January 3, 2017*

**Math**

try using letters like a,b,c instead of the special characters that are getting mangled.

*January 3, 2017*

**maths**

if x<0>=0 then 4/(x<0>)^2 is undefined. I'll assume that x<n+1> = 4/(x<n>^2+5) If so, then x<0> = 0 x<1> = 4/(0^2+5) = 4/5 = 0.8 x<2> = 4/(0.8^2+5) = 4/5.64 = 0.7092 ... The actual root ? 0.72408 If I got the formula wrong, then ...

*January 3, 2017*

**maths**

areas grow as the square of the linear factor. So the areas are A:B = (8:10)^2 = 64/100 = 16/25 A/160 = 16/25 A = 16*160/25 = 102.4 cm^2

*January 3, 2017*

**math**

tan(x) = h/d = 4d/d = 4 x = arctan(4) = 76° for part b, h decreases, because x decreases.

*January 3, 2017*

**math**

3^-n = 0.2 3^n = 1/0.2 = 5 (3^4)^n = (3^n)^4 = 5^4 = 625

*January 3, 2017*

**Maths**

V = 4/3 ? r^3 A = 4? r^2 = 3V/r = ?(36?V^2) Now just plug in your numbers.

*January 3, 2017*

**Algerbra**

He will need (253.5/6)^(3/2) * 2 = 549.25 kg of clay So, he will need 55 10kg bags

*January 3, 2017*

**Surfaces of cones**

?r?(r^2+h^2) = 2160? (12x)?(144x^2+256x^2) = 2160 (12x)(20x) = 2160 x^2 = 9 x = 3 so, the volume is ?/3 r^2h = (?/3) (24*3)^2 (16*3) = 82944? V = 82944 cm^3

*January 3, 2017*

**Maths**

just expand the left side. It will be the same as the right side. One way to start would be (3x-1)(x+5)(4x-3) = 3x(x+5)(4x-3)-(x+5)(4x-3) Now, (x+5)(4x-3)=4x^2+17x-15 so you can proceed with = 3x(4x^2+17x-15)-(4x^2+17x-15) = ...

*January 3, 2017*

**Math**

sorry - only one equation.

*January 3, 2017*

**Math**

useless problem. 6 apples - 2 oranges =

*January 3, 2017*