Sunday

July 5, 2015

July 5, 2015

Total # Posts: 32,100

**help**

x + 3x = 20 . . .
*May 20, 2015*

**physics**

No.
*May 20, 2015*

**math**

to find the # of children's tickets, you just solve 35(2c) + 20c = 1890 then double that for the adults.
*May 20, 2015*

**Good call, Reiny**

Yes, I neglected to mention excluded values. One cannot gloss over stuff like that. My bad.
*May 20, 2015*

**algebra**

x/(6x-x^2) = x / x(6-x) = 1/(6-x) (x^2 - 3x - 18) / (x + 3) = (x-5)(x+3)/(x+3) = x-5 (k+3)/(4k-2) * (12k^2+2k-4) = (k+3)/(2(2k-1)) * 2(2k-1)(3k+2) = (k+3)/(3k+2)
*May 20, 2015*

**math**

if there's a 20% discount, it sells for 80% of its original price. so, 0.80 p = 126.40
*May 20, 2015*

**math**

well, geez, just check your answer 11 dimes and 2 nickels does indeed add up to $1.20
*May 20, 2015*

**math**

add up the coins: n+d = 13 add up the values: 5n+10d = 120 Now just solve for n and d.
*May 20, 2015*

**Algebra**

(8m^7 - 10m^5) / 2m^3 = 8m^7/2m^3 - 10m^5/2m^3 = 4m^4 - 5m^2 -2/(x+3) - 4/(x-5) = [-2(x-5) - 4(x+3)] / (x+3)(x-5) . . . 7/(x-3) + 3/(x-5) = [7(x-5) + 3(x-3)] / (x+5)(x-3) . . . -1/(x-9) - -2/(x+7) = [-(x+7) + 2(x-9)] / (x-9)(x+7) . . .
*May 20, 2015*

**Trig**

correct.
*May 20, 2015*

**algebra**

yes, it appears to be an exercise in the distributive property. Do you have a question? I expect you need to use the property to get rid of the parentheses and then solve for the variables. What do you get?
*May 20, 2015*

**Statistics**

so, at 5% per period, after 6 periods, the account will have 2000*1.05^6 + 2000*1.05^5 + ... + 2000 = 2000(1.05^7 - 1)/0.05 = 16284.02
*May 20, 2015*

**geometry**

well, v = 1/3 πr^2 h so, unless you know something about the height, there's no way to pin down the radius. All you know for sure is that r^3 h = 960/π Maybe you can fix things and then solve for r. Please don't just come back and say h=5, so what's r? ...
*May 20, 2015*

**CSIR General Aptitude**

Just subtract the starting position from the ending position to get the distance traveled...
*May 20, 2015*

**pre-algebra**

9z+8+3(2-4z)-9=-40 9z+8+6-12z-9 = -40 -3z+5 = -40 -3z = -45 z = 15
*May 20, 2015*

**math**

I assume you meant exactly 2 standard deviations above the mean So, 2 std's = 8, making the mean 85-8 = 77
*May 20, 2015*

**Maths**

so, what's wrong with that as an answer? That's what I get, too. http://www.wolframalpha.com/input/?i=1%2F%281%2F%281%2B2i%29+%2B+1%2F%282%2B5i%29%29
*May 20, 2015*

**Math Check Answers Quickly Please**

both are correct.
*May 20, 2015*

**math**

If the base is a n-gon, then the base has n sides and n vertices. So, clearly the pyramid has n lateral faces and edges, no?
*May 20, 2015*

**Maths**

depends on the tools you have learned. Most easily, use a calculator. figure 18*3.14159/5.86 and then hit the sin button. Make sure you are in radian mode.
*May 20, 2015*

**math**

have you no calculator? If not, you can just enter that text into the google search box, and it will give you the result.
*May 20, 2015*

**science**

check the label to see the expected delay.
*May 20, 2015*

**Math**

5m+13
*May 19, 2015*

**algebra2**

That file exists only on your computer. No way for us to open it. Best suggestion: review the properties and work it out. Next best: tell us the question for some help.
*May 19, 2015*

**Math**

you can find the distance PQ using the law of cosines: x^2 = 10^2 + 6^2 - 2(10)(6)cos137° x = 14.959 for the bearing, locate P and Q in rectangular coordinates and then the bearing is 90-arctan(y/x) where x and y are the corresponding displacements from Q to P.
*May 19, 2015*

**Algebra 2 Honors**

you can get that without worrying about k. since r = kst, r/st = k and is constant. So, you need s in 140/(-5*20) = 7/(2.5s) s = -2 as you calculated
*May 19, 2015*

**algebra**

-1/(x-9) - -2/(x+7) -1(x+7) - (-2)(x-9) ------------------------- (x-9)(x+7) (-x-7+2x-18) / (x-9)(x+7) (x-25) / (x-9)(x+7)
*May 19, 2015*

**algebra**

a/b/c (a/b)/c = a/(bc) So just plug in your values and reduce.
*May 19, 2015*

**Calculus**

1. I get = π[(1/4)y^2 + y] from -2 to 1 = π((1/4+1)-(1-2)) = 9π/4 2. I get 9π/4 also making the total volume 9π/2
*May 19, 2015*

**Calculus**

suppose ∫ f(x) dx = F(x) Then ∫[0,8] f(x) dx = F(8)-F(0) = 4 ∫[0,2] f(4x) dx = ∫[0,2] 1/4 f(4x) d(4x) = 1/4(F(8)-F(0)) = 1 check: ∫[0,8] x^2 dx = 512/3 ∫[0,2] (4x)^2 dx = 16(8/3) = 128/3
*May 19, 2015*

**math please help**

the perimeter is just the sum of all the sides. So, for the large triangle, p = 4x+2 + 7x+7 + 5x-4 = 16x+5 Now do the same for the smaller triangle. For (B), just subtract one expression from the other.
*May 19, 2015*

**Calculus**

how about |x^2-cosx| or, 2*integral from 0 to 2, since both cosx and x^2 are even?
*May 19, 2015*

**Calculas**

the volume of a disc of thickness k is pi r^2 k If the radius is y, it means you are rotating around the x-axis, so the volume of each disc is pi y^2 dx If rotating around the y-axis, each disc's radius is x, so the volume is pi x^2 dy Try drawing a diagram, guys. As you ...
*May 19, 2015*

**Calculas**

pi y^2 is used when rotating around the x-axis.
*May 19, 2015*

**Calculas**

if you use discs, they have holes. Shells are just like nested cylinders, so there are no holes. Using discs (washers), the calculation is more complex, since the lower portion is bounded by the y-axis (solid discs), and the upper portion is bounded by the two curves (washers ...
*May 19, 2015*

**Calculus**

The curves intersect at (±√3,4) So, using the area in the first quadrant, we have, using shells, v = ∫[0,√3] 2πrh dx where r=x and h=(x^2+1)-(2x^2-2)=(3-x^2) v = 2π∫[0,√3] x(3-x^2) dx = 9π/2 Looks more like a bowl than a vase...
*May 19, 2015*

**Calculus**

because f(-x) = -f(x) There is equal area above and below the x-axis.
*May 19, 2015*

**Quadratic Functions and Calculus**

Holy crap! Have you done anything on this? Have you determined the three data points to use? Have you decided what to do using the ID number? Come back with some input, ok?
*May 19, 2015*

**Quadratic Functions**

Not sure what BiQuad is, but if you want to fit a parabola through those three points, you have, assuming that v(t) = at^2 + bt + c, 0a+0b+c = 10000 9a+3b+c = 2025 36a+6b+c = 6400 Somehow I suspect what you really wanted was v(3) = 1000-2025 = 7975, since most stocks don't...
*May 19, 2015*

**math**

well, shoot. Which is bigger: a ft^3 or a yd^3?
*May 19, 2015*

**math please help**

they are the same. It shows why 11^0 = 1 or why x^0 = 1 for any value of x. (except x=0)
*May 19, 2015*

**math help please!!**

3^3/3^6 = 3*3*3 ----------------- 3*3*3*3*3*3 Now you can see that three 3's cancel out in top and bottom, so you are left with 1/(3*3*3) = 1/27 This shows why x^n / x^m = x^(n-m)
*May 19, 2015*

**Math**

can you try that again in English?
*May 19, 2015*

**Math**

the cars are 360/16 = 22.5° apart. car 16 is 3 positions from car 13. So, a rotation of 67.5° will move car 16 to car 13's position.
*May 18, 2015*

**geometry**

in a triangle, angles are opposite sides. Two angles cannot be opposite each other. Using the tan function, the hypotenuse is not used. If the base 12 is opposite the 71° angle, then 12/d = sin 71° 12/c = tan 71° Better review your trig function definitions.
*May 18, 2015*

**Math**

you have a rectangle that is 100x(40+20) and a circle that has diameter 40+20=60 so, just add those two areas. If the area inside the circular ends but outside the rectangle does not count, then you have to subtract off the smaller circle of diameter 40.
*May 18, 2015*

**Math**

a^2 = 6 b^2 = 5 c^2 = a^2+b^2
*May 18, 2015*

**Math**

you want x minutes, where 12+.05(x-500) = 10+.04(x-300) x = 1100
*May 18, 2015*

**math**

each interval has width 2, so the midpoints are at 1,3,5,7,9
*May 18, 2015*

**math**

16-x^2 is concave down, so you want to use the right endpoints. Each interval has width 1, so you just have 2(f(1)+f(2)+f(3)) = 2(15+12+7) = 68 There are several good online Riemann Sum calculators. You can use them to verify your work.
*May 18, 2015*

**math**

see related questions below
*May 18, 2015*

**math**

just plug in the values for n: t1 = 7 t2 = 2t1-3(2) = 2(7)-3(2) = 8 t3 = 2(8)-3(3) = 7 ...
*May 18, 2015*

**math**

so, did you use the formula?
*May 18, 2015*

**algebra 2 and trig**

how many std's away is the 2.5% boundary? The mean is at (78+92)/2 = 85
*May 18, 2015*

**Calculus**

Assuming x is not zero, the fractions are zero when the numerators are zero. So, f'=0 when 8x+1 = 0, or x = -1/8 f"=0 when 4x-1 = 0, or x = 1/4
*May 18, 2015*

**Algebra 2**

revenue for x hats is y=15x cost is y=0.4(x-150)^2+9000 So, you want x when 15x >= 0.4(x-150)^2+9000 x >= 262.5 so, at least 263 hats
*May 18, 2015*

**Math**

if the chord length is 2x, then half the chord subtends half the angle: x/200 = sin 34°
*May 18, 2015*

**please help math**

it is that.
*May 18, 2015*

**math please help**

after n years, its value is 400(1.07)^n
*May 18, 2015*

**Calculus**

it reverses direction when its velocity changes sign. That is, when v(t) = 0. So, find t when v(t) = s'(t) = 3t^2 - 18t + 24 = 0 then plug that value into s(t) and a(t)
*May 18, 2015*

**Math**

he used 15.75-10.5 = 5.25 gallons of paint in 1.5 hours. That is 3.5 gal/hr Thus, using the point-slope form of the line, if y is the amount of paint left at x hours after 12:00, then y-10.5 = -3.5(x-2) So, find y when x=0. Or, knowing that he used 3.5 gal/hr, and knowing that...
*May 18, 2015*

**math please**

when dividing numbers expressed as powers, you subtract the exponents. Think about it 10^7/10^3 = 10*10*10*10*10*10*10 --------------------------- 10*10*10 Three of the factors cancel, and you are left with 7-3 = 4 multiples of 10. That is 10^(7-3) = 10^4 So, you want (C)
*May 18, 2015*

**Math**

216 = 6*6^2 96 = 6*4^2 so, you wind up with 6√6t + 4√6t = 10√6t = √600t
*May 18, 2015*

**Algebra**

a = w(w+4) . . .
*May 18, 2015*

**MATH if work is correct**

78125 = -5^7 So, starting with 5^-2, the 10th term is -5^7 Note that the 3rd term is 1=5^0. So, you have to multiply by -5 another 7 times to get to -5^7.
*May 18, 2015*

**Math**

the full volume is 60*60*45 2/3 of that is still empty. So, divide that volume by 6 L/min (6000 cm^3/min) to get the time needed.
*May 18, 2015*

**Math**

In how many different orders can you read 4 books
*May 18, 2015*

**science**

a body of mass 5kg is thrown to a height of 100m.what is its potential energy at the maximum height.?
*May 18, 2015*

**math Help!!!!**

There are 6 ways you can draw the two marbles to include one red: RB,RY,RW,BR,YR,WR There are 4P2 = 12 ways to draw any two marbles So, P(one red) = 6/12 = 1/2
*May 18, 2015*

**math...Please help!!!!**

I think B, but the language is garbled. The SD will not change. Read up on how the SD is figured, and you should see why.
*May 18, 2015*

**Geometry**

h^2 + 7^2 = 28^2
*May 18, 2015*

**Math**

as you recall, if the axes are rotated through θ, tan 2θ = B/(A-C) so, tan 2θ = 24/7 and cosθ = 3/5 sinθ = 4/5 Now just plug those values into your rotation matrix and crank it out.
*May 18, 2015*

**Math**

tan(2θ) = -10/0 2θ = -pi/2 θ = -pi/4 x' = (1/√2)x + (1/√2)y y' = (-1/√2)x + (1/√2)y 13((1/√2)x + (1/√2)y)^2 - 10((1/√2)x + (1/√2)y)((-1/√2)x + (1/√2)y) + 13((-1/√2)x + (1/√2)y)^2 = ...
*May 18, 2015*

**grade (5) math**

If the pen has dimensions x and y, with n gates, and assuming there are posts only at the corners and on both sides of the gate(s) and that the gate(s) do not share a corner post, then using up the whole budget, (2x+2y-3n)+2(4+2n)+5n = 50 x+y+3n = 25 and, since the gates are 3...
*May 18, 2015*

**Calculus**

since h is not continuous at x=0, h'(0) is not defined.
*May 18, 2015*

**Calculus**

cannot say what the limit is. If f(x) is continuous, though, the limit is 7.
*May 18, 2015*

**Calculus**

this is true only if f(x) is continuous. Otherwise, all bets are off.
*May 18, 2015*

**Math!!! Please help!!!**

x^7 is the 5th term in the expansion, so it is 11C4 x^7 (-3)^4 = 26730x^7 The coefficients are 1 5 10 10 5 1, so (2y-3x)^5 = (2y)^5 + 5(2y)^4(-3x) + 10(2y)^3(-3x)^2 + 10(2y)^2(-3x)^3 + 5(2y)(-3x)^4 + (-3y)^5 Now just expand all those values nCr = n!/[(n-r)!r!] nC(n-r) = n!/[(n...
*May 17, 2015*

**math**

compare the sums of the two geometric sequences: (2^30 - 1)/(2-1) vs (3^20 - 1)/(3-1) There's a difference, but either one will set you up for life!!
*May 17, 2015*

**math**

not quite. The area is base times height, divided by 2. So, (B)
*May 17, 2015*

**Math**

consult your various conversion formulae. One way would be 6.58559 sin(x+0.73169) There are, of course, ways to express that using cosine and other phase shifts.
*May 17, 2015*

**Calculus**

See the related questions below. They should help. If not enough, some back and let us know how far you got.
*May 17, 2015*

**Math**

for the cards, does the way you arrange your cards make any difference? If not, it's a combination. For the actors, of course it makes a difference who plays which part. So, a permutation. But, if you just want to get four guys, it's a combination.
*May 17, 2015*

**Math**

3!
*May 17, 2015*

**math**

compare the sums of the two geometric sequences: (2^30 - 1)/(2-1) vs (3^20 - 1)/(3-1) There's a difference, but either one will set you up for life!!
*May 17, 2015*

**Maths**

after n years, the interest will be 800(0.02n)
*May 17, 2015*

**math for college readiness**

since y=kx, just find k: -30 = k*6 k = -5 ...
*May 17, 2015*

**trigonometry**

use your half-angle formula to find sin(15°) Now use your sum formula to find sin(270°+15°)
*May 17, 2015*

**Math**

one possible answer: 48
*May 17, 2015*

**math**

if the sides are in ratio r, the areas are in the ratio r^2. So, the sides are in the ratio 12:9
*May 17, 2015*

**math**

for an angle θ, the chord is 2r sin(θ/2) the arc length is rθ the area of a sector is 1/2 r^2 θ Now use that to find your answers
*May 17, 2015*

**Math**

4 * 6 - 2 * 3/2 = 21
*May 17, 2015*

**Math**

you give no dimensions of the key's rectangle. Its sides must be in the ratio of 94:50 to be similar to the court. All circles are similar to each other.
*May 17, 2015*

**Math**

Well, the standard form is (y-k)^2/b^2 - (x-h)^2/a^2 = 1 and for an hyperbola, c^2 = a^2+b^2
*May 16, 2015*

**algebra**

Well, you have the formula. What's the trouble?
*May 16, 2015*

**Maths**

f^-1(x) = (x-1)/2 (g◦f)(-2) = g(f(-2)) = g(-3) = 23 (f◦g)(x) = f(g) = 2g+1 = 2(3x^2-4)+1 = 6x^2-7
*May 16, 2015*

**Math**

I'd say either can be true, but not all shapes can be inscribed in a rectangle. That is, so that all their vertices touch the sides of the rectangle.
*May 16, 2015*

**Math**

thousands of years ago it was learned that the ratio of the circumference to the diameter of a circle is pi = 3.14159265... This is true for all circles: c = pi * d or, c = 2pi * r
*May 16, 2015*

**math**

clearly T16 = 1464-1290 = 174 S16 = 16/2 (a+174) = 1464 a = 9 T16 = 9+15d = 174 d=11 S20 = 20/2 (2*9+11*19) = 2270
*May 16, 2015*

**calculus**

find y' at the given point (h,k). That is the slope of the tangent line. And then just plug in that slope (call it m) into the point-slope form of the line: y-k = m(x-h) Don't forget your Algebra I now that your'e doing calculus. So, what do you get for y=√(x...
*May 16, 2015*