Friday

July 31, 2015

July 31, 2015

Total # Posts: 32,558

**Mathematics**

we know that (5a+12b)^2 + (12a+5b)^2 = (13a+kb)^2 25a^2+120ab+144b^2 + 144a^2+120ab+25b^2 = 169a^2 + 26kab + k^2b^2 169a^2+240ab+169b^2 = 169a^2+26kab+k^2b^2 (240-26k)ab + (169-k^2)b^2 = 0 b((240-26k)a+(169-k^2)b) = 0 Since a and b are positive, either 249-26k < 0 or 169-k^...
*July 30, 2015*

**trig**

tanx = 3 so, sinx = 3/√10 cosx = 1/√10
*July 30, 2015*

**Math**

After t days, you have 100*(1/2)^t
*July 30, 2015*

**Math**

280(4mi)^2 = 280*16 mi^2
*July 30, 2015*

**maths**

(1) no algebraic way to do it. Check the graph (2) 2x+2y < 4 3x+4y > 6 Just plot the lines and shade the areas that intersect (3) clearly x=100
*July 30, 2015*

**math**

Why do you say "for some reason"? Didn't you solve the problem? Did you check your answer? 500-70+10 = 440 = 500 * 0.85 ok 300-30+75 = 375 = 345 * 1.15 ok Here's how I did it. I imagine you did it in much the same way. to start, 500 ducks 300 chickens then ...
*July 30, 2015*

**Calculus**

well, 1-sin^2 = cos^2, so you have cos^2θ - sin^2θ = cos 2θ cos^2θ * sin^2θ = 1/4 sin^2(2θ)
*July 30, 2015*

**Calculus**

I see some algebra and trig, but no calculus also, no question
*July 30, 2015*

**What did I did wrong MATH**

(3x-6)/(4x+12) = 1/3 it takes 12 fifty-cent coins to be "the same amount" as six dollar coins
*July 30, 2015*

**Algebra I I**

sure looks like substitution to me.
*July 30, 2015*

**math**

5^(x+2) = (5^x)(5^2) = 7*25 = 175 However, as you wrote it, 5^x+2 = 7+2 = 9 Parentheses are important!
*July 30, 2015*

**math**

clearly, -2001 <= x <= 0 Thus, 0 <= x+2001 <= 2001 So, how many numbers between 0 and 2001 are squares? √2001 = 44.7 So, There are 45 numbers such that 0 <= x <= 44, meaning that 0 <= x^2 <= 2001
*July 30, 2015*

**math**

√1000000 = 1000 √4000000 = 2000 any number less than 1000 has a square less than 1 million any number greater than 2000 has a square greater than 4 illion so, you want all the numbers n such that 1000 < n < 2000
*July 30, 2015*

**math, hard**

who's talking about square roots? These are absolute values. |a| = -a, so a = -|a| same for b. So, a-b = -|a| - (-|b|) = -|a| + |b|
*July 30, 2015*

**math! VERY hard!**

3/4 = 9/12 1/3 = 4/12 so, you already have 13/12lbs of fruit You still need 1 11/12 = 23/12 lbs That is 9/12 + 14/12 = 3/4 + 6/12 + 8/12 = 3/4 + 1/2 + 2/3
*July 30, 2015*

**Math**

tanx+cotx = sec^2x/tanx = 1/(sinx cosx) secx+cscx = (sinx+cosx)/(sinx cosx) so, we have sinx+cosx + 2/sin2x + 2(sinx+cosx)/sin2x See what you can do with that
*July 30, 2015*

**Trig**

sinx+cosx = 1/5 square both sides and you have sin(2x) = -24/25 so, cos(2x) = -7/25 Now you can use your half-angle formula for tan(x)
*July 30, 2015*

**Trig**

Take a look here: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibpi.html and scroll down to Sec. 6.2
*July 30, 2015*

**Math**

Since sinx is always less than 1, you need to find where x^2/625 >= 1 There will be no solutions past that. Looks like you don't need to look beyond x=25. So, how many periods does sinx have in that domain? Looks like 4, positive and negative, so I expect there are 15 ...
*July 30, 2015*

**Math**

If the pirate is at horizontal distance x from the building, at the hill has height h, then (400-h)/x = tan 4° h/x = tan 2° Eliminate h and you have 400 - x tan4° = x tan2° Now you can find x. Assuming the pirate can fly horizontally, that is the shortest ...
*July 30, 2015*

**trigo - eh?**

midway is in the middle; same distance on each side which is taller, the near or far? Is "c" supposed to mean angle?
*July 30, 2015*

**maths**

what the heck is the radius of the area of a semicircle
*July 30, 2015*

**Trig**

draw your triangle. If the legs of the triangle are x+1 and x-1, the hypotenuse is √((x+1)^2 + (x-1)^2)) = √(2x^2+2) so, sin(arctan((x+1)/(x-1)) = (x+1)/√(2x^2+2)
*July 30, 2015*

**Trig**

Recall your sum/difference formulas. You have sin(2x-y)+sin(y) = 2sin(x)cos(x+y) cos(2x-y)+cos(y) = 2cos(x)cos(x+y) Now just divide and you have tan(x)
*July 30, 2015*

**Math**

0 and 180π/(180+π)
*July 30, 2015*

**Math**

f(t^2+1) = t^4+5t^2+3 = (t^2+1)(t^2+4) - 1 = (t^2+1)((t^2+1)+3)-1 That means that f(t) = t(t+3)-1 So, f(t^2-1) = (t^2-1)((t^2-1)+3)-1 = t^4+t^2-3
*July 30, 2015*

**Math**

clearly, f(2003) = 0, since f(f(0)) = 0
*July 30, 2015*

**maths**

pi not pai!! d/dx u^v = v*u^(v-1)u' + lnu u^v v' You have u=x and v=x^10 ∫[-2,-1]∫[0,1] x^2y^2+cos(πx)+sin(πy) dy dx = ∫[-2,-1] 1/3 x^2 y^2 + cos(πx)*y - 1/π cos(πy) [0,1] dx = ∫[-2,-1] 1/3 x^2 + cos(πx) + 2/π ...
*July 30, 2015*

**Trigonometry**

recall the law of cosines. The distance z will be z^2 = 20^2 + 44^2 - 2*20*44*cos100°
*July 30, 2015*

**math**

why all the words? a = x^2+x-42 Factor that to get a = (x+7)(x-6)
*July 30, 2015*

**math - eh?**

and then? and is your + key not working?
*July 29, 2015*

**Algebra II**

correct
*July 29, 2015*

**calc**

so, is F(x) = ∫[1,3x] ln(t^2) dt or F(x) = ∫[1,3x] (ln t)^2 dt ? Either way, f(x) = F'(x) = ln(3x)^2 * 3 or f(x) = F'(x) = (ln 3x)^2 * 3 And F" = f' Recall that the 2nd Fundamental Theorem of Calculus says that if F(x) = ∫[a,g(x)] f(t) dt ...
*July 29, 2015*

**math algebra**

if p>2 is prime, p is odd So, p+999 is even, so it cannot be prime.
*July 29, 2015*

**Calculus - Rate of Change**

the cone has volume v = 1/3 π r^2 h = 1/3 πr^3 dv/dt = πr^2 dr/dt when v = 1000, r = ∛(3000/π) so, 10 = π(3000/π)^(2/3) dr/dt now you can find dr/dt, and since r=h, that's also dh/dt
*July 29, 2015*

**trig**

Good catch. a^2+b^2 = 64, not 8
*July 29, 2015*

**trig**

c^2 = a^2+b^2 = 64 ab/2 = 8 (a+b)^2 = a^2+2ab+b^2 = 8+32 = 40 p = a+b+c = 8+√40
*July 29, 2015*

**Calculus**

craft travel on a heading, not a bearing. 700 @ N23°E = (273.51,644.35) 60 @ E = (60,0) Add them up and you have (333.51,644.35) = 725.55 @ N27.36°E
*July 29, 2015*

**Math - My Bad**

Oops. I misread the diagonal part. Could not help the gibberish, however.
*July 29, 2015*

**Math**

area of parameter is gibberish area = (8√10)(3*8√10) perimeter = 2(8√10 + 3*8√10)
*July 29, 2015*

**calculus**

x^2+16x < 0 x(x+16) < 0 You know the graph crosses the x-axis at x=0,-16. So, x(x+16) < 0 if -16<x<0
*July 29, 2015*

**Geometry**

That's some field!
*July 29, 2015*

**maths**

You appear to want x^4+3x-2 -------------------------- (x^2+1)^3 (x-4)^2 You manage to mangle things badly, and the solution to that fraction has been posted in response to an earlier version of the poorly typed question. Here it is again. http://www.wolframalpha.com/input/?i...
*July 29, 2015*

**College Algebra**

ln x = 6 x = e^6 ln(e^2) = 2*ln(e) = 2*1 = 2
*July 29, 2015*

**College Algebra**

You probably have a yx button on your calculator. Use it. Or, note that (2/3)^x = e^(ln(2/3)*x) = e^(-0.4055x) Now you can use your e^x button to evaluate the expressions. And it's pi, not pie!
*July 29, 2015*

**College Algebra**

1/5 log(x+3)^5 = log(x+3) log(x^6)-log(x^2-x-12)^3 = 3log(x^2/(x^2-x-12)^3) = log((x^2)/(x^2-x-12)) so, adding those logs, we have log(x+3) + log((x^2)/(x^2-x-12)) = log(x^2(x+3) / (x+3)(x-4)) = log(x^2/(x-4))
*July 29, 2015*

**College Algebra**

log(a^b) = b*log(a) log(u/v) = log(u)-log(v) So, you have 5log(x)-log(y)-6log(z) where all the logs are base a.
*July 29, 2015*

**maths**

60sin(40°) The ship is east of the starting point. Your question is worded backwards.
*July 29, 2015*

**Solving Geometry Formulas**

this is just like the last one. What do you get?
*July 29, 2015*

**maths**

((6+3)(4+3)-(6*4))(0.1) = 3.9 m^3
*July 29, 2015*

**maths**

If npr=3024, r=3024/np y' = (secx-tanx)/(secx+tanx) y' = (secx-tanx)^2/tan^2x y' = csc^2x - 2cscx + 1 Those are straightforward integrals
*July 28, 2015*

**trig**

5pi/4 = pi/4 + pi So, if you can find the terminal point in QI (pi/4), just change the sign of x and y to put it in QIII.
*July 28, 2015*

**Solving Geometry Formulas**

2(7+L) = 56 7+L = 28 L = 21
*July 28, 2015*

**MATHS**

the width has grown by a factor of 4/3 so, the volume has grown by a factor of (4/3)^3
*July 28, 2015*

**Math**

I think this is an example of logistic growth, since the population will rise sharply until the food supply gives out, limiting further growth. I'm not really up on curve fitting for logistic growth, but the general curve is y = 1/(1+(1/a-1)e^(-rt)) where a is y(0)/y(&#...
*July 28, 2015*

**maths**

for area, our new area is 4pi(.99r)^2 So, if we divide that by the original area, we have 4pi(.99r)^2 / 4pir^2 = .99^2 1-.99^2 = 0.0199 That is, a 1.99% decrease in area Note that the area decreases by about twice the percentage as the radius. I think you can expect the volume...
*July 28, 2015*

**Algebra**

assuming that f(x) = ax^3+bx^2+cx+d, d=47 a+b+c+47 = 32 8a+4b+2c+47 = -13 27a+9b+3c+47 = 16 f(x) = (52x^3 - 201x^2 + 104x + 47)/3
*July 28, 2015*

**math**

How about 1√1000000 2√250000 4√62500 5√40000 8√15625 10√10000 and so on List 'em and count them
*July 28, 2015*

**Calculus**

well, it ain't enough. If the cross-sections are perpendicular to the x- (or y-)axis, then each thin plate has base of width 2-x, so v = ∫[0,2] (2-x)^2 dx = 8/3 If the cross-sections are parallel to the line x+y=2, then each thin slice has base √(x^2+y^2) = &#...
*July 28, 2015*

**Calculus**

How are the cross-sections oriented? perp. to an axis, parallel to x+y=2, or some other way?
*July 28, 2015*

**Geometry**

if the triangle is half of a square, its legs are √72. So, its hypotenuse is √72√2 = 12
*July 28, 2015*

**maths**

(9π/2 m^3)/((.09π m^2)(0.2 m/s))(1min/60s) = 4.1666 min = 4'10"
*July 28, 2015*

**Geometry**

Dang. I used x as the vertex angle. Go with Reiny.
*July 28, 2015*

**Geometry**

well, tan(x/2) = 2/√3 now just take 2*arctan(2/√3)
*July 28, 2015*

**physics**

1.00 @ E = (1.000,0.000) 0.75 @ E60°S = (0.650,-0.375) 0.50 @ E20°N = (0.470,0.342) Add 'em up and you end up at (1.62,-0.33) which is 1.65 @ E11.51°S
*July 28, 2015*

**math PS**

and there is no (x-4)^2 term, since your original fraction had none. If you really want that to be included, I can only refer you to http://www.wolframalpha.com/input/?i=%28x^4%2B3x-2%29%2F%28%28x^2%2B1%29^3%28x-4%29^2%29 Scroll down a bit to the partial fraction breakdown.
*July 28, 2015*

**math - partial fractions**

(x^4+3x-2)/((x^2+1)^3(x-4)) = 266 / 4913(x-4) - 266(x+4) / (x^2+1) + 23(x+4) / 289(x^2+1)^2 + (7-11x) / 17(x^2+1)^3 Problems like this are so tedious that they really test only patience, not knowledge.
*July 28, 2015*

**Maths**

I think you mean least number of marbles. 12 = 2^2 * 3 15 = 3 * 5 20 = 2^2 * 5 So, the LCM is 2^2 * 3 * 5 = 60
*July 28, 2015*

**D.A.V. public school**

If its value last year was v, then 387000 = 0.9v
*July 28, 2015*

**Physics**

max height when starting with velocity v is v^2 / 2g The mass does not matter.
*July 28, 2015*

**Math**

Consider the group of 3 dogs as a unit. That means there are 7 friendly dogs, and another "fighting unit" which can be anywhere in the line. So, there are 8! ways to arrange the dogs which will cause a problem, because the three fighters are all together. Any other ...
*July 28, 2015*

**maths**

since the weight varies as 1/d^2, if its weight is 1/4 as much, d is twice as big. That is, 1/(2d)^2 = 1/4 * 1/d^2
*July 28, 2015*

**English grammar**

How about He seems to love reading good books.
*July 27, 2015*

**Math**

12h+10a = 7 5h+15a = 4 ...
*July 27, 2015*

**Math**

.30A + .70(800-A) = .54(800)
*July 27, 2015*

**trigonometry**

If we let d = distance between buildings h = height of the taller building h/d = tan 43°2’ (h-25)/d = tan 21°11’ so, eliminate d and then evaluate for h h cot43°2’ = (h-25) cot21°11’ ...
*July 27, 2015*

**math**

60 divides N^2 60 = 2^2*3*5 That means that 3^2 and 5^2 must also divide N^2 So, only 16 might not be a factor of N^2 (2*3*5)^2 = 900 16 is not a factor of 900; then others are.
*July 27, 2015*

**math**

true, depending on the value of x.
*July 27, 2015*

**Calculus**

If the point on the ground is x ft from the 60' tower, then the cable needed is z = √(x^2+60^2) + √((100-x)^2 + 80^2) minimum z is where dz/dx = 0
*July 27, 2015*

**Math**

1/(1+r) = 1-.067 = .933 1+r = 1.0718 r = 7.2%
*July 27, 2015*

**Algebra**

Not sure what you mean by the data set for 13 is -1.7
*July 27, 2015*

**math**

If we are talking about integers, then the product must be even and not negative
*July 27, 2015*

**math**

2001-2099 Looks like 99 to me
*July 27, 2015*

**algebra**

nope. Check your value of f(5).
*July 27, 2015*

**Algebra**

As usual, there are several ways a garbled posting might be interpreted. Damon's way has the advantage of a rational solution. Mine and others, such as 5^4 - (8)x^2 have square roots.
*July 27, 2015*

**Algebra**

No, what you got was x^2(5x^2-8) = 0 so, x=0 or 5x^2-8 = 0 x^2 = 8/5 . . .
*July 27, 2015*

**pre-calculus**

recall that sin^2 + cos^2 = 1
*July 27, 2015*

**pre-calculus**

good insight, simpler solution. Shows that with trig, there is almost always more than one path to solutions.
*July 27, 2015*

**pre-calculus**

112.5 = 5π/8 = (5π/2)/4 so, use your half-angle formula twice: cos(x/2) = √((1+cos(x))/2) cos(5π/4) = -√((1+0)/2) = -1/√2 since 5π/4 is in QIII sin(x/4) = √((1-cos(x/2))/2) sin(5π/8) = √((1-cos(5π/4)/2) = √((1+1...
*July 27, 2015*

**Math**

The triangular numbers are 1,3,6,10,... So, the nth tetrahedral number is the sum of the first n triangular numbers. The nth triangular number is the sum of the first n integers (1+2+3+...) = n(n+1)/2 It is clear that the nth tetrahedral number will be a cubic expression in n...
*July 27, 2015*

**pre-calculus**

well, you have tan^2 x = 1/3 see previous posting for solving it.
*July 27, 2015*

**English 9**

I get that the only correct answers are 5,6,8
*July 27, 2015*

**pre-calculus**

well, you have tan^2 x = 1/3 which angle does not have tan x = ±1/√3 ?
*July 27, 2015*

**Math**

call the solutions A and B. If there are x gal of A, then there are 7-x gal of B. So, if we keep track of the alcohol amounts, we have 4/5 x + 1/6 (7-x) = 1/2 (7) Now just solve for x, the amount of solution A.
*July 27, 2015*

**Algebra**

well, did you check your answer? money spent at store #1 = 3/4 * 51100 + 75 = 38400, leaving 12700 at #2, spends 3/4 * 12700 + 75 = 9600, leaving 3100. at store #3, spends 3/4 * 3100 + 75 = 2400, leaving 700 Looks like you are correct. It seems you need to learn how to check ...
*July 27, 2015*

**pre-calculus**

you know that sin 45 = √2/2 225 = 180+45, so sin(225) = sin(180+45) now just use your addition formula to get the above result.
*July 27, 2015*

**Math**

Hmmm. I see I was mistaken. You are correct. My equations add up the wrong stuff. I see I should have checked my answer.
*July 27, 2015*

**Math**

Hmmm. If 2 kg of alloy 1 are used, then that means there are 2 kg of gold 10kg of silver 6 kg of lead You are already over 10 kg of the mixture. Add in to that any amount of the other alloys and things are even worse.
*July 27, 2015*

**Math**

If there are x,y,z kg of the alloys, then we want x+5y+3z = 10 2x+3y+4z = 10 5x+2y+2z = 10 Just solve those to find x,y,z. Since the x value is from the 1st alloy, in the ratios 1:5:3, that is how many kg of that alloy we need.
*July 27, 2015*

**Pre-calc**

we have sin(u+v) = sinu cosv + cosu sinv sin(u-v) = sinu cosv - cosu sinv add them up and you have sin(u+v)+sin(u-v) = 2sinu cosv now just finish it up.
*July 27, 2015*