Sunday

August 30, 2015
Total # Posts: 33,132

**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*

**Physics**

well one lap is 500m.
*July 27, 2015*

**math**

or, we could convert to polar first: 2+j4 = √20 cis 1.107 -3-j4 = 5 cis 4.068 multiply and you get Reiny's value.
*July 27, 2015*

**math**

I assume each equation is separate, so we have x = 2/π 3x+2 = 5x+7-2x 3x+2 = 3x+7 No solution 6x+2-4x=5+2x-3 2x+2 = 2x+2 infinitely many solutions
*July 26, 2015*

**chemistry**

The mol mass of KI is 166, so 3.5g = 0.021 moles KI .021mol/.0275L = 0.767 M So, you need to dilute the original solution from 5.95M to 0.767M That's a dilution by a factor of 7.757 So, the volume must increase by that factor, so you need .047*7.757 = 365mL
*July 26, 2015*

**math**

You can check your calculations here: http://www.wolframalpha.com/input/?i=solve+{{1%2C-1%2C3}%2C{2%2C1%2C2}%2C{-2%2C-2%2C1}}*{{x}%2C{y}%2C{z}}+%3D+{{2}%2C{2}%2C{3}}
*July 26, 2015*

**math- partial fractions**

so, fix it and follow the steps. If you enter your formula at wolframalpha.com, it will show the decomposition, so you can check your work. Watch to be sure you include necessary parentheses. wolframalpha will show how it interprets your input.
*July 26, 2015*

**math- partial fractions**

If, however, you meant 1/((s^2)(s^2 +2s +3)) then that is A/s + B/s^2 + (Cs+D)/(s^2+2s+3) = A(s*(s^2+2s+3)) + B(s^2+2s+3) + (Cs+D)s^2 = As^3+2As^2+3As + Bs^2+2Bs+3B + Cs^3+Ds^2 = (A+C)s^3 + (2A+B+D)s^2 + (3A+2B)s + 3B all over s^2(s^2+2s+3) So, to make the two sides equal, we ...
*July 26, 2015*

**algebra**

I think that's the symmetric property. Transitive is A=B and B=C ==> A=C
*July 26, 2015*

**Math 156**

You know that √N * √N = N So, if you pick pairs of factors, one factor must be less than √N and the other must be greater than √N. So, if you have tested all possible factors less than √N and found none, you are done.
*July 26, 2015*

**Math Functions**

You know that y=e^x goes through (0,1) and has an asymptote at y=0, so, y = e^x + 3 has its asymptote at y=3. Further, 5e^x goes through (0,5), so one function would be 5e^x + 3 Now, you can pick any number greater than 1, and the above statements hold true, so 5*2^x, 5*10^x ...
*July 26, 2015*

**math**

plates: no repeated letters: there are 26 choices for the 1st letter, leaving only 25 for the 2nd, and 24 for the 3rd: 26*25*24 = 15600 repeated letters: 26*26*26 = 17576 matrix: If the matrix of coefficients is A, you want (x,y,z) such that AX = B where B = (2 3 2) That means...
*July 26, 2015*

**MAth 109**

I assume your formula is d = √(3h/2) If the person is 80' above the water, just plug it in: d = √(3*80/2) = √120 = 2√30 = 10.95 So, the person on the deck can see 10.95 miles.
*July 26, 2015*

**Physics**

PE = mgh 3800 = (mg)(4.42) His weight is mg
*July 26, 2015*

**calculus problem I have tried 100 times**

The radius of the surface of the chocolate at height y is 4y/5. So, the volume of a disc of thickness dy at height y from the tip of the cone is v = π(4y/5)^2 dy That makes the mass of the disc m = 16π/25 y^2 dy * 1080 kg = 691.2πy^2 kg The height that the mass ...
*July 26, 2015*

**maths**

could be anything, since 7≠56
*July 26, 2015*

**Maths**

1/a + 1/b + 1/c = 1/10 1/a + 1/b = 1/20 1/a + 1/c = 2/25 Now just solve normally use GJ, but use 1/a, 1/b, 1/c as your variables. When you get the results, the values for a,b,c are the reciprocals.
*July 26, 2015*

**trig**

or, you can proceed by noting that you have tanθ = (1-sin36)/cos36 = 0.509
*July 26, 2015*

**St Kabir**

sin(θ+36) = cosθ sinθ cos36 + cosθ sin36 = cosθ sinθ cos36 = cosθ(1-sin36) That's exact, but you can approximate sin 36 = 0.6 cos 36 = 0.8 You can change sin to cos bu squaring and remembering that sin^2 + cos^2 = 1
*July 26, 2015*

**ssc..math**

since the ratio of height:shadow is the same for both, 30/7 = h/5
*July 26, 2015*

**src**

well, clearly, the 1st term is 15, and the difference is 9.
*July 26, 2015*

**Maths**

5p + 2(p+10) = 370
*July 26, 2015*

**Calculus**

so, what formula did you use? was it 10*(1/2)^(t/30) ? If so, just recall that 1/2 = e^(-ln2), so that is the same as 10e^(-ln2/30 t) Now just plug in your various numbers.
*July 26, 2015*

**Calculus**

I assume you have been shown how to derive f' when f' = e^x. f(x) = 5^x f(x+h) = 5^(x+h) = 5^x * 5^h f(x+h)-f(x) = 5^x(5^h-1) dividing by h, we have 5^x (5^h-1)/h = 5^x (e^(h ln 5)-1)/(h ln5) * ln5 Recall than lim(u->0) (e^u-1)/u = 1, so we wind up with ln5 5^x x^2 ...
*July 26, 2015*

**Math**

the surface of the water is a circle of diameter 5m, so its area is 25π/4 m^2 Since it takes 4 hrs to fill the 1m, pool, the water comes in at a rate of 25π/16 m^3/hr So, after t hours, the volume of water in the pool is v(t) = 25π/16 t m^3
*July 25, 2015*

**lesiba**

Actually, it is not. The surface area of a sphere is 4πr^2, so the area of a hemisphere is 2πr^2. As to why, you can google the topic and find various ways to derive the formula.
*July 25, 2015*