Wednesday

July 23, 2014

July 23, 2014

Total # Posts: 23,925

**Math - ill posed problem**

Today's date keeps changing, and if you are calculating days elapsed, you cannot just say "June 2008" for the start date.

**physics**

the distance fallen at time t seconds is 4.9 t^2 meters I leave it up to you to discover how high the tower is, and just what you mean by "position."

**Math**

2/3 + 1/6 = 5/6 spent. So, 1/6 saved. Probably better spent on math books . . .

**College algebra**

2738222/(69+119) = 14,565.01 per vehicle

**Math**

y = 3^x has asymptote y=0 and passes through (0,1) y = 1.5 * e^3x passes through (0,3.5) y = 3^x + 2 has asymptote y=2 and passes through (0,2) y = 1.5 * 3^x + 2 has asymptote y=2 passes through (0,3.5) y = 1.5 * 3^-x + 2 is the reflection. Same intercept, same asymptote. see ...

**PreCalculus**

well, just solve for t: 20 = 30/(1+29e^-1.35t) 1+29e^-1.35t = 1.5 e^-1.35t = 0.5/29 -1.35t = -4.060 t = 3

**QZIA**

for i=1..10 read name(i),weight(i) sum = 0 for j=1..5 read score(i,j) sum += score(i,j) next j avg(i) = sum/5 next i for i=1..10 print name(i),weight(i)*2.2,avg(i) next i

**Trig**

I've also preferred to do the complex multiply stuff using the fact that (a-b)(a+b) = a^2-b^2, so (a-bi)(a+bi) = a^2+b^2 (x-(5-√2i))(x-(5+√2i)) ((x-5)+√2i)((x-5)-√2i) (x-5)^2 + √2^2 x^2-10x+25 + 2 x^2-10x+27

**Alegra**

(b) 4(6) + 5(-4) = 24-20 = 4 (c) 2(6 + 3(-3)) = 2(6-9) = 2(-3) = -6 when asking what you did wrong, it's a lot better if you show what you did, so we can figure that out.

**algebra**

If you want AD, first check for rank compatibility A is 2x2 and D is 2x1, so AD will be 2x1. The solution is here: http://www.wolframalpha.com/input/?i={{0.1%2C+0.03}%2C{0.07%2C+0.6}}*{{5}%2C{10}}+

**finance**

275000 * 1.05^10 = 447946

**math**

there's nothing to solve. You have just presented a function of x. Now, if you want to find where f(x)=0, that's a problem you can solve. On the way, we'll discover how to factor the quadratic, which is another kind of problem. F(x) = (x-3)(x+1) so, F(x)=0 when x-3...

**Mathematics**

first, multiply by x and you get 315.414x = 2πx^3 + 860.16 then rearrange in descending order of exponent: 2πx^3 - 315.414x + 860.16 = 0 Now you have to solve a cubic, which is generally not easy. And in this case, I guarantee it is not easy. You will have to fall ba...

**physics**

the falling time can be found by: h-4.9t^2 = 0 t = √(h/4.9) echo time is h/342, so √(h/4.9) + h/342 = 3.0 h = 40.673 meters

**maths - ahem**

How can 112 be the smallest? 112-99 > 7

**math**

since time = distance/speed, and avg speed = totaldistance/totaltime, = (68+20)/(68/51 + 20/40) = 48 km/hr

**Physics**

conserve momentum, so (2000)(47) = (2000+40000)v v = 1/21

**Math**

(2+√2)/(2-√2) = x/(3+√10) just multiply by 3+√10 and you have x = (2+√2)(3+√10)/(2-√2) = (26+3√2+2√10)/(2-√2) now multiply top and bottom by 2+√2 and you have x = (26+3√2+2√10)(2+√2)/2 = 58+32&...

**math**

1500(1-.20)(1-.15)(1-x) = 969 x = 5

**math**

first step: see the first related question below, where the problem is solved.

**Calculus**

well, since the kite is flying horizontally, I think we can assume that it stays 40m high. And we usually go with a straight string in elementary problems like this. So, if the kite is x feet away horizontally, the straight-line distance to the kite is d^2 = x^2 + y^2 = x^2 + ...

**math (4)**

Scroll down a ways and see all the problems posted by "Vanessa" which have already been solved. Or, look at the related questions, where I think you will find links to them.

**math**

y-0 = 2(x-0) y = 2x

**Math (6)**

Take a look at http://www.jiskha.com/display.cgi?id=1398307458 where I discussed this problem.

**math**

consider the deck either as 4 suits of 13, or as 13 suits of 4. P(same suit) = 1 * 12/51 P(~pair) = 1-P(pair) = 1 - 1 * 3/51

**Calculus Please Help3**

Finishing Reiny's excellent evaluation, we get 49/8 + 392/27 = 4459/216 = 20.643 Now, if you want to use horizontal strips, there is no reason to divide up the region, since each strip is bounded on the left and right by a single curve. Our two boundary curves are now x = ...

**math**

no idea. How many words are to be printed on the 21 pages?

**Calculus Please Help**

well, you can integrate and then take the derivative, and get f(x) = 1/3 (x^6-x^3) f'(x) = 2x^5 - x^2 or you can use Leibnitz' Rule and get (x^2)^2(2x) - (x)^2(1) = 2x^5 - x^2 read up on it at http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign

**calculus help**

see http://www.jiskha.com/display.cgi?id=1395265529

**Math**

n^2p^2 / m

**Coordinates Math**

Looks like y = -4x, so yes.

**math**

nope. d a relation is not a function if for some x=h, there is more than one pair with that x-value. according to d, f(4) = 4 f(4) = 1 both cannot be true for a function. All the other relations have distinct x values for all the pairs. They are functions.

**math**

Which of the following relations is not a function? (8, 4), (-5, 2), (4, 1), (-6, 2) (-5, 4), (4, 2), (8, 1), (-6, 5) (-6, 4), (4, 3), (-5, 1), (8, 2) (4, 4), (-5, 2), (4, 1), (-6, 2) my answer is b

**Geometery help**

not at all. Nothing changed. reflection through the origin reverses the signs of x and y. (x,y) -> (-x,-y) You need to get some graph paper. Plot a point, and draw a line from it to the origin. Then extend the line an equal distance past the origin. At the end of that line ...

**Algebra**

is there a question in there somewhere?

**Math**

(n-1)^2 = 5 n-1 = ±√5 n = 1±√5

**science**

a 2,500 kg car at 15 m/s squared, 15 m/s squared is the what of the car?

**science**

what is the force that results from a 3,000 kg car accelerating at 20.00 m/s squared?

**Math**

(16a)/(a√(6a)) 16 / √6a

**Permutations**

Odd. you got 4! correct, but think 5! is 5^2? And then you got #5 correct, too! #6 ok as well.

**Help me solve??**

since "right" is in the positive x direction, and "up" is the positive y direction, just add 8 and 3 to the (x,y) values, respectively: X(5,6) -> (13,9) and the same for Y and Z.

**Math**

I get m=13 y = 2x+1

**Algebra**

the key is correct a^3 = a^2 * a, so √a^3 = √a^2 * √a = a√a similarly for b and c

**Geometery - incomplete**

depends on what the figure is supposed to be.

**math**

Assuming the slippage is continuous, his climbing rate is 4 meters in 25 seconds. So, he scrambles up 53 meters in 53/(4/25) = 331.25 seconds.

**math**

that is correct.

**math**

some loaded dice! Since 5 occurred 2 times in 20 throws, that's 10% so, what's 10% of 240 throws?

**algebra**

Assuming the box is closed, it has 6 faces, so 2x(x+3) + 2x(1) + 2(x+3)(1) = 130 x^2+5x+3 = 65 x = 5.76 Hmmm. If the box is open, with no top, then x(x+3) + 2x(1) + 2(x+3)(1) = 130 x^2+7x+6 = 130 x = 8.17 I guess x is not an integer.

**maths**

that would be (mn^2 + nm^2)/(m+n) = mn seems odd, don't it?

**algebra**

assuming that x is in degrees, then knowing that cos(x) = sin(90-x), we have sin(2x+30) = sin(90-(x-15)) sin(2x+30) = sin(105-x) so, 2x+30 = 105-x 3x = 75 x = 25 check: sin(80) = cos(10) yep.

**Locus- Calculus**

|z-5-12i| is the distance of z from 5+12i |z-5+12i| is the distance of z from 5-12i As we know, the locus of a point whose distances from two other points add to a constant, is an ellipse. The equation of the ellipse should be no problem, since you know the foci and the sum of...

**Algebra**

√(2m)^-3) = (2m)^(-3/2) = 1/(2m)^(3/2) = 1/(√8 m^(3/2)) 1/m^-1 = 1/(1/m) = m so, you have m/(√8 m√m) = 1/√(8m)

**Algebra**

we know that the dividend is (x+3)(x^3+x^2-4) + 8 Just expand that.

**maths**

xsinθ - ysinθ = √(x^2+y^2) cos^2θ/a^2 + sin^2θ/b^2 = 1/(x^2+y^2) not sure what you mean by the "correct relation" We can solve for θ or solve for x and y. (x-y)^2 sin^2θ = x^2+y^2 so, sin^2θ = (x^2+y^2)/(x-y)^2 and we have (1-...

**calculus help**

v = pi r^2 h, so h = v/(pi r^2) the material used is determined by the surface area a = 2pi r^2 + 2pi rh = 2pi r^2 + 2pi r(v/(pi r^2)) = 2pi r^2 + 2v/r minimum area is when da/dr = 0, so we want 4pi r - 2v/r^2 = 0 2pi r^3 - v = 0 r = ∛(v/(2pi)) See the curve (with v=1) a...

**calculus - incomplete**

how expensive is the special fencing?

**CALCULUS**

this is a related rates problem. When the plane is x km (horizontally) away from the station, the distance y from the plane to the station is y^2 = 1^2 + x^2 When x=2, y^2 = 5 So, we now know what we need. 2y dy/dt = 2x dx/dt 2√5 dy/dt = 2*2*500 dy/dt = 200√5 km/hr

**pre algebra**

hard to say -- are there any red ones in the bag?

**Physics**

since t = 2π√(r^3/GM) just plug in the numbers and solve for M

**math**

usually * is used for multiplication, not powers. How about you write it the way I did? It's right above here...

**math**

V = s^3 = (3cm)^3 = 3^3 cm^3 = 27 cm^3

**Algebra**

what, no calculator? 1113.67 I assume you wanted to know what to invest at 10% compounded semi-anually for 6 years, to end up with $2000?

**math**

ok 1, 1 1/8, 1 2/8,...,2 7/8, 3

**Algebra**

f^-1 = 1/f f^-3 = 1/f^3 brush up on negative exponents, right? They are evaluated as reciprocals. Just an extension of the normal addition and subtraction of exponents. since 1 = f^0, f^-3 = f^(0-3) = f^0/f^3 = 1/f^3

**Algebra**

since √f = f^(1/2), √(f^-3) = f^-(3*(1/2)) = f^(-3/2) = 1/(f√f) so, √3 / (1/f√f) = √3 * f√f = f√(3f)

**Math**

assuming the dimensions are the outside dimensions, then v = (1/2)(2.2-2*0.2)(2.4-2*0.2)(6-2*0.2) = 10.08

**algebra**

x = (1±√133)/22

**math**

If we call the respective bases and altitudes a,b for ABC and d,e for DEF, then we have a = b/2 e = d/2, so d = 2e So, let's find the ratio b/e Since the areas are equal, (1/2)(ab) = (1/2)(de) (b/2)(b) = (2e)(e) b^2/2 = 2e^2 b^2/e^2 = 4 b/e = 2 So, (D) 2:1

**Calculus - hmmm**

Hmmm. I don't get your integral at all. I think you have tripped over a shortcut.

**Calculus - oops**

That'd be 5/4 ʃ t^5 + 2t^3 + t dt

**Calculus**

well, just do the integral: v(t) = ʃ (5/4 t)(t^2+1)^2 dt = 5/4 ʃ t^3 + 2t^2 + t dt

**Algebra - incomplete**

hard to say, without any other clues.

**Algebra ll**

Seems like this one came through recently (2x-5y)^3 = (2x)^3 - 3(2x)^2)(5y) + 3(2x)(5y)^2 - (5y)^3 = 8x^3 - 60x^2y + 150xy^2 - 125y^3

**maths**

Since all you have is a bunch of integers, I don't see why you're worried about "rationalizing" the denominator. (5+2^3)/(7+4^3) = (5+8)/(7+64) = 13/71

**maths - incomplete**

not sure what to do with cos^2(theta)/a^2+sin^2(theta)/b^2 since it's part of no equation.

**Math**

p/6 - 2p/5 = (4p-5)/15 multiply through by 30, the LCM(6,5,15) 5p - 12p = 2(4p-5) -7p = 8p - 10 15p = 10 p = 2/3

**math(compunded interest )**

you know that the total amount for an interest rate of r compounded n times for t years is (1+r/n)^(nt) In this case, for #1, n=2 and r = .045, so the ending amount is (1+.045/2)^(2*1) then subtract off the original amount (1) to get just the interest. (1+r/n)^(nt) - 1

**math(compunded interest )**

(1+.045/2)^2 - 1 = 4.55% (1+.12/360)^360 - 1 = 12.75%

**math(compunded interest )**

for annual deposits, just work it out. At the end of year 1: 1000 2: 1000*1.08 + 1000 3: 1000*1.08^2 + 1000*1.08 + 1000 ... n: 1000(1.08^(n-1) + ... + 1) = 1000(1.08^n - 1)/(1.08-1) or, for an interest rate of r, the balance at the end of n years is 1000((1+r)^n - 1)/(r)

**Math**

1/12 + 2/6 = 5/12 spent so, how much was unspent?

**math**

4:05 is equivalent to 3:65 3:65 - 2:15 = 1:50 think of it as a subtraction problem, where each column is 60's instead of 10's (because of 60 minutes in an hour) 4 05 -2 15 you can't subtract 15 from 5, so you borrow a 60 from the hour column: 3 65 - 2 15 -------- 1...

**math**

if you mean 22% off, then p = 16.95(1-.22) = 13.22

**Math - incomplete**

can't tell, since you give no idea how much he started with.

**math**

316.50 * 3 + 425.25 = 1374.75

**math, calculus**

Since y < 0 for -4 < x < 3, ∫[-4,-4] y dx + ∫[-4,3] -y dx + ∫[3,5] y dx Now just plug in the polynomial for y, and evaluate each integral.

**Math**

25*12.49(1-.20) - 18*12.49 = 24.98

**math!!**

3000(75-59) - 975 = 47025

**math**

(11700-4300)/11700 * 100 = 63.25%

**Math**

since the center bed is a square, I will assume that in fact all four sides are 3m. So, subtract the areas to get 15*20 - 3^2 = 291 m^2

***math (4)**

Is this a trick question? I thought the equilibrium price is where supply = demand. That is, there is zero surplus.

**Algebra**

192 = 3*64, so √192 = 8√3 7/32 √192 = 7/32 * 8√3 = 7/4 √3 or, 7√3 / 4 be careful. It's not always clear whether √6/32 means √6 / 32 or √(6/32) If in doubt, use parentheses

**math**

80 degrees is 2/9 of a complete circle so, the area is 2/9 * pi * (8/2)^2 = 32/9 pi

**Math (3)**

If you plot the curves, as at http://www.wolframalpha.com/input/?i=plot+y%3Dx^3%2C+y+%3D+x%2B6%2C+y+%3D+-1%2F2+x it is clear that the area is ∫[-4,0] (x+6)-(-x/2) dx + ∫[0,2] (x+6)-x^3 dx = 12 + 10 = 22

**math (4)**

∫[0,1] -x e^(-x^2) dx let u = -x^2, and you have du = -2x dx, ∫[0,-1] (1/2) e^u du = (1-e)/(2e)

**math (2)**

if u=x^2+4, du = 2x dx and you have du/u

**Math (6)**

If you mean x = y^2, then ok. Otherwise, the two curves do not bound an area. It's better to integrate over y, and your integrand is just (y+2) - y^2 for -1 <= y <= 2 http://www.wolframalpha.com/input/?i=plot+x%3Dy^2%2C+y%3Dx-2

**math (9)**

It's clear from http://www.wolframalpha.com/input/?i=plot+y%3D-x^2%2B2x%2B4%2Cy%3D-x%2B4+ that the area is just the integral of f(x)-g(x) for 0<=x<=3

**math (8)**

just integrate (x^2+5)-1 for 3<=x<=5

**Math**

2(20+2*2 + 14+2*2) = 84 ft

**geometry**

5/6 pi or, showing the reasoning, (1 - 60/360) * pi * 1^2

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