Wednesday

September 3, 2014

September 3, 2014

Total # Posts: 24,573

**Math**

Perimeters are in the same ratios as the sides. You are unclear which rectangle is which, but I am assuming the larger one has perimeter 65, so the smaller one has perimeter 65 * 2/5 = 26 You are correct.
*September 2, 2014*

**Math**

Doesn't matter how many squares there are. The board is square, so each side is √11 = 3.3 cm must be a travel set. I suspect a typo.
*September 2, 2014*

**Calculus**

any line through (8,4) in the first quadrant is y-4 = m(x-8) So, if the line in question has slope m, the x-intercept is 8-(4/m) and the y-intercept is 4-8m So, L = √x^2+y^2 = √((8-(4/m))^2 + (4-8m)^2) = -4/m √((m^2+1)(2m-1)^2) Of course, there are other ways...
*September 2, 2014*

**algebra**

assuming any partial tiles are thrown away, the floor is 16 x 6 tiles So, 96 tiles are needed So, 5 boxes of 20 must be purchased, for $270
*September 2, 2014*

**Algebra 2**

If you will enter your expressions at wolframalpha, you can check your answers. Be sure to use enough parentheses so things are right. The web site will show you how it interprets your input.
*September 2, 2014*

**MATH**

start out with x x x x x minimum=5: 5 x x x x range=7: 5 x x x 12 median=9: 5 x 9 x 12 Now you can fill in any other values fitting in those intervals.
*September 2, 2014*

**math**

do you not have a calculator? Just add the numbers. Heck, you can even just enter 21456+9087 into google and it will calculate the sum.
*September 2, 2014*

**Algebra 2**

a - 2a/5 = 3 you have to multiply everything by 5: 5a - 2a = 15 You only multiplied the -2a/5 and the 3.
*September 2, 2014*

**Algebra1**

I assume you want the value for C. When you have an equation, as long as you do the same thing to each side, it remains true. So, C+4 = -24 You want C by itself, so if you subtract 4 from both sides, you have C+4-4 = -24-4 4-4=0, and of course, C+0 is just C, yo now you have C...
*September 2, 2014*

**math**

what's to explain? Write the facts in symbols: p^2 + (p+5)^2 = 625 Now just solve for p and find p+5
*September 2, 2014*

**Algebra**

$/ml is what we want. So, of course, that means that x $ / y ml = x/y $/ml = z $/ml z = x/y
*September 2, 2014*

**Algebra 1**

you are correct. Keeping the units around can help with this kind of problem. 15 gal * y mi/gal = x mi Note that the gal units cancel, leaving mi on both sides of the equation. If the units agree at the end, chances are good that your calculations are correct.
*September 2, 2014*

**Algebra 1**

just convert everything to decimal, and then the listing is easy, right? -√9 = -3 is a rational number. As such, it is, of course, also real.
*September 2, 2014*

**Algebra 2**

1: correct 2: (∞,3]U(5,11] all values less than or equal to 11, except those greater than 3 and less than or equal to 5.
*September 2, 2014*

**college algebra**

(x^2+x-6)/(x^2-4) = (x-2)(x+3) / (x-2)(x+2) Now divide that by (x+3) / 2(x+2) or, equivalently, multiply by 2(x+2)/(x+3) Lots of factors cancel.
*September 2, 2014*

**college algebra**

since x^2-x-12 = (x-4)(x+3), just multiply top and bottom by (x-4) x/(x+2) + (4x-3) / (x+2)(x-2) requires a common denominator of (x+2)(x-2). So, just multiply the first term top and bottom by (x-2) and then just add the numerators.
*September 2, 2014*

**math**

width: w length: w+4 2(w + w+4) = 84 Now just solve for w and calculate w+4.
*September 2, 2014*

**math**

c(1+.60) = 2000 Now just solve for c.
*September 2, 2014*

**math**

The side shortened least is longest, so (x-14)^2 + (x-13)^2 = (x-12)^2 Now, we all know that the 3-4-5 triangle is the only one with sides which are consecutive integers. So, x-14=3, making x=17. Solving the equation gives the same result.
*September 2, 2014*

**math**

what's to explain? Write the facts in symbols: p^2 + (p+5)^2 = 625 Now just solve for p and find p+5
*September 2, 2014*

**math**

(x+y)y = 70 (y-x)x = 12 now proceed
*September 2, 2014*

**algebra**

good work. Just for the future, recall that the domain of all polynomials is all real numbers.
*September 2, 2014*

**math**

221 = 13*17 you can use that to answer your question, rather murkily stated.
*September 2, 2014*

**Geometry**

clearly, HJ+JK=HK, so x/4 + 3x-4 = 22 now just solve for x and evaluate the distances
*September 1, 2014*

**KIPP Believe**

divide the 740 into 20 equal parts of x each. There are thus 17x students and 3x faculty. So, 17x+3x=740 20x=740 x=37 So, there are 17*37 = 629 students 3*37 = 111 faculty
*September 1, 2014*

**Math Proof**

we want to show that |x1 + x2 + ... + xn| <= |x1| + |x2| + ... + |xn| Let's take n=3 We know that the inequality holds for n=2, so now we have |(x1+x2)+x3| <= |(x1+x2)| + |x3| But, we already know that |x1+x2| <= |x1|+|x2|, so we now have |x1+x2+x3| <= |x1|+|x2...
*September 1, 2014*

**Math**

Only the student can tell what was intended. We've provided two solutions, so our work is done here.
*September 1, 2014*

**Math**

Not quite 9^(1/2) = 3 but (n^4)^(1/2) = n^2 so the answer is 3n^2
*September 1, 2014*

**Math**

looks good to me.
*September 1, 2014*

**MA 112**

for a proportion, the two ratios must be equal So, a = a+b which can only happen if b=0. But, we cannot divide by zero, so a/b and (a+b)/b would both be undefined.
*September 1, 2014*

**math**

the numerator is a whole number, and it is multiplied by the other whole number.
*September 1, 2014*

**Math**

I get x^2+x-6 in the numerator, so better double-check your math.
*September 1, 2014*

**Math**

If there were x liters to start, then we have the following table: month water at end of month 0 x 1 x/2 + 1 2 (x/2 + 1)/2 + 1 = x/4 + 3/2 3 (x/4 + 3/2)/2 + 1 = x/8 + 7/4 ... n x/2^n + (2*2^n-2)/2^n = (x + 2*2^n-2)/2^n So, at the end of the 5th month there were (x+62)/32 = 4 x...
*September 1, 2014*

**math**

A. You can factor out 3/√x to get 3/√x (x^2-3x+2) 3/√x (x-1)(x-2) Or, if you don't want that extra 1/√x, now you can divide by it to get 3(√x - 1/√x)(x - 2) or 3(x-1)(√x - 2/√x) B. xy(x^2-4) xy(x-2)(x+2)
*September 1, 2014*

**math**

what is #x ?
*September 1, 2014*

**algerbra**

okay - your turn. Check the solutions to your other problems, and take a stab at it.
*September 1, 2014*

**algerbra**

Hint: graph the two lines. Note where they intersect.
*September 1, 2014*

**algerbra**

Using elimination, double the first equation and rearrange things a bit to get 6x+4y = 8 5x-4y = 3 Now add them up and it's clear that 11x = 11 and the rest is easy
*September 1, 2014*

**math**

If x at 10%, the rest (8900-x) at 9%, then calculate the interest: .10x + .09(8900-x) = 831 x = 3000
*September 1, 2014*

**math**

You need the Binomial Theorem: (a+b)^n = a^n + n a^(n-1) b + ... So, you have (3x-2)^3 = (3x)^3 - 3(3x)^2(2) + 3(3x)(2^2) - 2^3 = 27x^3 - 54x^2 + 36x - 8 (4x+5)^2 = (4x)^2 + 2(4x)(5) + 5^2 = 16x^2 + 40x + 25 Subtract them to get 27x^3 - 70x^2 - 4x - 33
*September 1, 2014*

**MA112**

Look at the common factors on both sides d(a-b) = b(c-d) What you want to do now is unclear, but go for it.
*September 1, 2014*

**math**

as with the other, just plug in the value for x: -2(4^2) + 5(4) - 10
*September 1, 2014*

**math**

That would be 3(-2)^2 - 2(-2) + 5 = 3(4)+4+4 = 20
*September 1, 2014*

**algerbra**

If there are no typos, then 8 = 6x-4 means x=2 So, that means that -3(2)-4y=2 y = -2
*September 1, 2014*

**algebra**

the equation for the height is h = 80 + 960t - 16t^2 As you recall, the vertex (maximum height) is at t = 960/32 = 30 Now just figure h(30)
*September 1, 2014*

**math packet**

(4)(-3/5)(10/-9) = -120/-45 = 8/3
*September 1, 2014*

**math**

one of my favorite sequences. Thanks for sharing.
*September 1, 2014*

**math**

just solve for p in -2000p² + 2000p + 1700 = 5000 The problem is that D has a maximum of 2200 when p = 0.5 I suspect a typo somewhere. Is D the monthly demand, or the weekly demand?
*September 1, 2014*

**math**

x^2 + (x+4)^2 = 20^2 Think of a scaled-up 3-4-5 triangle.
*September 1, 2014*

**Algebra 2**

Josh, you can't even spell. And "gay fag" is redundant. Now, on to the math. Rearrange things a bit and you have 2m^2-8m + 6n^2-5n^2 2m(m-4) + n^2 It's kind of a strange problem; is there a typo in there somewhere?
*September 1, 2014*

**PRE CALC**

Just list the facts as given: u+f+t = 39 t = u-16 u+5f+20t = 208 Now just solve as usual
*August 31, 2014*

**math**

length=2w 2(w+2w) = 60 ...
*August 31, 2014*

**math**

If q quarts are to be drained, then .40(12-q) + 1.00q = .60(12) ...
*August 31, 2014*

**algebra**

S = P(1 - 0.25)
*August 31, 2014*

**stats**

since 2500 is exactly the mean (0 std away), look up P(Z < 0) in your standard table. Or, just consider the data and pick what seems reasonable in that case.
*August 31, 2014*

**math**

If x L are added, then .01(40) + .04x = .02(40+x) ...
*August 31, 2014*

**Math**

see related problems below
*August 31, 2014*

**physics**

(E).893J
*August 31, 2014*

**Math**

When you solve for a variable, you want it all by itself: P = I/rt r = C/2π
*August 31, 2014*

**Algebra**

#1 correct #2 If y=1, 4y-2=2 and 3*2 is not 18 You should have done 3(4y-2) = 18 4y-2 = 6 4y = 8 y = 2
*August 31, 2014*

**AP Calc. AB**

well, since -0 = 0 and -8 = -(8), it should not be hard to make the required change to the function I provided. I mean, you do remember your basic algebra, right?
*August 31, 2014*

**AP Calc. AB**

as x-> 0, 1/x -> ∞ 2^(-1/x) -> 2^-∞ = 0 when x>0 2^(-1/x) -> 2^∞ = ∞ when x<0 So, in those two cases, we have lim x->0+ = 8/(1-2^-∞) = 8/(1-0) = 8 lim x->0- = 8/(1-2^∞) = 8/-∞ = 0 http://www.wolframalpha.com/...
*August 31, 2014*

**AP Calc. AB**

a different problem was recently posted here where the left limit is 2 and the right limit is zero. Tweaking it a bit, we have (8)/(1-2^(-1/x)) which fits your criteria. If you can come up with any function where the left and right limits for some value of x are different, you...
*August 31, 2014*

**Math Compounded daily**

1450(1+.12/365)^(365*1) = 1634.84 Some problems specify that we use 360 days per year. In that case, change the expression as needed.
*August 31, 2014*

**trigonometry**

a circle of radius r has circumference 2pi r, so one lap = 6pi feet So, 20 laps = 120pi feet
*August 31, 2014*

**Algebra 2**

since distance = speed * time, (x+3.5)(5) = (x-3.5)(12) x = 8.5 Now you can easily show how far it went.
*August 31, 2014*

**Algebra 2**

distance = speed * time, so if the plane's speed is x and the wind speed is y, we have 4(x+y) = 5040 5(x-y) = 5040 x = 1134 y = 126
*August 31, 2014*

**Math**

to get the x-intercept, set y=0 and you have -5x = 10 to get the y-intercept, set x=0 and you have 2y = 10 Look at your graph.
*August 31, 2014*

**Algebra**

If the only possible value for x is 3, then the function must factor into (x+3)(x+3) = x^2+6x+9 When they say the only possible value, they mean the only value which makes the equation true. Naturally, x can have many other values, but in those cases, the value on the left is ...
*August 31, 2014*

**Calculus**

at time t hours, the distance d is d^2 = (60t)^2 + (35t)^2 when t=3, then, d=208.39 Since 2d dd/dt = 7200t + 2450t = 9650t at t=3, dd/dt = 28950/(2*208.39) = 69.46 km/hr
*August 31, 2014*

**math**

as usual, you need to find the common denominator. Assuming you meant 2/(m+3) + 1/m the LCD = m(m+3). So, you have (2m + m+3) / m(m+3) = (3m+3) / m(m+3) On the other, I think you have 2/((x+1)(x+2)) + 3/((x+2)(x+3)) so the LCD is (x+1)(x+2)(x+3). That gives you (2(x+3) + 3(x+1...
*August 31, 2014*

**Calculus 2**

Using shells, you have thickness dy, so v = ∫[1,3] 2πrh dy where r = y and h = x = 1/y v = 2π∫[1,3] y*1/y dy = 2π∫[1,3] dy = 2πy [1,3] = 2π(3-1) = 4π That strikes me as odd. Let's try discs (washers) of thickness dx. The ...
*August 30, 2014*

**algebra**

x + x-3 = 21 x = 12
*August 30, 2014*

**Algebra 1**

width: w length: 2w-3 after the changes, we have (2w-3-2 + w-1) = 24 Now just solve for w, the original width.
*August 30, 2014*

**math**

L+M=250 3/5 L = M-40 Now we can substitute for M, and we get 3/5 L = 250-L-40 8/5 L = 210 L = 131.25
*August 30, 2014*

**math**

rotation does not change lengths.
*August 30, 2014*

**Algebra**

12x^2+8x = 4x(3x+2) 9x^2-4 = (3x-2)(3x+2) I think you can see which factor cancels.
*August 30, 2014*

**Math**

If they get shares of x,y,z, respectively, then x = 2/3 y y = 1/4 z x+y+z = 1360 We want to know what y is, so substitute in for x and z: 2/3 y + y + 4y = 1360 17/3 y = 1360 y = 240
*August 30, 2014*

**physics**

dropped stone hits water: 24.4-4.9t^2=0 t=2.23 stone has fallen 6.6m: 4.9t^2 = 6.6 t = 1.16 So, Jill's thrown stone only has 2.23-1.16=1.07 seconds to hit the water: 24+1.07v-4.9*1.07^2 = 0 v = -17.19 m/s
*August 30, 2014*

**Physics Help!!!!**

total distance covered: 20+30+40=90m total time: 30s avg speed: 90/30 = 3m/s final displacement: 20-30+40=30m north avg speed: 30/30 = 1 m/s north
*August 30, 2014*

**Physics**

total distance covered: 20+13+10=43m total time: 45s avg speed: 43/45 = 0.96 m/s final displacement: 20-13+10 = 17m avg velocity: 0.38 m/s away from entrance
*August 30, 2014*

**maths - 2nd take**

what a lot of bother I went to above. The distance from (2,3) to (5,4) is √10. Since the circles are orthogonal at the intersections, the two given points and the centers of the circles form a square, with diagonal √10. So, the side of the square, which is the ...
*August 30, 2014*

**maths**

The line joining the two points is y-3 = (1/3)(x-2) Its midpoint is at (7/2,7/2) The centers of the circles lie on its perpendicular bisector, which is y-7/2 = -3(x-7/2) If the radius of the circles is r, the distance from (2,3) to the second line is r/√2 So, just figure...
*August 30, 2014*

**Algebra 2**

multiply the speed by the time taken. The freight train traveled for 15 hours at 42 km/hr.
*August 30, 2014*

**Algebra 2**

since distance = speed 8 time, we want 42t = 105(t-9) t = 15 So, at 5 am the next day they meet, after having gone 630km. Or, you can consider that the freight train has a 9*42=378km head start, so the express train, traveling 63km faster, makes up the distance in 6 hours, ...
*August 30, 2014*

**Physical Science**

hint: 1 degree is 1/360 of the circumference.
*August 30, 2014*

**Physical Science**

the usual formula s=rθ so, look up the radius of the earth, and multiply that by angle of 1'=0.00029 radians
*August 30, 2014*

**math**

3tanθ = 4, so tanθ = 3/4 That means that sinθ = 3/5 cosθ = 4/5 Now just plug in the numbers
*August 30, 2014*

**Algebra**

z^3 = 8p^3 cis2π z = 2p cis 2π/3, 2p cis 4π/3, 2p cis 0
*August 30, 2014*

**math**

If you label the points A,B,C from left to right, then just show that the slope of AB is the same as the slope of BC. Since B is on both line segments, they are part of the same line.
*August 30, 2014*

**math**

how do you get 306? The perimeter of the kite in cm is 9+9+17+17 Not sure how the inches come into play.
*August 30, 2014*

**math**

count the coins: q = 2h+3 d = q+10 count the values: 10d+25q+50h = 6205 Now you have three equations, and you can solve them to get 113 dimes 103 quarters 50 halves
*August 30, 2014*

**Calculus AB**

correct. The limit from the right is not the same as the limit from the left. But both limits are finite.
*August 29, 2014*

**Calculus AB**

Looking at the graph at http://www.wolframalpha.com/input/?i=%284%29%2F%281-2^%281%2Fx%29%29+ I'm sure you can justify your analysis.
*August 29, 2014*

**vector**

standard differential geometry stuff. You surely have formulas for this in your text. Any google search will provide many discussions of the topic. What do you come up with? We'll be happy to check over your answers.
*August 29, 2014*

**Math - any ideas?**

With that pesky 7 in there, I'm at a loss for any way to combine the terms.
*August 29, 2014*

**Math**

why is the Imperial system any worse than the Cheshire system? Fie on thee, sirrah - hie thee hence!
*August 29, 2014*

**Math**

Hmmm. It doubles every day. 6 = 4 * 1 1/2, so it must have taken the last 2 days to double twice from 1 1/2 to 3 to 6 inches. I say it took 28 days. I'd have stopped at .25 = 2^(t-30) -2 = t-30 t = 28 Not sure what went wrong after that.
*August 29, 2014*

**math**

didn't we just do one of these with different numbers? Follow the same steps.
*August 29, 2014*

**maths**

since distance = speed*time, if the still-water speeds of the men are a and b, (a-.5)(5) = 10 (b+.5)(3) = 10 5a - 5/2 = 10 3b + 3/2 = 10 b-a = (10 - 3/2)/3 - (10+5/2)/5 = 1/3
*August 29, 2014*

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