Wednesday

July 23, 2014

July 23, 2014

Total # Posts: 23,936

**geometry**

since -2/3 * 9 = -6, we want x such that -2/3 x = -10 x = 15

**physics (motion)**

check the units. (a) and (b) are obviously false. So, since (d) means constant velocity, (c) is the only possible choice.

**Physics--check answers please**

#1 ok #2 I get that the player is at (-22.5,-26.8) from the net, so his shot takes the puck to (2.5,-26.8). Making it 26.9 from the net. You can figure the angle. How did you get your solution? #3 ok

**math**

who said anything about coordinates? If you draw the diagram using the given values, you will have two right triangles. Their common leg will be the 20 yards from the sideline to the camera. The length of the field (plus the two end zones) will be 120 yards, divided 50:70.

**math**

Draw a diagram. The camera must pan to one side through an angle x, and to the other side, through an angle y, where tan(x) = 50/20 tan(y) = 70/20 Now just add up x and y.

**Physics**

This is a trick question. If the external load is x ohms, then the fraction of the voltage drop across the load will be x/(x+4), or 20x/(x+4) volts. You can see that if the external load is 0, then the current is 5 amps. But, that would take an infinite number of resistors in ...

**trigonometry**

convert radians to degrees, as usual. Then multiply by 60 and 60 on the decimal parts: 23.461° .461 * 60 = 27.66' .66 * 60 = 40" So, 23°27'40" Do the reverse to go the other way. Divide by 60 and 3600 to convert ' and " to °. Now use that...

**finite**

don't know the formula right off. Why don't you give the formula, and show how you evaluate it?

**Calculus**

I assume you mean (∛x^2)(2x+5)^3 ∛ has domain all reals all polynomials have domain all reals. f' = 2(2x+5)^2(11x+5) / 3∛x Now we see that f' is undefined for x=0 f'=0 for x = -5/2 and -5/11

**Calculus**

If u = 5x^2-3x+1, f(u) = csc^4(u) f'(x) = f'(u)*u' = 4 csc^3(u) (-csc(u)cot(u)) u' u' = 10x-3 Now just do the substitution.

**Calculus**

v = 1/3 pi r^2 h dv/dt = pi/3 (2rh dr/dt + r^2 dh/dt) Now just plug in dv/dt = 300pi dh/dt = -15 r=20 h=30 and solve for dr/dt

**Calculus**

v = 1/3 pi r^2 h From the shape of the cone, you know that r = h/6 so, dr/dt = 1/6 dh/dt dv/dt = pi/3 (2rh dr/dt + r^2 dh/dt) = pi/3(2(h/6)h(1/6 dh/dt) + (h/6)^2 dh/dt) = pi/18 h^2 dh/dt Now just solve for dh/dt when dv/dt = -1/4 h=4

**mat/117**

Not sure just what (9-5 4√6) means. But, let's ignore the 5 for now and pretend you had (9-4√6)(7- 4√8) Just expand like any polynomial: 9*7 - 4√6*7 - 9*4√8 + 4√6*4√8 63 - 28√6 - 36√8 + 16√48 Pulling out all the p...

**math**

Hmm. Obviously I misplaced a decimal point. I expect you can find the flaw in my reasoning.

**math**

look at the amounts of NaF involved: (.6/10^6)*100000 + 1.0*x = (1.75/10^6)(100000+x) x = 0.115 check: .6 ppm in 10^5L = 0.06g add 0.115 g and you have 0.175g, or 1.75ppm

**physics**

well, if it's incompressible, then its volume doesn't change under pressure, right? SO, since the mass also doesn't change, that must mean that the answer is (A), since density = mass/volume, both of which are constant.

**HELP!!!**

If h is the altitude from C to AB, we have 1/2 * ch = 3h = 7.52 So, h = 2.507 Now h/a = sin 40° a = 2.507/0.642 = 3.90

**Algebra**

first, you might fix your typos. Your formulas are bogus. Next, you can easily see that the values get bigger and bigger, so the 1/2 formula can't be right. Your first try is right, but should be written a(1) = 3 a(n) = 2*a(n-1) for n > 1

**math**

For the cost, we know that if there are n items, and each costs c, then the total cost is 330+nc For the revenue, if there are n items selling at price 25, the total revenue is 25n. So, just set things up to total revenue = total cost Not sure what your "break even" ...

**algebra**

-2x-2+7(x+1)=2x -2x-2+7x+7 = 2x 5x+5 = 2x 3x = -5 x = -5/3

**Pre-Algebra**

oops: tan 30° = 1/√3

**Pre-Algebra**

since the sides are in the ratio 1:√3:2 a,b) the sides are 41:41√3:82 c) as we all know (or should, because the come up repeatedly): sin 30° = 1/2 cos 30° = √3/2 tan 30° = √3 And since the co-functions are the functions of the complementary ...

**math**

I assume a typo, and that g(x) = -(x-1)^2 + 3 if -2 < x <= 2 just look at the data: since -2 < 1 <= 2, g(1) = -(1-1)^2 + 3 = 0+3 = 3 since -2 < 2 <= 2, g(2) = -(2-1)^2 + 3 = -1+3 = 2 since 5>2, g(5) = 3 Don't let all the symbols confuse you. Each just ...

**Math**

(8-4x)/(x+5) > 1-x If x+5 > 0, 8-4x > (x+5)(1-x) 8-4x > -x^2-4x+5 x^2+3 > 0 This is true for all x, but the initial condition was that x > -5. So, that is the domain: x > -5 If x+5 < 0, 8-4x < (x+5)(1-x) 8-4x < -x^2-4x+5 x^2 < √3 But, we ...

**math**

look at the amount of tar: 1.0*6 + .05*120 = x(126) x = 0.095, or 9.5% Or, consider that 5% of 120g = 6g So, now you have 12/126 = .095

**Algebra/Steve**

You're not sure about the zero? I'm not sure about the equation!

**algebra**

just do what you did with 13 and 6: wherever you see an x, substitute in p: f(p) = p-7 Not a numeric answer, because you haven't specified what value p has. f(bozo) = bozo-7 f(40) = 40-7 and so forth.

**math**

if the first term is 456, then the 17th term is 456+7*16 Tn = 456 + 7(n-1)

**programming oop**

sum12 = n1 + n2 diffs3 = sum12-n3 prodd4 = diffs3*n4 quotp5 = prodd4/n5 I think you can write the code to gather n1-n5 and display the values calculated above.

**Algebra**

Think of a number between 1 and 10: n Add 1; n+1 double the result; 2(n+1) = 2n+2 add 3; 2n+2+3 = 2n+5 subtract 4; 2n+1 add 5; 2n+6 halve the result; (2n+6)/2 = n+3 add 6; subtract 7; add 8; subtract 9; n+1 subtract the number you first thought of: 1 Don't see any inductio...

**geometry**

Using only reflections in vertical lines, then horizontal lines, you can get B by reflecting through x = -1/2: (-2,-1) -> (-1,-1) through y = -7/2: (-1,-1) -> (-1,-6) No idea what your operations were. Just move your coordinates an equal distance on the other side of the...

**calculus - unclear**

Man, there's a lot of missing information here. That leads me to conclude that the hammer spins in a counterclockwise direction, then flies in a straight, horizontal line from the circle to the screen. Apparently the screen is a circle of radius 50. Now, your solution of (...

**math**

If the current's speed is x, then 10+x is the downstream speed of the boat 10-x is the upstream speed. the downstream time is 90 minutes (3/2 hours) less than the upstream time. Hence, the given equation. 36/(10+x) = 36/(10-x) - 3/2 36(10-x) = 36(10+x) - 3/2 (10-x)(10+x) n...

**math**

since time = distance/speed, 36/(10+x) = 36/(10-x) - 3/2 x = 2 check: 36/12 = 3 36/8 = 4.5 ok

**math**

if his rowing speed is x and the stream's speed is y, then we have 4/(x+y) = 3/(x-y) 48/(x+y) + 48/(x-y) = 14 x=7 and y=1 check: x-y=6 x+y=8 It is clear these values fit the conditions.

**Math**

assuming you mean integers, 501 - 1499 will both round to 1000 any number can be rounded to the nearest thousand. You will have to be a bit more specific.

**math**

as written, I'd say $189.95 Now, if you mean that each phone costs $189.95, then 100 of them cost $18995.00

**math**

6 is the same as 5 8/8 So, 5 8/8 - 3 5/8 = 2 3/8

**mathematics**

can you see how little context you have provided?

**algebra**

c'mon. You know that 99 = 9*11 So, what might the two consecutive odd integers be? ...

**algebra**

Note that the terms come in repeating groups of 5. 38 = 7*5 + 3 So, S38 = 7(4+8+2+6+2) +4+8+2 = 168

**math**

4/7 * 3/8 = 12/56 or 3/14 Good luck finding that measuring cup!

**Math 11 - Rational Expressions**

the only restrictions are because you cannot divide by zero. So, check the denominator of any rational expression. Wherever it is zero, those values must be omitted from the domain. Otherwise, all real numbers are allowed. So, for #3 above, since 2x^2+5x+2 = (2x+1)(x+2) the va...

**math**

so, just fix the typo and solve -16t^2-32t+320 = 48 t = 3.24

**alegbra**

(3x-1)(x+2)

**Math**

Add up all the numbers: 632 There are 12 numbers, so the mean is 52.666 Now, you want to remove one value so that the remaining 11 have a mean of 56.09. Well, 11*56.09 = 617 So, if you remove one of the 15's, the remaining 11 numbers add up to 617, giving the desired mean.

**Pre-Calculus**

You show that it is true for n=1. Then you show that if it is true for n=k, then it must also be true for n=k+1. That means that the assertion is true for all values of n, starting with n=1, and thus for n=2,3,4,... For instance, a common proof is that the sum of the first n o...

**Math**

to lower the mean, remove the highest scores. To raise the mean, remove the lowest scores.

**finite Math**

17000 * 1.04^5

**physics**

same seat assuming that the windows are closed...

**math**

63 divided by (10-3) + 2 x 4 = 63/7 + 8 = 9+8 = 17 However, (63/7 + 2)x4 = (9+2)x4 = 11x4 = 44 Normally multiplication is done before addition

**math**

8*600 = 4800 radios produced. That leaves 2800 more to be done in 4 days. I think probably you can take it from there.

**Math**

I generally don't think of circles as having ends, so maybe the fence is rectangular or square. However, since you talk about using pi=22/7, I'll assume you want a circular fence, and the diameter is 10 ft. So, C = 10pi = 31.4 ft $440/31.4ft = $14.01/ft

**algebra**

first place is to check your text for some kind of distance formula. Next, get out some graph paper and plot the two points, then draw a line between them. Call them A and B, respectively. Now draw a horizontal line from A and a vertical line from B. Where the two lines inters...

**physics**

if the velocity changes from 25m/s to 0 in 5 seconds, that's an acceleration of -5m/s^2. So, the distance covered till the full stop is 25*5 + 25t + 1/2 (-5)t^2 with t=5, that's 25*5 + 25*5 - 125/2 = 187.5 m

**Math**

when (1/2)^(t/60) = .05 t/60 = log.05/log.5 t = 259.3 days

**chemistry**

HCl + NaOH -> NaCl + H2O each mole of NaOH reacts with one mole of HCl. so, how many moles of NaOH in 40g? How many grams of HCl in that many moles?

**Math**

25x^2 - 30x = -9 25x^2 - 30x + 9 = 0 (5x-3)(5x-3) = 0 ...

**Algebra**

mathematically, |b| ≠ 2 or b^2 ≠ 4

**physics**

on the ground, the weight is mg = 100*9.8 = 980N Now, g=9.8 because F = GMm/r^2 = mg, so g = GM/r^2 If we say that r=6371km, then at 1000km altitude, we will have 9.8 * (6371/7371)^2 = 7.32, so the 100kg mass will weigh only 732N

**Algebra direction how to do it**

As you know, y = 4x-4 is just a line. All the points on the line satisfy the equation. > means "greater than". So, if (2,4) is on the line, (2,5),(2,6),... are above the line, where y > 4x-4. So, the solution to an inequality is a whole region of the plane, not...

**Algebra**

1 - x^2/x = 1-x (1-x^2)/x = 1/x - x (2b-2)/(2b^2-8) = (b-1)/(b^2-4) = (b-1) / (b-2)(b+2) Not sure just what you're after

**math**

since a = bh/2, we have a = 14x/2 = 7x so, 7x <= 112 x <= 16

**College Alegbra**

(a) d(j(x)) = .75(j(x)) = .75(x-5) (b) j(d(x)) = d(x)-5 = .75x - 5 (c) go for it

**algerbra**

the difference between three-fifths and five is equal to the product of x and the sum of y and seven. Seems kinda dumb. Why all the nonsense with 3/5 - 5?

**math**

55+25-15 = ?

**geometry**

time to draw a diagram and review your basic trig functions: DE/10 = tan 25° It also helps to recall the alternate interior angles between parallel lines are equal.

**algebra**

well, you need a 1, so that would be 3/3 Now you need 30 more, using three 3's. Looks like 3^3 + 3 + 3/3

**Math**

a.bc a > 0 b = a+1 c = a+b 1.23, 2.35, 3.47, 4.59 Unless there are further conditions, just pick one.

**math**

Well, the roots are -1/2 and 1, so I hope some of the choices were near those values.

**math**

each of 5 people plays 4 others. That's 5*4 But, that counts each person twice, so you really have 5*4/2 = 10 games

**math**

all correct good work

**algebra**

If you mean 2y/3 + 2y - 6/15 then that is just 2y/3 + 6y/3 - 6/15 = 8y/3 - 2/5 = (40y-6)/15 If that's not it, then use some parentheses and show your own work.

**Math**

well, you have ax^2 = -h ax^2 = h In either case, a and h must have the same sign, since x^2 is always positive. Unless x=0, in which case, h=0. x = ±√|h/a| Naturally, if a=0, there is no solution unless h is also zero.

**math**

looks like you need to review your basic triangles: sin 60°/cos^2 45° - cot 30° + 15cos 90° √3/2 / 1/2 - √3 + 15*0 √3 - √3 0

**Math**

You have calculated the number of feet in 26 miles. Now divide that by (32/12), the number of feet in one stride. Or, you can multiply 137280 by 12 to get the number of inches in 26 miles, and then divide that by 32, the number of inches in one stride. Same difference.

**Trigonometry**

secx - √2 = 0 secx = √2 x = π/4 If this gave you trouble, it looks like you need to review your basic trig functions and common triangles.

**algebra**

x+y = 15000 .08x + .07y = 1100 Now just solve for x at 8% and y at 7%

**Physics**

(30,0) + (9.64,-11.50) = (39.64,-11.50) √(39.64^2 + 11.50^2) = 41.27 tanθ = -11.50/39.64

**Math**

750000 * 0.8^3 at the start of the 4th year. 750000 * 0.8^4 at the end of the 4th year you will have to decide when "during" the year you want its value.

**Math**

sinθ = 10/15

**Math**

(1*1 + 5*2 + 3*5 + 1*10 + 1*100) ------------------------------------ (1+5+3+1+1)

**finite math**

(# left field) / (# hits) = 6/59

**math**

Clearly, attendance on M-W is 33,34,35 So, attendance on Th-Sat is 36,37,38 with average=37

**algebra**

think 3-4-5 right triangle. or, do the math: √(50^2 - 30^2)

**Math**

481,924,821,947,217,481,274,832,748,971,298

**math - ouch**

Dang. Don't know what I was thinking.

**math**

what's the trouble? Just start working in the clues. The number is 2.abcdefg... c = a+b b = c-a (well, duh) b = c-1 It appears that the number is just 2.abc c = a+c-1 a = 2c+1 Now, since a < 10, c < 5 So, if a = 0,1,2,3,4 then c = 1,3,5,7,9 and b = 1,2,3,4,5 So, we c...

**Algebra 1**

the slope of the line is positive, so maximum y will occur where the maximum x occurs. So, y <= 5 Since lines extend to infinity, the range is thus (-∞,5]

**math**

The 48 is just noise. The cost (c) of the tickets for p pupils is c = 150p 150*43 = 6,450 So, the figure of 6,350 is incorrect. Whether or not it is wrong for the teacher to say otherwise is a moral question ...

**math**

1/(1+√2+√3) * 1/(1-(√2+√3))/(1-(√2+√3)) = (1-√2-√3)/(1 - (√2+√3)^2) = (1-√2-√3)/(-4-2√6) = (1-√2-√3)/(-2(2+√6)) = (1-√2-√3)/(-2(2+√6)) * (2-√6)/(2-√6) ...

**Math**

yes each value of x produces a distinct value of y.

**Math**

yes no maybe

**math**

If those "x" signs mean multiplication, then you surely know that 0x anything is zero.

**math**

n(n-1)/2 = 171 Since 342 = 18*19, there were 19 teachers.

**physics**

g = GM/r^2 at r+h, g1 = GM/(r+h)^2 at r-x, g2 = GM/(r-x)62 ∆g = GM(1/(r-x)^2 - 1/(r+h)^2) = GM/((r-x)^2(r+h)^2) ((r+h)^2-(r-x)^2) = GM/((r-x)^2(r+h)^2) (2rh+h^2 - 2rx+x^2) Since r-x and r+h are both approximately r, we have ∆g = GM/r^4 (2r(h-x) + h^2+r^2) = 2GM/r^3...

**finite mate**

anythingP0 = 1 There is only 1 way to choose nothing.

**Algebra**

You can factor out a^2 b, giving a^2 b (a^2 + b^2) Not sure that's any simpler.

**Math**

multiply RS top and bottom by cos2x and you have 2/(1-cos2x) = 2/(1-(cos^2 x - sin^2 x)) = 2/(1-cos^2 x + sin^2x) = 2/(2sin^2 x) = 1/sin^2 x = csc^2 x

**Algebra**

you want two factors of 45 which add to 18. Since 45 = 15*3, you get (x+15)(z+3)

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