Wednesday

August 27, 2014

August 27, 2014

Total # Posts: 24,450

**Hard Geometry Problem**

a negative scale factor changes size, and reflects through the dilation center. If it were positive, the whole reduced triangle would of course remain within ABC. But, by flipping through G as well as scaling, some of A'B'C' will lie outside of ABC.
*August 1, 2014*

**algebra**

c = p+37 i = 2p-38 c+p+i = 447 Now just solve for c,p,i using your favorite method.
*August 1, 2014*

**mat/117**

since the discriminant is negative, there are no real solutions.
*August 1, 2014*

**mat/117**

just start substituting for n: a1 = 2*1 = 2 a2 = 2*2 = 4 ...
*August 1, 2014*

**math**

what do we know? c = 2/3 p r = 1/2 c t = 100 = r-90 So, what is p? p = 3/2 c = 3/2 (2r) = 3r = 3(t+90) = 3(190) = 570
*August 1, 2014*

**math**

well, how many cubic yards in a single post? Expressing the dimensions in yards, the volume is (9/36)(9/36)(6/3) yd^3
*August 1, 2014*

**math**

Clearly that would be (5.75+8.2)-(6.2+9.4)
*August 1, 2014*

**math**

That would total up to 0.75 + 0.50 + 0.45 km
*August 1, 2014*

**Logarithms**

If the fraction is a/b c then you'd have +logc If, as I suspect, it is a/(bc) then it is indeed loga - (logb+logc) = loga - logb - logc
*August 1, 2014*

**Chemestry**

just look up element #50 on the periodic table. Protons determine the atomic number. It's short two electrons, so the charge will be +2.
*August 1, 2014*

**first chemistry**

just look up the formula for aspirin (acetylsalicylic acid) and add up the weights of all the atoms involved.
*August 1, 2014*

**chem**

Since the required amounts of the two reagents have almost the same mass, H2S will be the limiting one, since twice as many moles of that are required. So, it will be the limiting reagent. So, figure how many moles of H2S are in 87.1g, and calculate the mass of 3/16 that many ...
*August 1, 2014*

**Geometry**

Since the vertices are 36° apart, and X and R are 4 positions apart, it is indeed a central rotation of 144°. However, we want to rotate about U, not the center of the decagon. If you draw the figure, and if we call the center of the decagon A, then you can see that &...
*July 31, 2014*

**physics**

The horizontal speed does not change. It remains at 7.5 cos51° = 4.72 m/s The vertical speed is given by v = Vo -9.8t, so it ends up at 7.5 sin51° - 9.8*1.5 = -8.87 m/s So, the final speed is √(4.72^2 + 8.87^2) = 10.05 m/s
*July 31, 2014*

**Math**

use the two-point form of a line and you can see that (y-5)/(x+1) = (5+1)/(-1-2) y-5 = -2(x+1) y = -2(x+1)+5 So, y(x+1) = -2((x+1)+1)+5 = -2(x+2)+5 = -2x-4+5 = -2x+1
*July 31, 2014*

**Maths**

I'd say that would be 8*6 = 48
*July 31, 2014*

**Calculus**

Using the comparison test, the series diverges.
*July 31, 2014*

**AP Calc**

let the center of the cable be at (0,220). That becomes the vertex of the parabola, so the equation is y = ax^2+220 Since y(2100) = 750, that means that a*2100^2 + 220 = 750 a = 530/2100^2 = 0.00012 So, we now know that the height y = 0.00012x^2 + 220 Now, we know y' = 0....
*July 31, 2014*

**Translations**

what do you mean by "that's not right?" What calculation did you do? Without loss of generality, we can take the distances to be measured along the x-axis. We have two points, P and Q on the x-axis, where OQ = 4(OP) because of the dilation. Since PQ=18, that ...
*July 31, 2014*

**Translations**

all distances also scale by a factor of 4.
*July 31, 2014*

**Physics**

The trains approach at 260km/hr The distance to cover is 285km. So, it takes 285/260 hrs to meet. In that time, the train from Calgary covers 285/260 * 140 = 153 km
*July 31, 2014*

**geometry help!! answer all questions!! please**

Posting a whole homework assignment with no evidence of any attempt on your part will get you scant help here. Surely you have covered the relevant material, so you must have some ideas on at least some of the problems... The hints I posted earlier should get you started on ...
*July 31, 2014*

**Geometry**

P(x,y) -> P'(x+3,y-3) reflection over y=x is (x,y) -> (y,x) reflection over the y-axis is (x,y) -> (-x,y) reflection over the x-axis is (x,y) -> (x,-y) and so on. Where do you get stuck?
*July 31, 2014*

**Math**

distance is just the integral of velocity. So, just integrate, and evaluate at the ends of the interval.
*July 31, 2014*

**Math**

y = -(x-1)(x-4) so you want to integrate between the roots, in the interval [1,4] a = ∫[1,4] -x^2+5x-4 dx = -1/3 x^3 + 5/2 x^2 - 4x [1,4] Now just evaluate at the ends and subtract.
*July 31, 2014*

**trigonometry**

The problem states: Find the formula of the trigonometric function that models the height H of the weight t seconds after it reached its maximum height. Your function is supposed to model the height starting when it is at its maximum. So, let that be t=0.
*July 31, 2014*

**prealgebra**

you know that for such a triangle, the sides are in the ratio 1:√3:2 So, the short side is 41 The long side is 41√3 The hypotenuse is 41*2 The functions of 30° are standard values, which it would be well to learn.
*July 31, 2014*

**math**

All you want is the number of permutations for 3 out of 4: 4P3 = 4*3*2 = 24 You can pick the first in 4 ways. Now there are 3 choices for the 2nd, and 2 left for the 3rd.
*July 31, 2014*

**math(precalculus)**

You don't need any c. It can be absorbed into a and b. Actually, you have solved the problem. Pick any value for c, and plug it in. As long as a=c and b=-4c, R(x) will work. So, make things easy. Let c=1. As I worked it out, R(x) = ???/(x+1)(x-2) Since y=0 is the asymptote...
*July 31, 2014*

**Math**

heads: r+c = 35 feet: 4r+2c = 94 Now just solve for r and c, in any of several ways.
*July 31, 2014*

**Science/Physics**

weight (force) = mg Look up g if you don't know it. Now, divide that by the relative gravity of the moon. For example, if the moon has 1/2 the gravity of earth, you will need double the mass for the same weight.
*July 31, 2014*

**Physics**

The resultant velocity is v = (106,106) + (0,20) = (106,126) |v| = 164.7 The angle x obeys tan(x) = 126/106 The time taken is 500,000/164.7 seconds
*July 31, 2014*

**Math**

surely you can see that A=3 B=pi/20 C=0 D=7 I mean, just read it off. Now employ your skills to convert your cos(x) to sin(pi/2-x) A and D stay the same.
*July 30, 2014*

**Math**

since y is a max when t=0, we want a cosine function, since cos(0) = 1. y = 7+3cos(pi/20 t) because the period is 40 seconds, and the level varies between 7+3 and 7-3 feet.. C=0 because cosine is a max at t=0, so there is no offset. Naturally, you can now write the sine ...
*July 30, 2014*

**Alegbra1**

depends. How much of all that is under the radical sign? Once you specify that, just remember that the domain of √u is u>=0.
*July 30, 2014*

**Math**

20% + 15% = 35% given away. That leaves 65%, so .65x = 520 Now just solve for x.
*July 30, 2014*

**math - huh?**

I see several problems. who the heck is Mary? Is Mary really male?
*July 30, 2014*

**math**

f+l = 80 2/3 f = l-10 so, 2/3(80-l) = l - 10 160-2l = 3l-30 190 = 5l l = 38 check: initially there were 38 local and 42 foreign after removing 10 local and 14 foreign, there were 28 of each.
*July 30, 2014*

**PreCalculus**

just make the usual substitution. y = rsinθ x = rcosθ so, that means r^2sin^2θ/25 - r^2cos^2θ/9 = 1 r^2(9sin^2θ-25cos^2θ) = 225 r^2 = 225/(9sin^2θ-25cos^2θ) see the expected vertical hyperbola at http://www.wolframalpha.com/input/?i=plot...
*July 30, 2014*

**statistics**

Since there are 10 choices, I'd just use the nth digit of 3 numbers chosen from the random number table. which n depends on how many digits in each entry; just make sure all use the same n. Or maybe add up the digits and take the last digit of the sum.
*July 30, 2014*

**Maths**

1+2+4=7, so if we divide the sum into 7 equal parts of size x, 7x = 126 x = 18 So, the shares are 18,36,72
*July 30, 2014*

**Math**

find the distance between the two points, as usual. Now, since the area of a square is a = d^2/2, just figure that.
*July 30, 2014*

**math**

clearly, if the angle is x, tan(x) = 7.15/10.1
*July 30, 2014*

**Math - grrr.**

all that anti-derivative stuff is too much for me that early. Thanks for the fix. I'm sure Tim caught the error and fixed it though. . . . Right, buddy?
*July 29, 2014*

**Math**

since y" = 6x y' = 6x+a y = 3x^2+ax+b we know the slope is y'(1) = -3, so 6(1) + a = -3 a = -9 so, y = 3x^2-9x+b since 5-3x=2 at x=1, y(1) = 2, and we have 3-9+b = 2 b = 8 and y = 3x^2-9x+8 See the graphs at http://www.wolframalpha.com/input/?i=plot+y%3D3x^2-9x%...
*July 29, 2014*

**PreCalculus**

sin(x)sec(x) - 2sin(x) = 0 sin(x)(sec(x)-2) = 0 sin(x) = 0 sec(x) = 2 Those are pretty standard angles, so listing the solutions should be easy.
*July 29, 2014*

**geometry**

The diagonals cross in the center, so the acute angle is the vertex of an isosceles triangle with height=24 base=14 so the acute angle x obeys tan(x/2) = 7/24
*July 29, 2014*

**Geometry**

If the dimensions are x,y,z then we have 2xy+2xz+2yz = 22 4x+4y+4z = 24 xy+xz+yz = 11 x+y+z = 6 How about 1,2,3 diagonal is √14
*July 29, 2014*

**algebra**

I'd do it by taking the log of the values, and using a normal linear best-fit on the log values. Where log(15500/v) = log4, that's the number of years you want.
*July 29, 2014*

**math**

4 = 2^2 4/(2^x) = (2^2)/(2^x) = 2^(2-x) Now, just solve the equation. Hint: 2^(x-2) = 2^x / 4
*July 29, 2014*

**Trigonometry**

The individual displacements are 150 @ 220° = (-96.4,-114.9) 130 @ 130° = (99.6,-83.6) So, the final location is at (3.2,-198.5) Just find the distance from there to the origin.
*July 29, 2014*

**math**

On the other hand, if you meant 27√3, that makes b=7/2
*July 29, 2014*

**Finite Math**

clearly it's 500-x
*July 29, 2014*

**Math**

just start by listing the facts: width=w length=w+6 height=2 area is the 4 sides + bottom = 2(width*height)+2(length*height)+(length*width) = 2(2w)+2(2(w+6))+w(w+6) = 4w + 4w+24 + w^2+6w = w^2 + 14w + 24 So, now we know that w^2+14w+24 = 280 w^2+14w-256 = 0 Solve that for w, ...
*July 28, 2014*

**math**

after 1/8 were purchased, 7/8 were left. So, now we have 7/8 x + 10 = 52 7/8 x = 42 x = 48
*July 28, 2014*

**math**

a/p = 3/5 (2/5 a)/(p-40) = 2/5 a=60 p=100 so, she paid 60(.30)+100(.45) = $63.00
*July 28, 2014*

**math**

3278164781246
*July 28, 2014*

**math**

9/180 = 0.05 = 5%
*July 28, 2014*

**chemistry**

32 days is 4 half-lives. So, the mass is reduced to 1/2^4 = 1/16 of 40mg, or 2.5mg
*July 28, 2014*

**Math**

so far so good. Now just solve for b and plug that into your formula. Then check that (-3,-8) satisfies the equation.
*July 28, 2014*

**Math**

plot the point. Since the slope is 2/3, go right 3 and up 2 and plot another point. Draw the line connecting the two points.
*July 28, 2014*

**physics**

the difference is 4.9µC. So, move half that, or 2.45µC. Now just convert that to no. electrons.
*July 28, 2014*

**math/help**

4 1/6 = 25/6 1 1/6 = 7/6 so, (25/6)/(7/6) = 25/7 3 3/1 ??? fix that and work it as in the above one.
*July 28, 2014*

**math**

just as usual. Where is y" positive? e^-3t is always positive, so all you need is 2t-3 > 0 t > 3/2 e^3/2 = 4.48, and you can see from the graph that it is concave upward for x > 4.48 http://www.wolframalpha.com/input/?i=plot+x+%3D+e^t%2C+y+%3D+te^-t+for+1%3C%3Dt%...
*July 28, 2014*

**math**

x = e^t y = te^-t dy/dx = y'/x' = (1-t)e^-t/e^t = (1-t)e^-2t d^2/dx^2 = (x'y"-x"y')/x'^3 = ((e^t)(t-2)e^-t - (e^t)(1-t)e^-t)/e^3t = (2t-3)e^-3t or, do it directly x = e^t, so y = lnx/x y' = (1-lnx)/x^2 = (1-t)e^-2t y" = (2lnx-3)/x^3 = (2t...
*July 28, 2014*

**Science**

why do you include a third dimension? The taxi cannot travel vertically. The total displacements are 20+6=26 blocks south 15 blocks east. So, √(15^2+26^2) = √901 = 30.0
*July 28, 2014*

**physics**

15.7 + 17.5
*July 28, 2014*

**math**

1/2 * 1/2 * 1/6
*July 28, 2014*

**Algebra**

Start by expanding the parentheses: 3(x-1) -2(x^2+3x+4)-5(x+9) 3x - 3 - 2x^2 - 6x - 8 - 5x - 45 Then collect terms of like degree: -2x^2 + (3-6-5)x + (-3-8-45) -2x^2 - 8x - 56
*July 27, 2014*

**Math**

If their heights are k and r, then k/375 = sin 32° r/375 = sin 38° So, the difference is r-k = 375(sin38°-sin32°)
*July 27, 2014*

**math**

that's because you mixed up dollars and cents. It asked for only dollars.
*July 27, 2014*

**math**

3:40 to 6:20 is just 2 hours and 40 minutes. That should get you going.
*July 27, 2014*

**Math**

Assuming you mean "I" and not "O", just pick any two values for x and calculate the y: x=0: y=4 x=4: y=5 So, the points (0,4) and (4,5) are on the line. Plot them, and draw a line between the two points.
*July 27, 2014*

**math**

There are only 3 colors. Any group of more than 3 cards will have to have a duplicate color.
*July 27, 2014*

**Statistics**

0.066-0.033 < p < 0.066+0.033
*July 27, 2014*

**english**

That "ing" is a big clue for a gerund.
*July 27, 2014*

**chemistry**

when you say it was immersed, that implies there was some other liquid in the cylinder. If all you mean is that the sodium exactly filled the cylinder, then the density is obviously 405.5g/32.5ml = 405.5/32.5 g/ml
*July 27, 2014*

**Chemisrt**

What you have is just D = D0*x where x is all that exponential stuff. So, D0 = D/x Of course, R is not specified. If it's R you are looking for, then just work it out like this: exp[-Ea/R*T] = D/D0 -Ea/R*T = ln(D/D0) R = -Ea*T/ln(D/D0)
*July 27, 2014*

**maths**

if you mark off a length of 4, you can easily construct a length of 30. Double the 4 to 8, double the 8 to 16, double the 16 to 32. Halve the 4 to get 2, and subtract that from the 32. Now construct the leg BC, and the right angle at C. Then mark off a circle of radius 30 to ...
*July 27, 2014*

**algebra**

take a look at http://www.jiskha.com/display.cgi?id=1406381756
*July 27, 2014*

**math**

just plug in the values and see. For example, take (1,-2). Is 3*1 + 3*1 - 6 < -2 ? 0 < -2 ? No. Now try the others.
*July 27, 2014*

**math**

if you can recognize a quadratic and an inequality (as opposed to an equation), you should have no trouble picking the correct choice.
*July 27, 2014*

**math**

2(w+1.2w) = 88 4.4w = 88 w = 20 so, area = 20(1.2*20) = 480
*July 27, 2014*

**calculus 2**

dy/dx = y'/x' = (-3sint)/(2cost) = -3/2 tan t d^2y/dx^2 = (x'y"-x"y')/x'^3 = ((2cost)(-3cost) - (-2sint)(-3sint))/(8cos^3 t) = -3/4 sec^3 t Check: y = 3/2 √(4-x^2) y' = -3/2 x/√(4-x^2) y" = -6/(4-x^2)3/2 I'll leave it to ...
*July 26, 2014*

**calculus 2**

I think she means y = ∛x
*July 26, 2014*

**Math**

Correct. So, 5/11 - 7/11 = -2/11
*July 26, 2014*

**Math**

you're working with 11ths. So, what's 5-7?
*July 26, 2014*

**math**

s = ∫[0,6π] √((dx/dt)^2 + (dy/dt)^2) dt = ∫[0,6π] √((1-2cost)^2 + (2sint)^2) dt = ∫[0,6π] √(1-4cost+4cos^2t + 4sin^2t) dt = ∫[0,6π] √(5-4cost) dt
*July 26, 2014*

**trigonometry**

15rev/12sec = 5/4 rev/sec = 5π/2 rad/s That's the speed of the smaller pulley Assuming they are both attached to the same belt, the larger pulley's rotational speed is 6/20 the speed of the smaller one.
*July 26, 2014*

**Math - trig**

no, no. It's just a quadratic equation, in cosx, rather than just x using the quadratic formula, we have cosx = (1±√3)/4 or, -0.183, 0.683 So, just find the angles with those values as cosine.
*July 26, 2014*

**math**

the price, for n >= 150, is p(n) = 200-n Revenue is price*quantity, so r(n) = n(200-n) = 200n-n^2 That's just a parabola. The vertex is the point of maximum revenue. That's just algebra I, so no sweat. . . . right?
*July 26, 2014*

**math**

Looks like (a) to me. Just use the quadratic formula.
*July 26, 2014*

**trig**

draw a right triangle with the tops of the poles as the ends of the hypotenuse, and the bottom side perpendicular to the poles. You have a 5-12-13 triangle. It is useful to learn some of the basic triangles with integer sides -- they pop up frequently. 3-4-5 5-12-13 8-15-17 7-...
*July 26, 2014*

**College Algebra**

All you need is f(-1)*g(-1) = 4 f(x) = 2 g(x) = 2 will work, but they are rather trivial. I'm sure you can come up with two functions that work.
*July 25, 2014*

**Algebra**

Clearly E cannot exceed 5, since even with borrowing, A cannot exceed 10. That would mean something like FL50 -B55 ------- MO95
*July 25, 2014*

**Algebra**

1/3 of the mice are female, so 1/6 of the mice are white female 1/6 of the mice are brown female 3/5 are brown, so 3/5-1/6 = 13/30 are white male 1-(1/3+13/30) = 7/30 are brown male So, brownmale/whitemale = 7/13
*July 25, 2014*

**MATH WORD PROBLEMS**

to be centered, the extra space must be divided into two equal parts.
*July 25, 2014*

**College Algebra**

q(x) = 73(x+1)^2+2 h(x) = x^2 g(x) = (x-1)(x-2)(x-3) Since x^0=1 for any x, we have s(x) = 2*a^(kx) for some a and k. 2*1/2 = 1, so s(x) = 2*4^(-x/2) r(x) will involve something/(x-3) x-2 is zero at x=2, so r(x) = (x-2)/(x-3) will work
*July 25, 2014*

**College Algebra**

reflected: -|x| stretch vertically: -4|x| shifted: -4|x+3| see http://www.wolframalpha.com/input/?i=plot+y%3D|x|%2C+y%3D-4|x%2B3| You want t where 1(1+.075/4)^(4t) = 7200 The rule of 72 indicates it should double about every 10 years. Since 7200 is about 2^13, I expect it will...
*July 25, 2014*

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