Saturday

May 23, 2015

May 23, 2015

Total # Posts: 31,343

**calc**

f = 2x+15-19e^x f' = 2-19e^x f'(0) = 2-19 = -17 so, the tangent line is y+4 = -17(x-0) see the graphs at http://www.wolframalpha.com/input/?i=plot+y%3D2x%2B15-19e%5Ex%2C+y%3D-17x-4
*April 6, 2015*

**calc**

x = e^(8-p) revenue is x*p = pe^(8-p) profit = revenue-cost = xp-x = x(p-1) = (p-1)e^(8-p) so, if y is the profit at price p, dy/dp = (2-p)e^(8-p) set the price at p=2
*April 6, 2015*

**chemistry**

if you mean g/L, that would be 1mole/22.4L = 0.759 g/L
*April 6, 2015*

**Calculus**

well, geez, just plug in e^2 into f'(x) f'(e^2) = 4(ln(e^2))^3/e^2 = 4(2)^3/e^2 = 32/e^2 remember ln(e^u) = u e^(ln u) = u
*April 6, 2015*

**algebra**

what is that 4 doing there? is 4√y supposed to mean 4th root of y? x^(11/10) is 10th-root(x^11)
*April 6, 2015*

**math**

man, look up your point-slope formula: y-5 = 2/3 (x+1)
*April 6, 2015*

**geometry**

transitive: A≅B and B≅C ==> A≅C No idea what the diagram shows.
*April 6, 2015*

**geometry**

M=(A+B)/2 clearly AM=MB = 1/2 AB Look up your distance formula to figure length of AB.
*April 6, 2015*

**Calculus 2 (Series)**

see the solution here: http://www.math.ucla.edu/~hendricks/CalculusIISummer2011/calc2practicefinalsolutions.pdf
*April 6, 2015*

**math**

5f+4s = 14.80 (3/4)f = (3/5)s s = 1.85
*April 6, 2015*

**maths**

before: TTT after: TTT TTT so, how many T's?
*April 6, 2015*

**Differential Equation 4**

no ideas on these problems? Homework dumps are discouraged.
*April 6, 2015*

**Differential Equation 3**

why the stupid carets? These are not exponents. (x^2+1)y"+2xy'=0 Since y = c1 + c2 arctan(x) I suspect you can now come up with the series.
*April 6, 2015*

**algebra**

that copy/paste stuff will kill ya every time!
*April 6, 2015*

**calculus**

v(t) = ds/dt = 128-32t So, plug in t=4.
*April 6, 2015*

**math**

for m months, the cost c is c = 127 + 28m now plug in m=9 for the cost of 9 months. actually, c is the cost of joining and then using the club. The question is poorly worded.
*April 6, 2015*

**Precal**

so, if u=3^x, you have u^2 - 4u + 3 = 0 You can handle that, I presume?
*April 6, 2015*

**calculus**

wrong h = 130 + 35t - 16t^2
*April 6, 2015*

**ET 1210 FUNDAMENTALS**

E/R -> (2E)/R = 2E/R
*April 6, 2015*

**ET1210 DC/AC FUNDAMENTALS**

I=E/R so, the new I will be E/(R/2) = 2E/R
*April 6, 2015*

**math**

since the speed is the same, 7/60 = x/24
*April 6, 2015*

**Physics**

well, what was the date 250 years ago? A light-year is the distance light travels in a year. So, it took the light 250 years to get to earth.
*April 6, 2015*

**Precal**

Dan and Reiny just did one of these for you. Just make sure everything is expressed as powers of 3, then equate the powers to solve for x.
*April 5, 2015*

**physics**

momentum is mv surely you can figure the starting and ending values. The time taken makes no difference.
*April 5, 2015*

**math**

the slope is the constant of variation. In this case, 11. This is just a special case of the slope-intercept form y = mx+b where b=0.
*April 5, 2015*

**PHYSICS**

work = force * distance so, how far does it travel in 1 second? F=ma, so a=F/m s = 1/2 at^2 = 1/2 (F/m) t^2
*April 5, 2015*

**Precal**

Since 25 > 21, the ellipse is long on the y-axis. So, the vertices are (0,5) and (0,-5) You should have immediately seen your answers were incompatible. The foci must lie on the major axis. Your foci are correct.
*April 5, 2015*

**Math**

12000(1+.08/4)^(4*4) = 16473.43 900*(1.02^16+1.02^15+...+1.02^1) = 900(1.02^16 - 1)/(1.02-1) = 16775.36 Have I missed something? My answers are about 1/2 the expected values.
*April 5, 2015*

**AP CALC**

#1. solve for t in 0.985^t = 0.8 #2 dy/dx = 3√x √y dy/√y = 3√x dx 2√y = 2x^(3/2) + c at (1,4) we have 2√4 = 2*1^(3/2)+c 4 = 2+c c = 2 2√y = 2x^(3/2)+2 at x=0, 2√y = 2 y = 1 check: y = (x^(3/2)+1)^2 y' = 2(x^(3/2)+1)(3/2 &#...
*April 5, 2015*

**trig**

since sin^2 + cos^2 = 1, cos(23) = √(1-p^2) cos(113) = cos(23+90) = cos(23)cos(90) - sin(23)sin(90) = -p sin(46) = 2sin(23)cos(23) = 2p√(1-p^2) Time to review your basic sum/double-angle formulas
*April 5, 2015*

**maths**

sin(2x) = 1/2 tan(x) 2sinx cosx = 1/2 sinx/cosx 4sinx cosx = sinx/cosx Clearly, sinx=0 is one intersection Elsewhere, we can divide through by sinx, and we have 4cos^2x = 1 cosx = ±1/2 Now you can easily find the points you want.
*April 5, 2015*

**Math**

what? You can't take a number, multiply it by -2 and subtract 4? y(0) = -2*0-4 = -4 Clearly, only C and D are possible answers. Now find y(1) the same way and you should get the correct choice.
*April 5, 2015*

**algebra**

If the sons are age a and b, with a<b, f+a+b = 50 f-4 = 8(b-4) f+4 = 4(a+4)
*April 5, 2015*

**Geometry help**

if the perimeter is p and the apothem is a, then the area is 1/2 ap #1. correct: 12/√2 #2. hint: the triangle's altitude is 18. #3. side is 12
*April 4, 2015*

**Differential Equation 1**

well, the solution of the homogeneous equations is y = c1 cos(2lnx) + c2 sin(2lnx) + c3/x^2 Not sure what to do with the t's, though.
*April 4, 2015*

**Physical Science**

(12/120) * 500
*April 4, 2015*

**Physics**

I think you might get a clearer picture of the situation by reading the discussion here: http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/earthole.html
*April 4, 2015*

**math - incomplete**

42
*April 4, 2015*

**Consumer Math**

correct
*April 3, 2015*

**Algebra 2**

pick any value for x, say 20 Then 2*20+y = 8 40+y = 8 y = -32 So, (20,-32) is one pair Or, pick any value for y, say 0 Then 2x+0=8 x=4 So, (4,0) is another pair
*April 3, 2015*

**Math**

if the circumference is 1/40 as big, then so is the diameter.
*April 3, 2015*

**Physics**

If the amplitude is A, then it goes from -A to +A, or a distance of 2A. But during one period, it goes up and back again, for a distance of 4A.
*April 3, 2015*

**geometry**

or, you can use a compound inequality. 34-22 < x < 34+22 12 < x < 56
*April 3, 2015*

**Geometry**

correct
*April 3, 2015*

**Geometry**

A pilot sees a guy at an angle of depression of 50 degrees. If the pilot is 200 degrees from the guy, how high is the pilot? Would it be sin(50) x 200?
*April 3, 2015*

**pre calculus**

oops (b) [100*20 + x(20-5)]/(100+x)
*April 3, 2015*

**pre calculus**

(a) ?? unless the pricing varies somehow the cost for x skiers will always be the same (b) [100*20 + x(20-5)]/x ...
*April 3, 2015*

**PHYSICS**

If the parts are x and M-x, then F = Gx(M-x)/r^2 = G/r^2 (Mx-x^2) This is just a parabola, with vertex at M/2. That is, the force is maximum when the masses are equal. It's like the area of a rectangle with constant perimeter is greatest if it is a square.
*April 3, 2015*

**Math**

if x sell at $40, then the rest (6000-x) are $26, so 40x + 26(6000-x) = 190500
*April 3, 2015*

**Geometry**

the slope of AB is -5/3 so, which if the choices has that same slope?
*April 3, 2015*

**math**

try using a common denominator again. Then pick the smallest one.
*April 3, 2015*

**m ath**

2^-2 = 1/2^2 = 1/4
*April 3, 2015*

**math**

rewrite them all with denominator of 8: 4/8 5/8 7/8 6/6 now just sort them by numerator.
*April 3, 2015*

**Calculus**

if xi is the right endpoint, then 3 cannot be x1. That leaves only c and d for choices. Since the interval length is 2, and the number of divisions is 4, each interval is .5 in width. Looks like (c) to me. BTW, (d) is the set of midpoints.
*April 3, 2015*

**Math**

thank you
*April 3, 2015*

**Math**

If I know all 3 angles of a right triangle and the hypotenuse how do I figure out the short side length of the triangle?
*April 3, 2015*

**6th Grade Math ASAP!!!**

you're right -- it is complex. Especially so since you don't seem to have any questions about it.
*April 3, 2015*

**Math**

rotation through 180° changes both signs. So, (3,-5) -> (-3,5) Did you actually plot your solution? You would see that it was nowhere near the rotated point. an equilateral triangle has axial and rotational symmetry.
*April 3, 2015*

**programming**

for i=1 to 100, read a(i) read n for i=1 to 100, {if a(i)==n, print i} n = a(1) for i=2 to 100, if a(i)>n, n=a(i) print n See whether you can use the above section to repeatedly find the smallest element and place it in the next position. That is, sort the array.
*April 3, 2015*

**Algebra**

I don' see no steenking inequalities! For the equations, 6y = 9(x-6) 3(2y+5x) = -6 6y = 9x-54 6y+15x = -6 so, since we have 6y in both equations, set them equal: 9x-54 = -6-15x 24x = 48 x = 2 now you can find y.
*April 3, 2015*

**6th Grade Math Help**

excuse me? Every parallelogram is a regular parallelogram. They all have two sets of parallel sides. Pick any side and call it a base. The altitude is the perpendicular distance between that base and the parallel side.
*April 3, 2015*

**6th Grade Math Help**

the slanted side cannot be less than the height. That's like saying the hypotenuse of a right triangle is shorter than one of the legs. At any rate, if you can figure out what the height is (the perpendicular distance between two of the parallel sides), then the area is ...
*April 3, 2015*

**algebra**

you can go the e^kt route, but it's more transparent to use base 2, since we're talking about half-lives. 2^-1 = 1/2, so A = Ao*2^(-t/10) after 5 years, you have A = 20*2^(-5/10) = 14.14g Of course, since 2 = e^ln2, that means you could use (e^ln2)^(-t/10) = e^(-t*ln2/...
*April 3, 2015*

**math**

you are correct 62kg/ft^3 / 8.3kg/gal = 7.46gal/ft^3
*April 3, 2015*

**math**

well, xy=k, so 12*6 = k
*April 2, 2015*

**Math 222**

I assume you are trying to find excluded x values. If so, you are correct. You know, we like to try and help find answers. Don't make us come up with the questions, too!
*April 2, 2015*

**Math 222**

because I have been exposed to similar careless mistakes displayed by others. Tell me. Just looking at your posting, could you easily tell what was being asked? The questions and solutions are all mixed up together.
*April 2, 2015*

**Math 222**

Leaving aside the virtually impenetrable formatting, your math appears to be ok.
*April 2, 2015*

**forth**

you really had a hard time coming up with 27? Did you try just listing the numbers between 22 and 30 and looking at them?
*April 2, 2015*

**algebra**

LCD(20,30) = 10, so the desired LCD is 1/10 x^3y^4 Nothing larger can be factored from both items.
*April 2, 2015*

**math**

is the swap simultaneous, or did Rich give her his 1/6 after he received her 1/4?
*April 2, 2015*

**Steve/Ms.Sue**

since the triangle is equilateral, the radius of the circle is 18. so, your area is 5/6 of a circle plus the area of the triangle, which is 18^2/4 √3 = 81√3
*April 2, 2015*

**Algebra 1**

7-24-25 is one of the standard integer-sided right triangles. Use the Pythagorean Theorem to verify that. Just FYI, it's often helpful to remember the others: 3-4-5 5-12-13 8-15-17 7-24-25 9-40-41
*April 2, 2015*

**Algebra 1**

A right triangle's legs are 7 inches and 24 inches. What is the measure of the angle opposite the 24 in leg?
*April 2, 2015*

**Math**

well, 3 lbs is 48 oz. ...
*April 2, 2015*

**math**

a=2 http://www.wolframalpha.com/input/?i=solve+%7B%7B1%2C3%2C2%7D%2C%7B2%2C-1%2C-3%7D%2C%7B5%2C2%2C1%7D%7D*%7B%7Ba%7D%2C%7Bb%7D%2C%7Bc%7D%7D%3D%7B%7B3%7D%2C%7B-8%7D%2C%7B9%7D%7D
*April 2, 2015*

**Algebra 1**

how can you possibly not be sure? Look at your calculation! If the side you want is x, then x/10 = sin 40°
*April 2, 2015*

**Algebra 1**

A right triangle has a 40 degree angle. The hypotenuse is 10cm long. What is the length of the side opposite the 40 degree angle? I got 6.4cm but not sure of how I got it.
*April 2, 2015*

**Precalc**

(a) when 5t + 17t = 6*5280 (b) when 5t = √(5000^2+(6*5280-17t)^2)
*April 2, 2015*

**trig**

well, you could start out with a little algebra: cos^4-sin^4 = (cos^2-sin^2)(cos^2+sin^2) so you have cos^2-sin^2 ...
*April 2, 2015*

**MATH**

since the triangles are similar, the angles are all the same as in ABC.
*April 2, 2015*

**Calculus**

37y^36 y' = 8x^7 y' = 8/37 x^7/y^36 Now, since y^37 = x^8, y^36 = (x^8)^(36/37) y' = 8/37 x^7 x^(-288/37) = 8/37 x^(-29/37) Doing it explicitly, using the power rule, y = 8/37 x^(8/37 - 1) = 8/37 x^(-29/37) Kind of a dumb problem to use to illustrate implicit ...
*April 2, 2015*

**math**

2l + 2πr = 660 a = 2lr + πr^2 = 2(660-2πr)/2 * r + πr^2 = r(660-2πr) + πr^2 = 660r - πr^2 Now just set da/dr=0 to find r, then l.
*April 1, 2015*

**calculus**

To answer the 1st question, just plug in t=2 and solve for k: 68+30.6e^(-2k) = 96.6 Now just use the amended formula for the rest. If you get stuck, come on back and show us whatcha got so far.
*April 1, 2015*

**Geometry**

since PQR = 60°, PQX = 30° So triangle PQX is an isosceles triangle with base angles of 30° and base=12 Thus, its altitude is 6/√3 and its area = 36/√3 = 12√3
*April 1, 2015*

**Algebra2 Help please**

There are 10 scores. The 90th percentile is the score with 9 below it. Similarly, the 30th percentile is the score with 3 below it. Did you check the related questions below? Since the 3rd score is 105 and the 4th score is 113, I guess any score x where 105 <= x < 113 ...
*April 1, 2015*

**geometry**

√(250^2+200^2) = 50√(5^2+4^2) = 50√41 = 320.2
*April 1, 2015*

**Jonase the moocher**

c'mon. just sub in 3x^2 + 4x(2-x) + (2-x)^2 = 6 Now just collect stuff and solve the quadratic equation.
*April 1, 2015*

**Mathematics - garbled**

Want to fill in the missing words?
*April 1, 2015*

**Sciene**

looks like you need to gather your materials and stop complaining.
*April 1, 2015*

**Math**

(x,y) -> (x-4,y+1)
*April 1, 2015*

**Exponential Equations**

Just solve for t in 2500(1+.0275)^t = 3500 1.0275^t = 1.4 t log1.0275 = log1.4 t = log1.4/log1.0275 = 12.4
*April 1, 2015*

**Maths**

see the previous solution in the related question below.
*April 1, 2015*

**MATH**

can you say homework dump? I knew you could. Still, here's a freebie. Show some effort on the rest, ok? v = (48-16)(36-16)(8) = 5120 m^3 a = (48-16)(36-16) + 2((48-16)(8) + (36-16)(8)) = 1472 m^2
*April 1, 2015*

**Algebra 2**

first off, it's theta, not treta! cot(θ-π/2) = (1+cot(θ)cot(π/2))/(cot(π/2)-cot(θ)) = -1/cotθ = tanθ
*April 1, 2015*

**Math**

looks good to me
*April 1, 2015*

**algebra**

avg velocity is (total distance)/(total time) so, that would be (s(4)-s(2))/(4-2)
*April 1, 2015*

**precalculus help**

z = 2cis(2π/3) ∛z = ∛2 cis(2π/9)
*March 31, 2015*

**precalculus**

sin2θ = 2sinθcosθ sin(θ+π/2) = cosθ cos(θ+π/2) = -sinθ play around with that some, and verify your algebra with the graph: http://www.wolframalpha.com/input/?i=plot+r%3D2sin2%CE%B8
*March 31, 2015*