Thursday

April 24, 2014

April 24, 2014

Total # Posts: 21,976

**Math**

-√60 = -√(4*15) = -√4√15 = -2√15

**Algebra 2**

there are 7 zeros in the first number, so it is 24*10^7 there are 4 leading zeros in the next number, so it is .24*10^-4 the denominator is .5*10^-2 Expressing each of those in scientific notation, where the number is between1 and 10, we have 2.4*10^8 * 2.4*10^-5 / 5.0*10^-3 =...

**math**

1/x = 1/3 + 1/6 x = 2 So, (B) Extra credit: how long does it take if all three work together?

**math**

13.3/3.2 * 72 = 299.25

**math**

Assuming the usual carelessness with parentheses, I read this as (10t-640)/t^2 = 0 a fraction is zero if the numerator is zero. So, 10t-640=0 t = 64 Now, if you meant 10t - (640/t^2) = 0 then clear the fraction and we have 10t^3 - 640 = 0 t^3 = 64 t = 4

**algebra 2**

The pay on the nth day is 2^(n-1) So,we want 2^(n-1) = 10000 n-1 = log10000/log2 n = 14.28 So, on the 15th day the pay is 2^14 = $163.84

**algebra 2/trig**

1/x + 1/8 = 1/3 x = 24/5

**Algebra**

5*x*x*(3x+5) divide up the factors any way you choose

**algebra 2**

256 + 2*256*.75 + ... + 2*256*.75^4 = 2*256(1-.75^5)/(1-.75) - 256 = 1306

**algebra 2**

In the nth year he makes 32000+300(n-1) so, when does that exceed 34700? 32000+300(n-1) > 34700 300(n-1) > 2700 n-1 > 9 n > 10

**GEOMETRY**

volume per cylinder: pi*9*8 = 226.19 2000/226.19 = 8.84

**algebra 2**

sigma upper limit 28 algebraic expression (8i-13) lower limit 1 index=i 28 ∑ 8i-13 i=1 You can see from the formula that the sum increases by 8 with each additional term (8i) The first term is 8*1-13 = -5 Now recall that the sum of the 1st n terms of an arithmetic sequen...

**Calculus - amendment**

just recall that the period of tan(t) is pi, not 2pi, so tan(2t) has period pi/2

**math**

if there were n pieces, you know that 1/6 n = 2 n = 12

**Math**

inequality sign is garbled. But P(n≠3) = 5/6

**Calculus**

no ideas? Try reading the theorem.

**math**

50 in^2 * (2.54cm/in)^2 = 322.58 cm^2 Note how the units cancel, leaving the desired units at the end.

**math**

since 832 >= 400, 832/400 > 1, and the 1st digit is in the hundreds position 217 < 700, but 217 >= 70, so 1st digit of quotient is in the tens position.

**Math**

both. vertical angles are congruent.

**precal**

Hi I know is is probably basic but can anyone assist with the following : Find all real Roots, expressing as whole # and # under radical sign, not as fraction. and factor completely f(x)= X^3-2x^2-15x+30 I got roots at 2 (x-2) and +/- √5 If someone solves and gets factor...

**Calculus Help Please!!!**

a I assume you can sketch graphs b yes f(0) = 0 and limit=0 from both sides c yes f'(0) = 1 and limit of f' = 1 from both sides

**algebra**

Im using this for trig functions ....so I cannot use fraction or decimal I have to simply use π. Im getting : π-3/2 (2)π-3/2 ***I don't know how to add 2π +(-3)? thank you for helping

**algebra**

No I mean how do you add them using common denominator? not using decimal or calculator

**algebra**

How do you add π -3/2?

**ovarian and uterine cycles**

Im sure this is in the book on this section.. but i think (not sure) FSH is 1.

**Gr 10 math - finding vertices**

whatever you do for one line, do for the other. If you have two points, they define a line.

**algebra**

can someone show line by line step by step factor x^2-15

**Precalculus /Algebra**

Hi I know is is probably basic but can anyone assist with the following : Find all real Roots, expressing as whole # and # under radical sign, not as fraction. and factor completely f(x)= X^3-2x^2-15x+30 I got roots at 2 (x-2) and +/- √5 If someone solves and gets factor...

**math**

Jamal writes the letters K-A-N-S-A-S on cards and then places the cards in a hat. What are the odds against picking an S?

**alg.**

recall that in the general quadratic, max y occurs at x = -b/2a. In this case, when t = 17/32. Just plug that in to get maximum h.

**college alg.**

recall that for the general quadratic, max y is at x = -b/2a. In this case, at t=7.1/9.8 Just plug that in to get s.

**math multiple choice**

13.75 + 55000(.60/200) = 178.75

**Calculus - oops**

oops --- y' = -y, so the tangent line is y-13 = -13(x-1) y = -13x+26 as shown at http://www.wolframalpha.com/input/?i=plot+y+%3D+e^%28x%2Barccos%28x-1%29%29%2Fx%2C+y%3D-13x%2B26

**Calculus Help Please**

when x=1, ln(y) - pi/2 = 1 y = e^(1+pi/2) at x=1, y' = -y so, now you have a point and a slope. y-e^(1+pi/2) = -e^(1+pi/2) (x-1) and convert that to slope-intercept.

**Calculus Help Please**

ln(xy) - cos^-1(x-1) = x 1/(xy) (y+xy') + 1/√(1-(x-1)^2) = 1 Now just solve for y'. You should get y' = y(1 - 1/x - 1/√(2x-x^2))

**Calculus Help Please!!!**

the normal line is perpendicular to the tangent line. Since the slope of the tangent line at x is y', find y'. 2x + 4y + 4xy' + 2yy' = 0 y' = -(2x+4y)/(4x+2y) = -(x+2y)/(2x+y) So, at (2,1) the tangent line has slope -4/5 Thus, the normal line has slope 5/4....

**Math**

D

**college alg**

well, just plug in your numbers and solve for t when the height is zero: -16t^2 + 184t = 0 just a good old quadratic equation.

**Math**

#2 f? x? I get 5f^2 + 11f - 2 #3 B #4 coefficients are the numbers multiplying the powers of x. So, B the final -3 is not a coefficient; it's just a constant.

**Math**

1A what did you get for the others?

**physics**

the man changed his velocity by 6.59 m/s in 3.92 ms, so his deceleration was a = 6.59m/s / 0.00659s = 1000m/s^2 and, since F = ma, the force should now be obvious.

**maths Pls help trig**

but oops: 25/√50 = 5/√2, not 1/√2

**maths Pls help trig**

the denominator is 50√50

**PROGRAMMING**

you can probably adapt this little perl ditty to your needs. Other languages may be more verbose, however. my %authors,%ages,%counts; while (<>) { last unless $_; my ($count,$age,$name) = split(/\s+/,$_,3); $authors{$name} = [$count,$age]; $ages{$age}++; $counts{$count}+...

**Math**

well, there's algebraic algebraically those are the only words I can think of right off where I can "form the word algebra". Git

**Math**

too late for your spelling and punctuation, apparently...

**Math**

well, it's clear that if there is 1 bus, the numbers work out: 40+16 = 36+20 If there are more buses, then since LCM(40,36) = 360, any multiple of 360 + 56 will also work, with an extra 9 buses if 40/bus, and an extra 10 buses if 36/bus, for each extra 360 kids.

**Calculus Help Please!!!**

just use the quotient rule: y" = (2xy^2 - 2x^2y y')/y^4 = 2x(y-xy')/y^3 = 2x(y-x(x^2/y^2))/y^3 = 2x(y^3-x^3)/y^5 or, use implicit differentiation twice: x^3-y^3 = 3 3x^2 - 3y^2 y' = 0 6x - 6y(y')^2 - 3y^2y" = 0 y^2 y" = 2x - 2y(y')^2 = 2(x -y...

**math**

x + x+30 = 90 now solve for x

**pre calculus**

y = 12-√x looks like a function to me, though the domain is restricted.

**Math (help)**

don't know what steps you want. The area of a cone is a = 2πr(r+s) just plug in the values for r and s

**Math**

correct since the legs are hypotenuse/√2, √6/√2 = √3 check: √3^2 + √3^2 = √6^2 3 + 3 = 6

**Calculus Help Please!!! URGENT!!!!**

A(t) = Ao * 2^(-t/5700) If you want that to use e instead of 2, then since 2 = e^ln2, A(t) = Ao * e^(-ln2/5700 * t) and k = ln2/5700. Now just plug in your numbers to answer the other questions.

**calculas**

if the box is all cardboard, and AB = 193 and h=2, then something doesn't add up, since the surface area is 4(A+B)+2AB

**math**

room: Vr = 11.35*9.5*8.3 = 894.9 box: Vb = 4.25*6.5*4.0 = 110.5 now, subtract and round.

**Calculus Help**

y = sin(π/√(θ^2+5)) weird, but just apply the rules y = sun(u) y' = cos(u) u' u = π/v u' = -π/v^2 v' v = √(θ^2+5) v' = θ/√(θ^2+5) put it all together and it spells MOTHER. Or, more mathematically, π&...

**Calculus Help Please!!!**

hmmm. well, I guess you can't use Taylor series either, eh? (sin2x)/(2x-tanx) I assume you've shown that lim (sinx)/x = 1, so if our limit is L, 1/L = (2x-tanx)/(sin2x) = (2x)/sin(2x) - tanx/sin2x = 1 - (sinx/cosx)/(2sinx cosx) = 1 - (1/cosx)/(2cosx) limit as x->0 i...

**Calculus Help Please!!!**

The key here is l'hospital's rule lim f/g = lim f'/g' so, the limit is the same as 2cos2x / (2 - sec^2 x) = 2*1/(2-1) = 2

**Math**

1n+7p = 5.55 3n+2p = 12.28 that's the same as 2n+14p = 11.10 21n+14p = 85.96 now subtract to get rid of the p's 19n = 74.86 n = 3.94

**Geometry**

Note that this is just a trapezoid with bases 3 and 8, and height 8. So, the area is (3+8)/2 * 8 = 44

**Calculus**

ok. If you mean (-1/3 t) - 1 then we have for the two tanks: Va = 48 - ∫16/5 dt = 48 - 16/5 t Vb = 48 - ∫1/3 t + 1 dt = 48 - t - 1/6 t^2 So, Va(5) = 48 - 16 = 32 Vb(5) = 48 - 5 - 25/6 = 38 5/6 Now just solve for when Va and Vb = 0 to find which drains first. If I s...

**Calculus - eh?**

Cannot parse -1/3t-1 Is that (-1/3)(t-1) or -1/(3t)-1 or -1/(3t-1)? none of those choices makes sense at t=0.

**MATH ✿╣Check╠✿**

looks good to me

**Math**

well, we have cosa = 12/13 sinb = 4/5 So, just plug them into cos(a+b) = cosa cosb - sina sinb

**Calc2**

well, we know that L{1} = 1/s and, we know from our handy table of transforms that L{t^n f(t)} = (-1)^n F(n)(s) so, L{t^2} = 2/s^3

**MATH - Algebra**

yes, except you had 22x, not 11x So, the answer is x^2 + 2x also, be sure to use parentheses to make it clear that you mean (11x^2+22x)/11 and not 11x^2 + (22x/11)

**Algebra II -Please help**

*whew* I had no idea how to apply the matrix to your calculations.

**math**

the two legs are a and b, and the hypotenuse is √(a^2+b^2). So, assuming a,b > 0, So, in QII, sin2x = b/√(a^2+b^2) cos2x = -a/√(a^2+b^2) sin^2(x) = (1-cos2x)/2 = (1+a/√(a^2+b^2))/2 = (a+√(a^2+b^2)) / 2√(a^2+b^2) and similarly for cosx

**math**

all those angles are the standard ones you should know. cot pi/3 = 1/√3 csc pi/3 = 2/√3 cos pi/4 = 1/√2 plug it in and you have (1/√3 + 2/√3)/(1/√2) = √6

**Math**

r/h = sin(a) v = 1/3 pi r^2 h = 1/3 pi h^3 sin(a)

**Geometry**

D

**Dance Therapy**

Best first step would be to talk to someone doing what you plan to do, and see what is best to study that is available, and where to pursue further education.

**calculus I**

gotta remember that v(t) = h'(t) = 24.5 - 9.8t now just plug in t=2 or 4

**Calc 2**

the 2nd one is just (3/2)^(k/2) so it of course diverges. so, what do you get for arctan?

**math**

how about .15 * 63.00 = ?

**calculus**

you do know that cos(z) = 0 when z is an odd multiple of pi/2, right? So, cos(2t) = 0 when 2t is an odd multiple of pi/2; that is, t is an odd multiple of pi/4. 45 degrees is an odd multiple of pi/4, right?

**calculus**

you have the product of three factors (2.5)(e^-t)(cos 2t) = 0 2.5 is never zero e^-t is never zero cos 2t = 0 when t = (2k+1)(pi/4) for integer k.

**Discrete Math**

just make your usual Venn diagram and start labeling the intersections.

**Math**

Thinking of the doubling rate, we can see that after t hours, the population is P(t) = 50 * 2^(2t) = 50*4^t So, we want 50*4^t = 1000000 4^t = 20000 t log4 = log 20000 t = log20000/log4 = 7.14 hours

**advanced algebra**

y = 10 - 2x

**physics**

5x10^6 rad/s = 795774 Hz Z = √(R^2 + (Xl-Xc)^2 Xc = 1/ωC = 1/(5*10^6 * 100*10^-12) = 1/(500*10^-6) = 2000 Xl = ωL = 5*10^6 * 100*10^-6 = 500 So, now you know R and Z. Go for it, and recall that I = E/Z

**Math**

since all dimensions are 1.6 times as big, the volume (multiplying all three dimensions) is increased y a factor of 1.6^3. 92 * 1.6^3 = ?

**Calculus II**

actually, Γ(n) = ∫[0,∞] x^(n-1) e^(-x) dx which is unusual, since n is usually used for constants. Usually it is written Γ(t) = ∫[0,∞] x^(t-1) e^(-x) dx As you probably should know, Γ(n) = (n-1)! for integer n. So, Γ(1) = 1 Γ(2)...

**Calculus II**

Using long division, we know that x^4(1-x)^4/(1+x^2) = x^6-4x^5+5x^4-4x^2+4 - 4/(x^2+1) So, the integral is just x^7/7 - 2x^6/3 + x^5 - 4x^3/3 + 4x - 4arctan(x) evaluated at 1 and 0, we have 1/7 - 2/3 + 1 - 4/3 + 4 - π = 22/7 - π now add π to that and you wind u...

**math**

cost is rate * distance. If the distance from home to A is x, then from home to B is 6-x. So, since the costs are the same, 50x = 75(6-x) 50x = 450-75x 125x = 450 x = 3.6 So, (C)

**Math**

there are only 3 colors, so the 4th sock must match one of the first three.

**cape1 physics**

thanks for the example.

**Math/Physics**

when Tyson left at 11:00, Rashad had already gone 200km, so the two were 580 km apart. They approached each other at 180 km/hr, meaning they met in 3 2/9 hours. So, by the time they met at 2:13:20, Tyson had gone 257.8 km.

**Math**

1000000 people/66000 mi^2 = 1000/66 people/mi^2 That's a lot less than any of the proposed densities, so (A) assuming the cited figures actually apply to Florida...

**Math**

correct

**Math**

1*1*C(6,2) = 15 extra credit: why did I multiply by 1 -- twice!? also, proofread before posting. I assume you meant the same topping can't be used twice for 1 pizza.

**geometry**

Draw the picture r^2 = 5^2 + 21^2

**geometrey**

the locus of points equidistant from a point called the center.

**math**

x + x+10 = 90 just solve for x

**math**

add up the amount of copper in each part. It must total the amount of copper in the final alloy .10*25 + .18x = .13(25+x) now just solve for x

**math**

how many ways can you choose 3 dishes from a menu of 8? my answer is 56

**Math**

√(x+2) = x x+2 = x^2 x^2-x-2 = 0 (x-2)(x+1) = 0 x = 2 or -1 Now just plug in the roots. -1 does not fit because √1 = 1, not -1. -1 is the solution to -√(x+2)=x because both the plus and minus roots when squared come out positive.

**Math**

since each dimension is 1/3 as big, the volume (product of all three dimensions) is 1/27 as much.

**math**

y = (cos 2x)^x log y = x log(cos 2x) 1/y y' = log(cos 2x) + x/cos2x (-2sin2x) y' = (cos2x)^x (log cos2x - 2x*tan2x) It is interesting to note that if y = u^v y' = vu^(v-1) u' + (logv)(u^v)(v') If v is a constant, v'=0, so we have the familiar power rule...

**Pre calculus**

thx

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