Tuesday

April 28, 2015

April 28, 2015

Total # Posts: 30,858

**math**

since both A and B are at the same longitude, they are separated by 60° latitude = π/3 the distance s = Rθ where R is the radius of the earth. If the plane flies to C along the 30° latitude line, it flies along a circle with radius Rcos30°, through an ...
*March 16, 2015*

**Math**

(-3)^2 = (-3)(-3) = +9 -3^2 = -(3^2) = -9 Review the order of operations. -23 subtracted from 17 is 17-(-23) = 17+23 = 40 -23-17 is 17 subtracted from -23 = -40 Get out your number line and review how to do subtractions and negative values.
*March 16, 2015*

**calculus**

I assume I will find it where it landed. That would be where the height is zero. y = tanθ x - g/(2(vcosθ)^2) x^2 In this case, that's y = 0.4877x - 0.0012088x^2 y=0 at x=403.5 So, I'd start looking at 403.5 feet from where it was shot. Did you have something ...
*March 16, 2015*

**algebra**

|-4-5| = |-9| = 9 |7+3| = |10| = 10 9-10 = -1
*March 16, 2015*

**chemistry**

0.178 * 300/1000
*March 16, 2015*

**chemistry**

2H2 + O2 = 2H2O 2Na + 2H2O = 2NaOH + H2
*March 16, 2015*

**Science**

line up 4 AA cells, end to end, + to -. That is an example of 4 cells in series. Your 12V car batter is another good example of cells in series. I think each cell can charge up to 2.4V, so there are 5 or 6 cells. See http://hyperphysics.phy-astr.gsu.edu/hbase/electric/leadacid...
*March 16, 2015*

**Geometry**

If the altitude h from A divides a into two parts, x and y, then since any side of a triangle is less than the sum of the other two sides, h < b+x h < c+y 2h < b+c+x+y but, x+y = a, so 2h < b+c+a h < (a+b+c)/2
*March 16, 2015*

**math**

I think this is answered in the related questions below.
*March 16, 2015*

**math**

There are 24C5 = 42,504 ways to pick 5 people. Only one of those ways includes the 5 youngest.
*March 16, 2015*

**Trigonometry**

(tanx-8)(tanx+1) = 0 tanx = 8 or -1 arctan(8) = 1.446 or 4.588 arctan(-1) = 3π/4 or 7π/4
*March 16, 2015*

**consumer math**

start off with the gross. 12 hrs @ 7.45/hr = 89.40 I will assume that you deduct your expenses before taxes. In that case, the taxable income is 89.40 - 3*5.70 = 72.30 Now for the tax bite. Total tax rate is .0765 + .1065 + .057 = .24 tax is thus .24*72.30 = 17.35 Subtract ...
*March 16, 2015*

**Math**

you can convert all to decimals, then it's easy to order them: .20,.375,.40,.45,.60 As you can see, they're already in order Or, you can convert all to fractions with a common denominator. In this case, that would be 40: 8/40, 15/40, 16/40, 18/40, 24/40
*March 16, 2015*

**Math Help**

as you know a^2 + b^2 = c^2 so, plug in your numbers 7^2 + b^2 = 11^2 49+b^2 = 121 b^2 = 72 b = √72 = 8.5
*March 16, 2015*

**math**

the real world has dimensions 1000 times as large as the map, so 4cm -> 4000cm = 40m 3cm -> 3000cm = 30m the map area is 3cm * 4cm = 12 cm^2 the actual area is 40m * 30m = 1200 m^2 or, you can use the scale factor to figure the real area. Since area is length * width, ...
*March 16, 2015*

**Math**

pick 4 numbers (bottles) multiply each by the unit cost, 1.50 Now plot the pairs: (bottles, cost)
*March 16, 2015*

**Math**

come on. which numbers are between 4 and 6? as you may have noticed, I forgot 5x12=60 One you have the height, the other factor is the width.
*March 16, 2015*

**Math**

the factors of 60 are 1x60 2x30 3x20 4x15 6x10 Now pick the pairs that meet your requirements.
*March 16, 2015*

**maths**

30*20cm = 600 cm
*March 16, 2015*

**MATH**

the heading is now 135° period. All headings are measured clockwise from due North. This can also be expressed as relative headings, such that the angle is at most 90°, as in E 45° S S 45° E These are both abbreviated as just SE.
*March 16, 2015*

**math**

(8/9) * ((64/243)/(8/9))^(1/3) = 16/27
*March 16, 2015*

**Math**

so, how many diagrams have you seen here?
*March 16, 2015*

**Math**

in 27 days they will double 3 times. So, what is 2550*2*2*2 ?
*March 16, 2015*

**maths**

I assume you mean the angle of the sun's rays. The altitude of the sun is 93 million miles. tanθ = 1/2
*March 16, 2015*

**math**

Q2. points with equal y-coordinate are all in the same horizontal line: y = 1.3 Q3. slope = -1/3, so it slopes down. the other line has slope -1/2 So, is -1/3 greater or less than -1/2? Q4. y-intercept is at x=0 x-intercept is at y=0 check the points to see whether they fit ...
*March 16, 2015*

**Calc**

well, of course. (it is kind of a silly question) You know after 5 seconds the balloon has gone up 5*50 = 250 feet. Now you know x and you can get z.
*March 16, 2015*

**Calc**

as you say, tanθ = x/50 so, sec^2 θ dθ/dt = 1/50 dx/dt You know x at t=5, and you know dx/dt=20, so you can find θ. The distance z is given by z^2 = 50^2 + x^2 z dz/dt = x dx/dt so, find z at t=5 and plug in the numbers.
*March 16, 2015*

**math**

ok. now what?
*March 16, 2015*

**Calculus**

well, f' = 5e^x - 4e^(2x) so, plug in x=1
*March 16, 2015*

**math(5th grade)**

steps area perimeter 1 1 4 2 3 8 3 6 12 4 10 16 ... n n(n+1)/2 4n so, 10 steps will have area 55 and perimeter 40
*March 16, 2015*

**math**

There are 5C2 = 10 ways to select the weights. However, there are some duplicates, such as 1+5 and 2+4 I suppose you can find how many, and subtract them from 10.
*March 15, 2015*

**algebra**

C = 5/9 (F-32) = 5/9 (45) = 25
*March 15, 2015*

**algebra**

clearly not C. 3 < 7, so √3 < √7
*March 15, 2015*

**algebra**

-(-3)^2 - 4(-3) -9+12 3 you are correct
*March 15, 2015*

**business math**

1460(1+.039/12)^(12t) = 1552.84 t = 1.583 years, or 19 months
*March 15, 2015*

**Statistics**

as always in a normal distribution, 50% are above the mean, regardless of std.
*March 15, 2015*

**math**

If the number doubles every 17 minutes, then that means it doubles 1440/17 times in 24 hours. That means there will be 100*2^(1440/17) = 3.15x10^27 zombies I guess that's not reasonable. So, assume that the number grows by 100 every 17 minutes. That means there will be 100...
*March 15, 2015*

**trig**

since the arc length is radius * angle, arc speed (linear speed) is radius times angular speed: 32 * pi/9 cm/s
*March 15, 2015*

**precal**

-1+i = √2 cis 3π/4 raise that to the 7th power and you have (√2)^7 cis 7*3π/4 (√2)^7 cis 21π/4 = 8√2 cis 5π/4 = -8-8i
*March 15, 2015*

**Algebra**

kinda weird, but if you've written it correctly, log(4/x) (x^2 - 6) = 2 raise 4/x to both sides and you have x^2-6 = (4/x)^2 x^2 - 6 = 16/x^2 x^4 - 6x^2 - 16 = 0 (x^2+2)(x^2-8) = 0 x = ±√2 i or ±2√2 (x+√(x^2-5))/5 = (x+√(x^2-5))(x-&#...
*March 15, 2015*

**math**

(800000*r^5)(.96) = 480000 r = .91 So, the car depreciated at 9% per year
*March 15, 2015*

**Calculus 2 Trigonometric Substitution**

I think your 2nd line is bogus, and the brackets are unbalanced. (1-tan^2)/sec^2 = 1/sec^2 - tan^2/sec^2 = cos^2 - sin^2 = cos(2x) integral of cos(2x) = 1/2 sin(2x)
*March 15, 2015*

**Algebra**

-2(2x + 9) > -4x + 9 -4x - 18 > -4x + 9 -18 > 9 Never
*March 15, 2015*

**Calculus**

we want the slopes to be equal 4cosx = x That occurs only 3 places. See http://www.wolframalpha.com/input/?i=x+%3D+4cosx
*March 15, 2015*

**Calculus check**

D Consider the function y = |sin(x-1)|
*March 15, 2015*

**Programming in Java**

check out MATH.sin and MATH.cos If you have the hypotenuse and an angle, these will help. If all you have is the hypotenuse, you're stuck. There is no way to figure the legs with just that.
*March 15, 2015*

**Please check my calculus**

since h = 3r, v = 1/3 pi r^2 h = pi/27 h^3 dv/dt = pi/9 h^2 dh/dt -50 = pi/9 * 9 dh/dt dh/dt = -50/pi C is correct
*March 15, 2015*

**calculus check**

siny = cosx cosy y' = -sinx -y' = -1 C is correct
*March 15, 2015*

**calculus**

#2 since lnx is not defined for x<0, I'd say False #3 since ln 7x = ln7 + lnx, I'd say True #4 7 (why?)
*March 15, 2015*

**calculus**

y = ((x+3)/e^x)^lnx wow! lny = lnx(ln(x+3)-ln(e^x)) = lnx(ln(x+3)-1) 1/y y' = 1/x (ln(x+3)-1) + lnx*1/(x+3) y' = y(ln(x+3)/x - 1/x + lnx/(x+3)) Now you can massage that in many ways.
*March 15, 2015*

**math**

If the scale factor is r, then the area grows by r^2 the volume grows by r^3 So, if the area scales by 20/45, the volume scales by (20/45)^(3/2). The smaller cylinder has cross-section area of .945*(20/45)^(3/2) = 0.28 cm^3 If x cm is transferred to the smaller container, then...
*March 15, 2015*

**Calc**

do you mean y = x ln(ln x)? if so, then y' = ln(lnx)) + x * 1/lnx * 1/x = ln(lnx) + 1/lnx y = log_x(7) = ln7/lnx if u = lnx, then y = ln7 u^-1 y' = -ln7 u^-2 u' ... since cosh^2 = 1+sinh^2, sinh^2-cosh^2 = -1 so the derivative is zero
*March 15, 2015*

**Math Geometry - Please check my answers!**

C is not correct. The point (16,0) clearly indicates that 16 is in the domain. It has to be A.
*March 15, 2015*

**Math Geometry - Please check my answers!**

#1 ok #2 ok #3 The only points we can use are those given. That would make the domain {10,13,16}. But, they have said line is a parabola, so the function is continuous. The domain is thus [10,16]. I guess you could consider that they started counting at t=0, when the dolphins ...
*March 15, 2015*

**precal**

√2/2^2 = √2/(√2*√2*2) = 1/(2√2) If that is what you meant... I suspect it is not. In that case, use some parentheses so we know just what the expression is. (√2/2)^2 √(2/2)^2 ...
*March 15, 2015*

**Calculus check**

clearly (3,3) is no kind of intercept!! Since f'=0 and f"<0, it is a relative max.
*March 15, 2015*

**precal**

r^2 = (2√3)^2 + 2^2 = 16, so r=4 (2√3 + 2i) = (4,π/6) (2√3 + 2i)^5 = (4^5,5π/6) = (-512√3,512)
*March 15, 2015*

**math**

the plate has a volume of (1050g)/(8.4 g/cm^3) = 125 cm^3 Since v = s^2 * h, .2s^2 = 125 s^2 = 625 s = 25 cm
*March 15, 2015*

**Please check my calculus**

looks good to me.
*March 15, 2015*

**math**

θ = 5/3, so a = 1/2 r^2 θ = 1/2 * 3^2 * 5/3 = ?
*March 15, 2015*

**Calculus check**

looks good to me.
*March 15, 2015*

**math**

for the smaller segment, sin θ/2 = 2/3.3 the larger segment's angle is thus 2π-θ for each segment, a = 1/2 r^2 θ
*March 15, 2015*

**AP CALC**

the average value of f' on the interval is ∫ f' dx = (f(4)-f(2))/(4-2) = -3/2 Since II and III say the same thing, I pick (D) we know nothing about the average value of f.
*March 15, 2015*

**calculus**

well, y" = 20x^3 - 2 - sinx where does y" = 0?
*March 15, 2015*

**Calculus**

f(-2) and f(2) do not exist. Thus, the theorem's conditions are not satisfied -- f must be continuous on the closed interval. This should have been clear, since none of the choices for c is in (-2,-2), except 0, and f(0) = 0. However, D and E both give the wrong reason. So...
*March 15, 2015*

**Calculus check**

y' = cosx - 2sinx y'(pi/2) = 0 - 2 y-1 = -2(x-pi/2) or 2x+y = 1+pi I get B. http://www.wolframalpha.com/input/?i=plot+y%3Dsinx%2B2cosx%2C+y+%3D+1%2Bpi-2x+for+x+%3D+0+to+3
*March 15, 2015*

**Calculus**

the value given is the definition of f'(5). So, II and III are true, making E the correct choice.
*March 15, 2015*

**math**

can you not look up the formulas? a = 1/2 r^2 θ, so 1/2 (36^2) θ = 54 pi (not pie!) θ = 108π/1296 = 0.2618 now for the arc length, s = rθ = 36 * 0.2618 = 9.42
*March 15, 2015*

**math**

40/x = tan 60°
*March 15, 2015*

**math**

geez, what's the problem? Just plug and chug: A(1 + e^kt + e^2kt + ... + e^(n-1)kt) is just a geometric series where a = 1 r = e^kt and there are n terms. So, Just before the 5th dose, there have been for doses, so n=4, and Sn = A(1-r^n)/(1-r) = 0.3 (1-e^(-.862*4*4))/(1-e...
*March 15, 2015*

**Calculus**

clearly, f"=0 at {0,1/2,-1/3} However, for an inflection point, f' must not change sign. (Think of the graph of x^3) The function for f'(x) is a nasty polynomial, so I assume that (A) is correct. If you actually do the calculations, you will see that f'(x) >...
*March 15, 2015*

**Calculus**

√x(√x+1) = x + √x surely you can do that...
*March 15, 2015*

**maths**

since the medians intersect 2/3 of the way from vertex to opposite side, and each median is also an altitude, the altitude is 60, making the sides 40√3
*March 15, 2015*

**Calculus**

assuming you can interpret the garbled text, did you follow the tip?
*March 15, 2015*

**math**

we have x^2 + y^2 = 25 x^2 + (x+1)^2 = 25 2x^2 + 2x - 24 = 0 x^2+x-12 = 0 (x+4)(x-3) = 0 So, the points are (-4,-3) and (3,4)
*March 15, 2015*

**Calculus**

the 5th derivative of a 4th degree polynomial is zero. But since they gave you f' rather than f, the 5th derivative of f is the 4th derivative of f'. SO, f^(5) = 2*3^4 * 4! = 3888
*March 14, 2015*

**Calculus**

since dC/dt = 2π/5, dr/dt = 1/5 a = πr^2, so da/dt = 2πr dr/dt = 2π(5)(1/5) = 2π
*March 14, 2015*

**Calculus check**

d/dx (x^2 f(x)) = 2xf(x) + x^2 f'(x) = 2*5*3 + 25(-2) = -20 you are correct
*March 14, 2015*

**Calculus**

y = √(8-x^2) y' = -x/√(8-x^2) y'(-2) = 2/2 = 1 so, the normal has slope -1
*March 14, 2015*

**calculus trig substitution**

since sec^2 = tan^2+1, let x = 3tanθ dx = 3sec^2 θ dθ x^2+9 = 9+9tan^2θ = 9sec^2θ Now the integrand is (3tanθ)^3 secθ (3sec^2θ) dθ = 81 tan^3(θ) sec^3(θ) dθ See what you can do with that.
*March 14, 2015*

**Calculus**

since cos^2θ = 1-sin^2θ, let x = 4sinθ 16-x^2 = 16-16sin^2θ = 16cos^2θ dx = 4cosθ dθ Now you have an integrand of (4sinθ)^3 (4cosθ) (4cosθ dθ) = 1024 sin^3(θ) cos^2(θ) dθ see what you can do with that
*March 14, 2015*

**Calc**

you can use partial fractions. (2x+1)/(x+1) = 2 - 1/x+1 so, the integral is 2x - ln(x+1) + C Note that my C is different from Jai's, because it includes the extra 2.
*March 14, 2015*

**Calculus 2**

tan^2 (sec^4) = tan^2(1+tan^2)sec^2 = (tan^2 + tan^4)(sec^2) dx Now note that if u = tanx, you have (u^2 + u^4) du See what you can do with the others.
*March 14, 2015*

**Calculus**

moving right when s is increasing. That is, when ds/dt > 0 a is increasing when da/dt is positive. That is, 48t-24 > 0 t > 2
*March 14, 2015*

**Calculus**

compare the graphs of x^2, x^3, x^4, x^5 Or, just consider the fact that an inflection point occurs when f"=0 and f' does not change sign.
*March 14, 2015*

**Calculus**

ever heard of the chain rule? y = 3/u y' = -3/u^2 u' Now just plug in u = sinx+cosx
*March 14, 2015*

**Pre-Calculus**

a does not change y-k = a(x-h) has vertex at (h,k) regardless of a.
*March 14, 2015*

**Pre-Calc.**

recall the basic definition of the logarithm: b^(logb(x)) = x logb(b^x) = x 5^x = 125 so, just take log5 of both sides: log5(5^x) = log5(125) x = log5(125)
*March 14, 2015*

**Calculus check revised.**

y = -1(x^2+1)^(-1/2) y' = 1/2 (x^2+1)^(-3/2) (2x) = x/(x^2+1)^(3/2) you are correct you can always check your answer at wolframalpha.com: http://www.wolframalpha.com/input/?i=derivative+-1%2Fsqrt%28x%5E2%2B1%29
*March 14, 2015*

**Calculus**

(1-3x) -> -11 (x-4)^2 -> +0 so, looks like D to me
*March 14, 2015*

**Pre.Calc**

you want e^10r = 2 10r = ln2 = 0.693 r = 0.0693 so, a rate of 6.93% will do the trick
*March 14, 2015*

**math**

.9982^-60 = 1.1142, so .9982^(x-60) = 1.1142*.9982^x So, just split 3.9657 into two parts, such as 1+2.9657: g(x) = f(x) + 2.9657/1.1142 f(x-60)
*March 14, 2015*

**algebra - incomplete**

Since I have no idea what the data were, I cannot tell which is the line of best fit. The $ number is the result you get when plugging some value into the equation. Presumably it is the price of some given quantity of gasoline.
*March 14, 2015*

**Calculus**

y' = 2cosx y" = -2sinx y' is max when y" = 0, at 0,π,2π y'(π) = -2 y'(2π) = 2 So, the point (2π,0) has max slope. y = 2(x-2π) Note that x=0 is not in the given domain. b is clearly not 4π See the graphs at http://www...
*March 14, 2015*

**Vectors**

3x-6y-9z = -5 x-2y-3z = -2 are clearly parallel planes.
*March 14, 2015*

**Math**

http://www.wolframalpha.com/input/?i=plot+2x%5E2+%2B+y+%E2%89%A5+2%2C++x+%E2%89%A4+2%2C++y+%E2%89%A4+1
*March 14, 2015*

**math**

T = (x,y) -> (x+4,y-5) so, do that to (-4,5)
*March 14, 2015*

**math, scale drawing**

the scale is 1.5cm/2ft = 0.0492 the area scales as .0492^2, so a = (120ft^2)(30.48cm/ft)^2 * .0492^2 = 270 cm^2
*March 14, 2015*

**math**

as written, there is no solution. If the product of some numbers is 0, then one of the numbers must be zero. 3.9657 ≠ 0 ln 0.9982 ≠ 0 0.9982^x is never zero Looks like you have written d/dx (3.9657 * 0.9982^x) care to clarify what you really want?
*March 14, 2015*

**Math**

tanθ = 395/148
*March 14, 2015*