Saturday

May 23, 2015

May 23, 2015

Total # Posts: 31,336

**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*

**maths**

28 min = 1680 seconds 1.4 L/s * 1680s = 2352 L = 2352000 cm^3 so, the side of the cube is 133 cm
*March 14, 2015*

**math**

If there were x people in all, then men = (3/4)(2/5)x = 3/10 x 3/5 x = 140 + (1 - 3/5 - 3/10)x = 140+x/10 6x = 140+x 5x = 1400 x = 280 check: 3/5 of 280 = 168 3/4 of 280-168 = 84 women = 168-140 = 28 168+84+28 = 280
*March 14, 2015*

**algebra2**

see why you should wait for one question to be answered before posting another just like it? ∛(n+8)-6 = -3 ∛(n+8) = 3 n+8 = 3^3 ...
*March 14, 2015*

**algebra**

.07x + .15(9) = .13(x+9)
*March 14, 2015*

**math**

As you can see from the graphs at http://www.wolframalpha.com/input/?i=plot+y+%3D+sin+x%2C+y+%3D+cos+x the region is roughly triangular. The graphs intersect at x=π/4. Using vertical strips of width dx, you have the area a = ∫[0,π/4] sinx dx + ∫[π/4...
*March 14, 2015*

**Physics**

just solve for t in 5t^2 + 100 = 6t^2
*March 13, 2015*

**Calculus**

y = ab^x ln y = lna + x lnb 1/y y' = ln b y' = y ln b The instantaneous rate of change is just (ln b) times the current value. That is the very definition of exponential change. The rate of change is proportional to the current value.
*March 13, 2015*

**math**

a = 25 - 9t^2 v = 25t - 3t^3 + c v(0) = 4, so c=4 and v = 25t - 3t^3 + 4 now just find v(2)
*March 13, 2015*

**math**

I can't figure out what A,B,C have to do with P,Q,R. PR = r-p = 3q-p/2 - p = 3q - 3/2 p RQ = q-r = q-(3q - 1/2 p) = -2q + 1/2 p I don't see that they are collinear. If we let p = 2j q = i then p and q are not collinear, and r=3i-j is not collinear with either p or q. ...
*March 13, 2015*

**Math**

You quoted the wrong formula, but used the correct one in your calculations. Those give the future value, given her payments. But the question is, what amount will she receive to deplete the funds at age 80? So, use the formula in reverse to find the PV=0 at age 80. The ...
*March 13, 2015*

**Math (Pre-Calc)**

sinC/c = sinB/b = sin58/74 = .01146 sinC = 84*0.1146 = .9626 So, C=74.29 or 105.71 Since A+B+C=180, A = 47.71 or 16.29 Now use a/sinA = b/sinB = 1/.01146 = 87.259 a1/sin47.71 = 87.259 or a2/sin16.29 = 87.259 Don't forget your Algebra I now that you're taking trig...
*March 13, 2015*

**physics(Please respond)**

after t seconds, the rock has dropped 1/2 g t^2 Use the value of g for the units you desire.
*March 13, 2015*

**Math - Geometry**

Lilly, you are correct. "therom," you are out of order.
*March 13, 2015*

**Math - Geometry**

recall that the vertex (maximum height) of at^2+bt+c is at t = -b/2a it hits the ground when h=0, so just solve the quadratic as usual. h(0) = 20 while the function has domain (-∞,+∞), we have not yet achieved time travel, so in the real world, the domain is [0,&#...
*March 13, 2015*

**Math**

x/40 = sin 30°
*March 13, 2015*

**Pre-Calculus**

I mean, that for each value of c, there is a corresponding value of b. Since B can be either 88.26 or 13.74, b1/sin(88.26) = 54.0265, so b1=54.10 or b2/sin(13.74) = 54.0265, so b2=12.86
*March 13, 2015*

**Pre-Calculus**

sinC/c = sinA/a sinC/43 = sin39°/34 = 0.01851 sinC = .7959 so, C = 52.74° or 127.26° Since A+B+C=180, B=88.26° or 13.74° Now use those two values for B to get B, using b/sinB = 1/0.01851 = 54.0265
*March 13, 2015*

**Math - Geometry**

surely, giben the vertex form of the parabola, you can read off the vertex? for the zeros, just solve normally the quadratic -2(x+5)^2 + 4 = 0 the y-intercept is where x=0 The parabola opens downward, so there is no minimum The vertex is the maximum, which you have already found.
*March 13, 2015*

**Algebra 2**

if 1+i is a root, so is 1-i. So, the factors are (x+1)(x-(1+i))(x-(1-i)) (x+1)((x-1)^2 - i^2) (x+1)(x^2-2x+2) x^3-x^2+2
*March 13, 2015*

**Math**

f = 2log5(2y-8) + 3 f = log5(2y-8)^2 + log5 125 f-3 = log5((2y-8)^2) 5^(f-3) = (2y-8)^2 5^((f-3)/2) = 2y-8 2y = 8 + 5^((f-3)/2) y = 4 + 1/2 √(5^(f-3)) so, the inverse function is 4 + 1/2 √(5^(y-3)) logx(x/y) = logx(x) - logx(y) = 1 - logx(y) logy(y/x) = 1 - logy(x...
*March 13, 2015*

**math**

the sum of all 39 of the numbers is 39/2 (17+55) so, divide that by 39 to get the average.
*March 13, 2015*

**Math**

384/6 = 64 There are three terms, so r = 4 T6 = T4*r^2 = 6*16 = 96 r^3 = 8!/7! = 8 so, r=2 T1 = T5/r^4 = 7!/16 = 315
*March 13, 2015*

**Statistics - Can anybody help me?**

after t hours, starting with A mg, the amount remaining is A*.76^t So, to find the original dose, solve A*.76^5 = 74 To get to 25%, you want .76^t = .25
*March 13, 2015*

**Steve - Stat**

9.0508% 2250 * 1.0905077^33 = 39258 showing the proper growth rate. Maybe they don't want a percent, but a factor?
*March 13, 2015*