Monday

July 6, 2015

July 6, 2015

Total # Posts: 32,121

**Earth Science**

the magnification of a telescope is changed by changing what? my answer is the eyepiece
*May 3, 2015*

**Math**

well, just plug in 180 = pi
*May 3, 2015*

**Math**

Since you are actually performing the action, I'd have to go with Experimental. If asked just to calculate the probability, then it would be theoretical. The actual observations might or might not agree with the calculation. In fact, for any finite number of trials, the ...
*May 3, 2015*

**Math**

why all those words? Try using a little actual notation: If tan A = a / b, a>0, b>0, and 0<A<pi/2, then what is cos A? a. a / b B. b / a C. a / √(a^2 + b^2) D. b / √(a^2 + b^2) E. √(a^2 + b^2) / b given tanA = a/b, you know that opposite side is a...
*May 3, 2015*

**Algebra**

add up the coins, and add up the values: n+d = 29 5n+10d = 230 Now just solve for n and d.
*May 3, 2015*

**Algebra**

done. see related questions below
*May 3, 2015*

**Calculus 2**

This looks suspiciously like 1/5 e^x which converges everywhere.
*May 3, 2015*

**Calculus - more data**

Is the intersection of the planes tangent to the curved edge of the cylinder? Which axis is the axis of the cylinder? the y-axis? If so, why mention it by name?
*May 3, 2015*

**Calc 1**

the base has area 2w^2 The volume is 10, so the height h=5/w^2 So, the cost is 13(2w^2) + 10(wh + 2wh) = 26w^2 + 10w(3h) = 26w^2 + 10w(5/w^2) = 26w^2 + 50/w
*May 3, 2015*

**Math**

cost: 4.50*14 revenue after spoilage: 10p we want 10p = 4/3 (4.50*14) p = 8.40
*May 3, 2015*

**Calc 1**

oh, please. Surely you can describe the diagram. As a calculus student, you have surely developed some skill with language as well.
*May 3, 2015*

**Calc 1**

no diagram. How about a description? I can guarantee that the maximum area is achieved when the fencing is divided equally among lengths and widths.
*May 3, 2015*

**Math**

Equal divisions: x receives $500 Proportional: x receives 4500/(4500+4500+1000) * 1500 = $675
*May 3, 2015*

**Math**

Just look at the ratios: 1/2 = 0.5 .5/1 = 0.5 .25/.5 = 0.5 ... So, the next number is .125 * .5 = .0625 It might have been easier using fractions. The sequence is thus seen to be 2, 1, 1/2, 1/4, 1/8, ...
*May 3, 2015*

**math**

I think the related questions below will provide the method of solution.
*May 3, 2015*

**Math**

a/b = 3/4 c/b = 1/2, so b/c = 2/1 and thus a/c = a/b * b/c = 3/4 * 2/1 = 3/2
*May 3, 2015*

**Math. first and last repost. please help**

done. see earlier post.
*May 3, 2015*

**maths**

8C5 = 8*7*6*5*4 ------------- 1*2*3*4*5 Note that this es equal to 8C3 = (8*7*6)/(1*2*3) because the 4*5 just cancels out. to figure nCm, just start with n, and count down for m factors in the top, and starting with 1, count up for m factors in the bottom. If m > n/2, you ...
*May 3, 2015*

**Math please help.**

Only the graph of a function has a slope. This and similar questions are extremely poorly worded. y has a greater rate of change. There is only one slope: that of the function. Its slope is 3. Only the function has a slope. x and y do not have slopes. They just have rates of ...
*May 3, 2015*

**math**

I only see one function: f(x) is (-1,0),(0,1),(1,2),(2,3) x and y change at the same rate. When x increases by 1, so does y.
*May 3, 2015*

**trigonometry**

120651/47 π = 2567.04 π So, tan(-120651/47 π) = -tan(120651/47 π) = -tan(2567π + 2/47π) = -tan(2/47 π) arctan(-tan(2/47 π)) = -2/47 π You are correct. x = a/√(25-a^2) son tan(arcsin(a/5)) = a/√(25-a^2) Always use the ...
*May 3, 2015*

**mth121**

2x+3y = 3 x-2y = 5 has solution (3,-1) This point is also on on the third line, so it is the solution to all three equations.
*May 3, 2015*

**Math 222**

a good place to start would be here. Does your text not have any examples of graphing functions? http://www.wolframalpha.com/input/?i=plot+y%3D-x%5E2%2C+y%3D%E2%88%9A%28x-1%29
*May 3, 2015*

**math**

y = -(10-x)^2+75 This is just a parabola with vertex at (10,75) So, the max height is 75 meters
*May 3, 2015*

**Maths**

The 4 points define a tetrahedron. What do you mean by "boxes" as related to the tetrahedron?
*May 3, 2015*

**trigonometry**

done. see earlier post.
*May 3, 2015*

**trigonometry**

just pull out your calculator. cos(1.50) = 0.07 Since cos x > 0 in QI and QIV, the other solution is 2π-1.50
*May 3, 2015*

**trig**

Trying to get past the copy/paste of strange fonts, I interpret the question as finding the trig functions of the angle whose terminal side is (√15,8) You have x=√15 and y=8, so r = √(15+64) = √79 Now just use the definitions sinθ = y/r cosθ...
*May 3, 2015*

**Math**

You don't need it. arctan(-x) = -arctan(x) So, now you want arctan(-tan(7/15π)) = -arctan(tan(7/15π)) = -7/15π
*May 3, 2015*

**Math**

tan(-22/15π) -tan(22/15π) -tan(π+7/15π) -tan(7/15π) not a recognizable value. As long as you have to use your calculator anyway, why nut just plug in the original value? You get -9.514 Or is there more to the question than what you posted?
*May 3, 2015*

**math**

Try this, which is a better rendering of the given statements: x+23 = 2y y+23 = 3z z+23 = 4x
*May 3, 2015*

**Physics Help Please**

just review your formulas. In particular, hi/ho = f/f-do 32mm/17m = 40mm/(40-do)mm 32/17000 = 40/(40-do) do = 21210mm = 21.21m
*May 3, 2015*

**Math**

let u = arcsin(x) then du = 1/sqrt(1-x^2) dx now your integrand is just 1/u du Think you can handle that one?
*May 3, 2015*

**physics**

since there is no friction, there is no horizontal force. Thus, no acceleration. So, think of Newton's 1st law of motion.
*May 3, 2015*

**Chemistry**

Given the very nature of the question, you must have studied some examples. Care to share your ideas on the topic?
*May 3, 2015*

**Trigonometry - PS**

However, since inverse trig functions have principal values (because they are multi-valued), you have to be careful. arctan(tan(3π/4)) = arctan(1) = π/4, not 3π/4. So, you need to reduce your fraction to see how many multiples of π you can discard, and then...
*May 3, 2015*

**PHYSICS**

That would be zero, of course. Care to rephrase the question?
*May 2, 2015*

**algebra 1**

with the exception of calling the vertex the axis of symmetry, yes. The axis always passes through the vertex, but it is not the vertex; the vertex is just a point. The axis has no minimum point; it is a line.
*May 2, 2015*

**algebra 1**

axis is the line x=0
*May 2, 2015*

**algebra--1 question**

recall that (a+b)^2 = a^2 + 2ab + b^2. So, (2√5+3√7)^2 = (2√5)^2 + 2(2√5)(3√7) + (3√7)^2 = 4*5 + 12√35 + 9*7 = 83 + 12√35
*May 2, 2015*

**calculus**

see related questions below
*May 2, 2015*

**Optimization Calculus**

so, now go for it. You know that 2x+y = 96, so y = 96-2x the area is xy = 2x(96-2x) = 192x - 4x^2 That's just a parabola; its vertex gives the maximal area.
*May 2, 2015*

**Calculus**

something wrong with Reiny's solution? See http://www.jiskha.com/display.cgi?id=1430504193
*May 2, 2015*

**Maths(URGENT ANSWER please )**

you have sum[1,∞] 2n/n! = 2 sum[1,∞] 1/(n-1)! = 2 sum[0,∞] 1/n! recall that e = 1 + 1/1! + 1/2! + 1/3! so, your sum is 2e
*May 2, 2015*

**Calc 1**

y' = 2(x-5) Let x0=1 x1 = 1 - y/y' = 1 - 16/-8 = 3 x2 = 3 - 4/-4 = 4 x3 = 4 - 1/-2 = 6 x4 = 6 - 1/2 = 5.5 and so on for six decimal places.
*May 2, 2015*

**Maths(URGENT ANSWER please :**

sum[1,∞] 2n/(n+1)! = 2 sum[1,∞] n/(n+1)! = 2 sum[1,∞] (n+1-1)/(n+1)! can you take it from there?
*May 2, 2015*

**Algebra**

recall that a^3-b^3 = (a-b)(a^2+ab+b^2) You have 3^3 - (m+2n)^3 = (3-(m+2n))(3^2 + 3(m+2n) + (m+2n)^2) = (3-m-2n)(9+3m+6n+m^2+4mn+4n^2) I think I like the original way more . . .
*May 2, 2015*

**Calc 1 - ouch! (duh)**

I sure flubbed that one. Your later solution was spot on, Reiny.
*May 2, 2015*

**Calc 1**

Naturally, the tangent with the largest slope will be the one closest to the origin. So, we want -6cosx = -6sinx/x x = tanx solve that using Newton's method, and you have x = 4.93 so, your line is y = -6cos(4.93)(x-4.93)+6sin(4.93) or y-5.98 = 0.52(x-4.93) See the graphs ...
*May 2, 2015*

**Physics Help Please --- Urgent**

since the rays from the building may be considered parallel, we have f = 24. So, to find the distance of the insect's image (from the lens), 1/16 + 1/x = 1/24 Use that to arrive at your final answer.
*May 2, 2015*

**Math**

since t=0 when y is a max, we will have y = a cos(kt) the radius is 5, so y = 5cos(kt) the period is 1/25 min, so k/2pi = 1/25 k = 2pi/25, and thus y = 5cos(2pi/25 t)
*May 1, 2015*

**Math- Trigonometric Functions**

since the curve starts at the top when t=0, y = a cos(kt) the diameter is 10, so the radius is 5: y = 5 cos(kt) The period is 1/25 min, so 2pi/k = 1/25, making k = 2pi/25 y = 5 cos(2pi/25 t) Not sure what you mean by the blade's distance from the center. I think you meant ...
*May 1, 2015*

**Algebra**

b + 2(2b-0.3) = 4.9 That gives the base; now figure the legs.
*May 1, 2015*

**college precalculus**

a = 1/3^6 r = 1/3^2 S = a/(1-r) = (1/3^6)/(1-1/9) = (1/3^6)(9/8) = 1/648 a = -100/9 r = -3/10 S = (-100/9)/(1+3/10) = -1000/117
*May 1, 2015*

**Trigonometry**

Assuming that the first-named point on the opposite side (P) is the point directly opposite from the point on his own side (Q), then we have, if the observer is point O, m<POQ = 28 m<OQP = 128, so m<OPQ = 24 Now, using the law of sines, if the width is w, we have w/...
*May 1, 2015*

**math ( help please instantly )**

just a little synthetic division trying to find a root. Cubics are not always easy to factor by inspection.
*May 1, 2015*

**math ( help please instantly )**

m^2-7m+12 = (m-3)(m-4) m^3-2m^2-2m-3 = (m-3)(m^2+m+1) so, LCM = (m-3)(m-4)(m^2+m+1) Now try the others. What do you get?
*May 1, 2015*

**Maths**

p = √(32/6)
*May 1, 2015*

**Math**

It's always easiest to convert mixed numbers to improper fractions. for adding, find the lowest common denominator for multiplying, just multiply/divide 9 9/10 × 8 8/9 = 99/10 * 80/9 = 88 15 11/15 + 14 5/9 + 7 3/5 = 708/45 + 655/45 + 342/45 = 1705/45 = 341/9 = 37 8/9...
*May 1, 2015*

**Math**

3^3 = 27, not 18 4 * 27 -54÷6+25 Just as multiplication/division are done left to right, so are addition and subtraction 108-9 +25 99+25 124 Besides the typo of 26, you did 72-(9+26), not 72-9+26 314-3 [2^5 -(9- 4 ) ^2 +3]^2 314-3 [2^5 - 5^2 +3]^2 314 - 3[32-25+3]^2 314...
*April 30, 2015*

**Math**

so, show us your work, and we'll check it.
*April 30, 2015*

**Calculus**

the area of a triangle of side s is √3/4 s^2 So, if the triangles have side x and the tent has length y, we have (√3/4)x^2 * y = 2.2 the area of cloth used is a = 2(√3/4)x^2 + 3xy and the cost c is c = 1.4(2(√3/4)x^2 + xy)+xy Now just find dc/dx, after ...
*April 30, 2015*

**algebra please help**

I only see one polynomial.
*April 30, 2015*

**Trigonometry**

hey, come on. The y-intercept is where x=0! So, just plug it in y(0) = -2 sin(0-pi/2) = -2(-1) = 2 getting x-intercepts can be hard. Getting the y-intercept is trivial! Don't forget your Algebra I now that you're in trig.
*April 30, 2015*

**Trigonometry**

the origin is an x-intercept if f(0) = 0. Is that true here?
*April 30, 2015*

**Advanced Functions**

sin4x/(1-cos4x) (2 sin2x cos2x)/(2sin^2 2x) cos2x/sin2x (2cos^2 x - 1)/(2sinx cosx) cosx/sinx - 1/(2sinx cosx) cotx - 1/2 cscx secx
*April 30, 2015*

**Math**

the diameter is 14, so the radius is 7 the area is π*7^2 = 49π
*April 30, 2015*

**Math**

One way: divide the big volume by the small volume: (3 3/4)(5)(4 1/2) / (1/4)^3 = (675/8) / (1/64) = 5400 Another way: decide how many cubes it takes to fit each of the larger dimensions (3 3/4) / (1/4) = 15 5 / (1/4) = 20 (4 1/2) / (1/4) = 18 So, the prism is 15x20x18 cubes...
*April 30, 2015*

**Math**

y=x+(x/4) ? Why not just 5/4 x? Once you get your curves figured out, r = the distance to the inside curve R = the distance to the outside curve You need to solve for x, so you know r. Clearly, R=3, since that's the far boundary. The limits of integration (along dy) are ...
*April 30, 2015*

**Math**

assuming the cylinder also has diameter 7, then if the cone has height h and the cylinder has height k, you need π/3 * r^2 h = π * r^2 k h/3 = k The cone needs to be 3 times as tall as the cylinder
*April 30, 2015*

**math**

Draw a diagram. You have a 30-60-90 triangle, with the tree at the right angle. You want the altitude from the tree to the road. The distance of the tree from B is 200 The distance of the tree from A is 200√3 So, h = 100√3
*April 30, 2015*

**Calculus**

with 3 unknowns and only two equations, there will be no unique solution. It should be 13y+5z = 39 so, pick any y and z that fit, and then figure x. Or, you can evaluate y and z in terms of x: y = 5x-7 z = 13(2-x) Now you can pick any value for x, and then figure y and z. In ...
*April 30, 2015*

**Math**

1982: (3.35*40+80)(50) = 10700 2015: 10700*1.045^(2015-1982) = 45732.12
*April 30, 2015*

**Math**

317.24(1.085)(1.15)+25.00
*April 30, 2015*

**Calculus**

just add the vectors. That is the resultant, call it r. The equilibrant is -r: that vector which cancels the resultant, keeping things in equilibrium.
*April 30, 2015*

**Calculus**

vectors are just a compact way of describing three independent sets of numbers. You just have 3a-6 = -3 9-2b = 1 6-2b = -2 clearly, a=1, and any value of b will work. since (1,3,2) and (3,b,b) are not collinear, they span R2.
*April 30, 2015*

**Math**

pq-pr-qp+qr-rp+rq or, arranging factors alphabetically, pq-pr-pq+qr-pr+qr -2pr+2qr 23ab-12ab - 12bc^2-13bc^2 11ab - 25bc^2
*April 30, 2015*

**Math**

just plug and chug: 2(-7/2)^2 - 9(-7/2) + 7 ...
*April 30, 2015*

**Math**

nothing to solve. just evaluate the expression 3(-1.3)(2/5) - |-1.3-2| -1.56 - |-3.3| -1.56 - 3.3 -4.86
*April 30, 2015*

**Math**

(3245.50-143.15)(1-0.10) = 2972.115
*April 30, 2015*

**Calculus**

dy/dx = (1+x)/xy y dy = (1+x)/x dx 1/2 y^2 = lnx + x + c since y(1) = -4, ln 1 + 1 + c = 8 c = 7 1/2 y^2 = lnx + x + 7 . . .
*April 30, 2015*

**math**

pi * 9^2 * 4
*April 30, 2015*

**math**

if the digits are a,b,c we see that a=4 Then we are doomed. If b and c are both odd, they cannot sum to 5, which is also odd. And how about a little punctuation? That run-on sentence is very annoying.
*April 30, 2015*

**Algebra 2**

Is there a question in there somewhere?
*April 30, 2015*

**math**

just subtract what was used from the original amount: 9100 - 321*28.286 However, that assumes that all the leftover parts could be combined to form new circles; not very likely. A better way to look at it is to decide how many rows and columns of discs could be cut. 130/6 = 21...
*April 30, 2015*

**SOLVE -MATH**

9C6 * 6C3
*April 30, 2015*

**physics**

we have, since the focal length is constant, 1/f = 1/20 + 1/x = 1/16 + 1/(x+2.7) Just solve for x, and then find f.
*April 30, 2015*

**Calculus**

good work. Confirm with the graphs at http://www.wolframalpha.com/input/?i=plot+x%5E2%2B3x%2B2%2C+y%3D3x%2B2+for+-1+%3C%3D+x+%3C%3D+1
*April 30, 2015*

**maths**

you have a common ratio, so p/-6 = -32/p p^2 = 192 ...
*April 30, 2015*

**math**

No, P(black) = 9/10 1 - 9/10 = 1/10 B is the odds of drawing a white, not the probability.
*April 30, 2015*

**Physics**

1A is 1C/s You have 73mC/9s = 8.11mA the net charge is not relevant here, since all we need is the number of Coulombs.
*April 30, 2015*

**MAT 142**

keeping all the measurements in yards, (15/3)(8/3)(8/36)($30)
*April 30, 2015*

**Math**

clearly you can drop the trailing zeros, getting 504/5940 both are even, so you can factor out 2 top and bottom: 252/2970 and again: 126/1485 now 3's: 42/495 and again: 14/165 That's as far as you can go. Or, you can factor both numbers. 5040 = 2^4 3^2 5 7 59400 = 2^3 ...
*April 30, 2015*

**Math**

(8*7)/2 - π/4*3^2
*April 30, 2015*

**math**

hmmm. None of the polynomials factors over the reals, so I don't see any help there. Looks like GCF=1 and LCM = (1+4x+4x^2-16x^4)(1+2x-8x^3-16x^4)(16x^4+4x^2+1) if x=-1 we have -17,-7,21 x=0 we have 1,1,1 x=1 we have -7,-21,21 x=2 we have -231,-315,273
*April 30, 2015*

**math - eh?**

what's with the .. ?
*April 30, 2015*

**Math**

first expand: 4x^3y+30x^2y-35xy^2-72x^2y+80xy^2-11xy^2 then combine: 4x^3y-42x^2y-34xy^2
*April 30, 2015*

**Math**

(4)(14)+(-1)(3)+(-6)(8) = 56-3-48 = 5 Pretty grim score, eh?
*April 29, 2015*

**English**

What do you do when the author can't be found? How do you cite the book? Would use just the title of the book or explain that the info is missing?
*April 29, 2015*