Wednesday

May 25, 2016
Total # Posts: 40,790

**math**

3/200 = 1.5/100 = 1.5%
*May 3, 2016*

**Math**

tanx/sinx = 1/cosx cosx=0 at odd multiples of pi/2. So, that is where you have asymptotes. cosx changes sign at those points, so also does your graph.
*May 3, 2016*

**math**

depends on the area of the base. Divide the volume by the area to get the height.
*May 3, 2016*

**math - eh?**

huh?
*May 3, 2016*

**math - Correction.**

that is an obtuse angle. I'd say 0° < d < 90°
*May 2, 2016*

**Maths**

The three numbers will therefore be close to 21.98. So, I'd say they'd be 21,22,23 whose product is 10626
*May 2, 2016*

**math**

480ft/30s * 60s/min = 960 ft/min
*May 2, 2016*

**math**

not so. you have found d/dt (dy/dx), but you want d/dx (dy/dx) = d/dt(dy/dx) / dx/dt = -1/(1+t)^2 / e^t = -1/((1+t)^2 e^t)
*May 2, 2016*

**TRIGONOMETRY2**

can't use google? 1st hit led to http://www.cut-the-knot.org/proofs/2Colors2Points.shtml#solution
*May 2, 2016*

**TRIGONOMETRY**

√(x+b)-√(x-a) >= √(x+a)-√(x-b) since x>=a>=b, we have x+b >= 2b x-a >= 0 x+a >= 2a >= 2b x-b >= x-a see what you can do with that
*May 2, 2016*

**Math**

the ratios are: length: 11/33 = 1/3 width: 7/28 = 1/4 so, not similar
*May 2, 2016*

**math**

not quite. log(x^2) = 2logx 150 = 3*100/2, so log 150 = log3 + 2log10 - log2 75 = 3*(10/2)^2, so log75 = log3 + 2log10 - 2log2 or, since 75 = 150/2 log75 = log150-log2 as above
*May 2, 2016*

**Probability**

1/15000 (1000 + 2*300 + 20*10) = 3/25
*May 2, 2016*

**Math**

there is no limit. There must be some other considerations.
*May 2, 2016*

**Geometry**

well, just check. a sphere with diameter d has volume π/6 d^3 3*(π/6)(5^3) = 375π/6 (π/6)*8^3 = 512π/6 π/12 * 10^3 = 500π/6
*May 2, 2016*

**calculus**

94 ∑ k k=3 assuming you meant ... 144, 12 ∑ k^2 k=3
*May 2, 2016*

**Algebra**

y=5x
*May 2, 2016*

**pre-cal**

as they said, the total area is xy as for the print area, that is (x-2)(y-3) = 26
*May 2, 2016*

**calculus**

maximize p=x(30-x)-100-6x subject to x >= 10 x <= 20 the profit is revenue-cost, so p(x) = x(30-(x-10)) - 100 - 6x = -x^2+34x-100 dp/dx = 34-2x dp/dx=0 at x=17
*May 2, 2016*

**calculus**

this is just the chain rule in reverse. G'(x) = cos(x^2) (2x)
*May 2, 2016*

**calculus**

change all the t's to x^2, then multiply by (x^2)' G(x) = cos(x^2)(2x)
*May 2, 2016*

**math**

it is a function whose graph is s straight line.
*May 2, 2016*

**calculus**

Sketch the region. It has vertices at (0,0), (2,0), (4,2) You can get the area by using vertical or horizontal strips. Using vertical strips, they all lie below the curve √x, but the lower boundary changes from y=0 to y=x-2 at x=2. So, the area is ∫[0,2] √x ...
*May 2, 2016*

**math**

as usual, v = Bh = 20*10 = 200 cm^3 note that the area of the base must be in cm^2, not just cm.
*May 2, 2016*

**Math**

You have only listed 4 side lengths, so I'm not clear. But, to get the area, divide it up into triangles, and maybe rectangles, if there are any. Then just add up the areas of the pieces.
*May 2, 2016*

**math**

the base is 139.2/9.6 = 14.5 since all sides are the same, the perimeter is 4*14.5 = 58
*May 2, 2016*

**algebra**

(17-1)/(2-(-2)) = 16/4 = 4
*May 2, 2016*

**calculus too hard help**

Let's pick up at lny = x ln(y) 1/y y' = ln(y) + x/y y' y' (1/y - x/y) = ln(y) y' = y*ln(y)/(1-x) I'd suggest not looking too hard for a pattern. Pop over to wolframalpha.com and try entering x^x, then x^x^x, then x^x^x^x, and so on. The derivatives ...
*May 2, 2016*

**Math**

Thank you Reiny. That makes perfect sense to me. Next time I post a question I will think a little harder to possible ways of working questions like this. 😀😀
*May 2, 2016*

**Math**

When a telephone exchange is established, there are 10,000 new phone numbersto use(i.e. 722-0000 to 722-9999). Of these 10,000 phone numbers, how many do not contain the digit "9"? I know I can do it this long way, but how do I do it in a shorter more reliable way. ...
*May 2, 2016*

**Calculus - urgent**

Consider the mass as a bunch of horizontal strips, each with area (x2 - x1) dy and thus mass (x2-x1)y dy = y(√(1+2y/3) - (2y^2-y)) dy Now just integrate from 0 to 1.0989
*May 2, 2016*

**trigonomentry...Very hard**

theta, not tita. let x = 3tanθ x^2 = 9tan^2θ 9+x^2 = 9+9tan^2θ = 9sec^2θ √(9+x^2) = 3secθ so, draw a right triangle. Since tanθ = x/3, those are the two legs, and the hypotenuse is √(9+x^2) Now you can write down sinθ and cosθ
*May 2, 2016*

**Geometry (answer please)**

#1. well, each wire is the hypotenuse of a right triangle, with legs 16 and 10√2 back later for #2.
*May 2, 2016*

**Math**

surely, 3+3+3 feet, no? assuming the ribbon is to go once around the edges of the blanket...
*May 1, 2016*

**Math**

since u.v = |u||v|cosθ we must have cosθ = -1. In other words, v = -u
*May 1, 2016*

**Algebra 1**

B is good.
*May 1, 2016*

**math**

If you're using logs base 10, recall that log(32) = log(10*3.2) = log(10)+log(3.2) = 1+log(3.2) So, just look up log of 3.2 and add 1. Whoa - who uses tables these days?
*May 1, 2016*

**Plane trigonometry**

I don't know what "the five points" are, but here's the graph: http://www.wolframalpha.com/input/?i=2sin%28pi*%28x%2B1%2F2%29%29%2B1
*May 1, 2016*

**trig**

I'll use x for ease of typing tan(x)/(tan(2x)-tan(x)) tan(x)/((2tanx)/(1-tan^2(x))-tan(x)) tan(x)/((tanx(2/(1-tan^2(x)) - 1) 1/(2/(1-tan^2(x)) - 1) (1-tan^2(x))/(2 - (1-tan^2(x)) (1-tan^2(x))/(1+tan^2(x)) (1-tan^2(x))/sec^2(x) (1-tan^2(x))cos^2(x) cos^2(x)-sin^2(x) cos(2x)
*May 1, 2016*

**Algebra**

you are correct.
*May 1, 2016*

**MEASUREMENTS**

15m^2 * (3.28ft/1m)^2 * 1bag/20ft^2 * $2.00/bag = $16.14 you said M^2, then used YD^2. I used meters, but you can fix it. Your units do not work out right, and you lost the 15
*May 1, 2016*

**Exponential Modeling**

on the other hand, given the topic of the posting, 0.10(1+r)^40 = 1.25 (1+r)^40 = 12.5 1+r = 1.065 r = 6.5% per year
*May 1, 2016*

**MATH**

The scores are x x x x x x x x The median is 5, so they are x x x 5 5 x x x or x x x 4 6 x x x or 3 3 3 3 7 7 7 7 In the first case, 5 goals are scored. Otherwise, we need not have any scores of 5.
*May 1, 2016*

**math**

this is discussed at http://mathforum.org/sum95/suzanne/whattess.html google is your friend. Use it.
*May 1, 2016*

**MATH- HELP PLEASE. IM LOST**

the other side is either 3.7Ø or 3.7/Ø
*May 1, 2016*

**Geometry ( helllppp pleaaassse )**

#1. How can working with a single point on the ceiling tell anything about the length of the ceiling? #2. Drop the ladder to the floor, and mark where it hits. The use the ladder to construct a hexagon. Two of the main diagonals intersect at the center of the hexagon, just ...
*May 1, 2016*

**Exponential Modeling**

100 * (1/2)^7
*May 1, 2016*

**Exponential Modeling**

2000 * (1/2)^x
*May 1, 2016*

**Exponential Modeling**

p = 1013 * .88^x check for typos
*May 1, 2016*

**Math**

subtract 32 and multiply by 5/9 C = 5/9 (F-32)
*May 1, 2016*

**Algebra**

try this useful site: http://www.zweigmedia.com/RealWorld/LPGrapher/lpg.html
*May 1, 2016*

**Math**

5/6 * 1/2 = 5/12 you might as well ask, what is 1/2 of 5/6?
*May 1, 2016*

**Math**

a + a+d + a+2d = 24 a+3d + a+4d + a+5d = 51 now simplify the equations and solve for a and d.
*May 1, 2016*

**Geometry**

If the foot of the ladder started out at x ft away from the wall, then we know that the length of the ladder is 2x ft. With a 45° angle, then the two legs are x+8 ft, and the length of the ladder is (x+8)√2. So, now we have 2x = (x+8)√2 x(2-√2) = 8√...
*May 1, 2016*

**Solid Mensuration**

circles can be packed in many ways. If they are in a rectangular lattice then they may be in a 2x3 array, in which case, each circle's diameter is 1 foot. Surely by now you have made some attempt to solve the problem...
*May 1, 2016*

**Matthhh**

the straight-line distance is √(60^2+6^2) = 60.3 ft, so ...
*April 30, 2016*

**Algebra**

since there is a walkway on both sides, each dimension is increased by 2x. A(x) = (20+2x)(15+2x)
*April 30, 2016*

**Algebra**

since time = distance/speed, 6/(90-x) + 6/x = 4.5 x ≈ 1.35 km/hr
*April 30, 2016*

**math**

1 2 3 5 6 6 6 so, it can be 1. or even 1 5 5 5 5 5 6
*April 30, 2016*

**Discrete Mathematics**

26!/(8!12!6!) wow - who'd have thought that the entire set of books would contain only three different titles?
*April 30, 2016*

**CES**

1*22 + 9x = 8(x+22)
*April 30, 2016*

**MATH**

volume = 50*10*20 cm^3 1 liter = 1000 cm^3
*April 30, 2016*

**algebra**

I suspect the subtraction problem is 2x^2+5x-10 3x^2-x+9 ----------------- -x^2+6x-19
*April 30, 2016*

**math**

42 is the answer to the Ultimate Question of Life, the Universe and Everything -- surely it answers your feeble inquiry! Actually, I have no idea what you're talking about. which columns are you talking about? The only thing I can think of is that the line of cars looks ...
*April 30, 2016*

**Math**

well, 1 meter is 100 cm. So, how much is still needed?
*April 30, 2016*

**rio claro west secondar**

0.4^2 * 3.7 = 0.16 * 3.7 = ?
*April 30, 2016*

**math**

bzzt. The new value is 120% of the old one, so it increased by 20%.
*April 30, 2016*

**math**

3 of 8 = 3/8 = .375 = 37.5% t2o killed leaves 6 to share, so adjust the fraction. then subtract to find the change.
*April 30, 2016*

**math**

well, what are the nearby multiples of 500? ... 29,000 29,500 30,000 ... which is closest?
*April 30, 2016*

**maths**

you kidding me? You can't count by twos?
*April 30, 2016*

**Maths**

LCM(30,60,x) = 1500 30 = 2*3*5 60 = 2^2*3*5 1500 = 2^2*3*5^3 x could be 5^3 = 125 or any multiple of 25 consisting only of 2,3,5 as factors GCF(30,60,x) = 10 so, 125 is out, but 250 is in LCM(30,60,250) = 1500 GCF(30,60,250) = 10
*April 30, 2016*

**Math**

what is the difference between 6 and 10? How did you find it? Do the same step with the numbers above recall how to subtract negative numbers: 10 - (-5) = 10+5 = 15
*April 30, 2016*

**Solid Mensuration**

The diameter of the circle is 12 cm. The quadrilateral consists of two congruent triangles, with base 12 and altitude 4. So, its area is 48 cm^2
*April 30, 2016*

**Maths**

what is the difference between 6 and 10? How did you find it? Do the same step with the numbers above recall how to subtract negative numbers: 10 - (-5) = 10+5 = 15
*April 30, 2016*

**GEOMTRY**

1 yard = 3 feet 1 yard = 0.9144 meters 1 foot = 12 inches 1 foot = 0.3048 meters now just do the conversions
*April 30, 2016*

**Calculus**

If the rectangle has height x, then it has vertices (x,0), (x,x), ((20-x)/3,x), ((20-x)/3,0) so, its area is a = x((20-x)/3-x) = 20x/3 - 4x^2/3 da/dx = 20/3 - 8x/3 da/dx=0 at x=5/2, and a(5/2) = 25/3
*April 30, 2016*

**Homework Dump**

No ideas on any of these??
*April 29, 2016*

**Homework Dump**

no ideas on any of these?
*April 29, 2016*

**Algebra**

that is correct.
*April 29, 2016*

**Algebra**

nope. Subtract the exponents. 10^7/10^3 = (10^3 * 10^4)/10^3 = 10^4 or, more brute-force, 10*10*10*10*10*10*10 -------------------------- = 10*10*10*10 10*10*10
*April 29, 2016*

**College Algebra**

4-3(x+1)=2x 4-3x-3 = 2x 4-3 = 5x 1 = 5x x = 1/5 15x^2 + x = 2 15x^2 + x - 2 = 0 x = (-1±√(1+120))/30 = (-1±11)/30 = -12/30,10/30 = -2/5, 1/3 √(4x+5) = x 4x+5 = x^2 x^2-4x-5 = 0 (x-5)(x+1) = 0 x = -1, 5 only x=5 is a solution, since √(-4+5) = &#...
*April 29, 2016*

**Algebra**

6/(√3+2) = 6(√3-2)/-1 = 12-6√3 or, B
*April 29, 2016*

**Algebra**

6/(√3-2) recall that (a-b)(a+b) = a^2-b^2 multiply top and bottom by (√3+2) and you have 6(√3+2) ---------------- (√3-2)(√3+2) = (6√3+12/(3-4) = -6√3-12 check your text for typos. It looks like B except for the extra minus sign.
*April 29, 2016*

**Math/Computer**

try google, something like graphs ms word or something. (3,0) is just a point. Seems like a strange project.
*April 29, 2016*

**Math**

Since you have described no experiment, I will have to rely on the theoretical probability of 1/6, making my answer 50.
*April 29, 2016*

**Algebra**

Nope. What did you do? And use some parentheses: -3/(x+2) - 5/(x+3)
*April 29, 2016*

**math**

list the possibilities: P(WWW) = 2/4 * 1/3 * 0/2 P(BWW) = 2/4 * 2/5 * 1/4 ... P(BBB) = 2/4 * 3/5 * 4/6 then add up all the ones where the 3rd ball is black.
*April 29, 2016*

**@Need Help**

The subscripts are a trick of HTML. Most of the codes do not work on this site, but the sub and /sub tags work. For superscripts, the sup and /sup tags work. enclose the words in <> brackets to make them into HTML tags.
*April 29, 2016*

**@Reiny**

Some of these blokes use log6^(1/36) to mean log6(1/36), which is indeed -2 #2 ln(1/e^4) = ln(e^-4) = -4 #3 ok #4 ok #5 since the other logs specify a base, this must be just a common log, and log 10^8 = 8 #6 log3√3 = log33^(1/2) = 1/2
*April 29, 2016*

**math -.-**

just divide each term in the numerator by the denominator: (x^4+5x^3)/(3x^2) = x^4/(3x^2) + 5x^3/(3x^2) = x^2/3 + 5x/3 or, 1/3 x^2 + 5/3 x
*April 29, 2016*

**Calculus**

πr^2h = 200, so h = 200/πr^2 The cost is thus c = .02*2πrh + .06*2πr^2 = .04πr(200/πr^2) + .12πr^2 = 8/r + .12πr^2 dc/dr = -8/r^2 + .24πr dc/dr=0 when r ≈ 2.2 Now just crank out h and c
*April 29, 2016*

**Math Check**

Surely there should be π somewhere in the answer. And what is the shape of the solid? You have given a semi-circular base, but no other info.
*April 28, 2016*

**Math**

n+d+q = 22 5n+10d+25q = 175 n = 3d now crank it out
*April 28, 2016*

**Math**

just solve for t in -16t^2 + 120t+ 50 = 10 probably the quadratic formula will be helpful.
*April 28, 2016*

**Calculus**

Draw a side view. If the cylinder has radius r, and height h, then r^2+h^2 = 6^2 the volume of the cylinder is thus v = πr^2h = πr^2√(36-r^2) now just find dv/dr=0 and thus the maximum volume's radius.
*April 28, 2016*

**Calculus**

if the squares are of side x, then v = x(12-2x)^2 = 4x^3-48x^2+144x dv/dx = 12x^2 - 96x + 144 = 12(x-2)(x-6) so, v has a max at x=2.
*April 28, 2016*

**Geometry**

if the lateral area is 232pi, then pi rs = 232pi rs = 232 s^2 = r^2+h^2, so s = √(r^2+441) r√(r^2+441) = 232 r = 9.97 So, let's say r=10. Then a = 10√541 = 232.6 If the total area is 232pi, then use the adjusted formula for that.
*April 28, 2016*

**Maths**

the scale is 1cm:5km the area scale is thus 1cm^2:25km^2 multiply that by 405, and you have 405cm^2 : 10125km^2
*April 28, 2016*

**homework**

I assume the axis of rotation is the line y=49. So, the shells have thickness dy, and v = ∫[0,49] 2πrh dy where r = 49-y and h = (7+√(49-y))-(7-√(49-y)) v = ∫[0,49] 2π(49-y)((7+√(49-y))-(7-√(49-y))) dy = 26891.2π you can check...
*April 28, 2016*