Sunday

November 29, 2015
Total # Posts: 35,841

**math**

Nope. Draw the quadrants. QI is where things are simplest: x>0 and y>0. That is the upper right. Then, the quadrants are numbered counter-clockwise. google is your friend. use it. It will show lots of images, all neatly labeled.
*October 12, 2015*

**Logarithm help**

you know that the domain of log(n) is n>0 So, the domain of log(2-x) is 2-x > 0 x < 2 similarly for the even easier 2nd one.
*October 12, 2015*

**math**

(22/17) * 436
*October 12, 2015*

**math**

since 15 = 5*3, 3:5 = 5*3:5*5 = 15:25
*October 12, 2015*

**math**

since time = distance/speed, d/5 = d/6 + 1/6 Now just find d.
*October 12, 2015*

**math**

you want the line through (0,5) that has slope 2/3. So, y-5 = 2/3 x
*October 12, 2015*

**math**

The slopes must be different to have a unique solution. So, we need -k ≠ -1/k k^2 ≠ 1 To find that solution, multiply the 1st equation by 2k and you have 2k^2 x + 2ky = 10k 2x + 2ky = 3 Now subtract to eliminate y, and you have (2k^2-2)x = 10k-3 x = (10k-3)/(2k^2-2...
*October 12, 2015*

**geometry**

If the angle θ between the sides is acute, then (9 sinθ)(10)/2 = 36 θ = 53.13° and the third side s comes from s^2 = 9^2+10^2-180cos53.13° s = √73 The angle might be obtuse. A little thought should show how to handle that case.
*October 12, 2015*

**Chemistry**

well, I looked up the mass of a neutron, and found 1.675*10^-24 g. So, using that and your volume, we have (1.675*10^-24 g)/(4/3 π *10^-39 cm^3) = 4.00*10^14 g/cm^3 so, a .01cm ball would have a mass of (4/3 π * 10^-6 cm^3)(4.00*10^14 g/cm^3) = 1.68*10^6 kg You must ...
*October 11, 2015*

**Math**

there is only one cubic which passes through all four of those points: p(x) = x^3-3x^2-4x+5 no need to approximate p(1), since you can see that it is -1. As for p(x) = 0, you'll have to approximate that, using a graph, or some other numeric tool you have studied, such as: ??
*October 11, 2015*

**physics**

v(t) = at s(t) = 1/2 at^2 at = 17, so a = 17/t 1/2 (17/t)t^2 = 1.1, so t = 0.129
*October 11, 2015*

**math**

Ok. How about this. If the e.c. is tipped up at an angle x, we need 48cos(x) + 48sin(x) <= 35 48√2 cos(π/4 - x) <= 73 Now, the e.c. just fits across when x = 10.4° So, any angle higher than that will still fit, since the top corner will never be higher ...
*October 11, 2015*

**math**

Oh, yeah. I didn't think of tipping it on its side first. Well, try the same equations, but substituting in 48 for 60. Good catch.
*October 11, 2015*

**math - ooops**

Well, that solution won't work, because the top corner of the tipped e.c. is higher than 73". So, I think it will not fit.
*October 11, 2015*

**math**

Of course you know where to start. Draw a picture! You're clearly going to have to tip the entertainment center. So, suppose the bottom corner is x from one side of the door, and y from the other side. On the x side, the base of the e.c. reaches up to a height z. Since the...
*October 11, 2015*

**algebra**

sorry - files on your C: drive are not available to us.
*October 11, 2015*

**Algebra 2**

No solution. If you rearrange things a bit, you have 6x-2y=10 3x-y=2 Double the 2nd equation and you have 6x-2y=4 There's no way 6x-2y can be both 10 and 4.
*October 11, 2015*

**physics**

you omitted it, but if the initial velocity is v0, then the height and velocity at time t are h(t) = v0*t - 4.9t^2 v(t) = v0 - 9.8t You left out several words, but with the above equations, you should be answer whatever questions were asked.
*October 11, 2015*

**Physics**

the horizontal component is v cosθ So, plug in your v and θ.
*October 11, 2015*

**Geometry**

First step: PLOT THE POINTS! DOF must form a straight line with OE, which is the line y=x/5 So, just look at the points where that holds, and decide which angles are adjacent to DOE. Hint. <DOF, with F (10, 2) does not work, because that is just the same angle as DOE, not ...
*October 11, 2015*

**physics**

I don't see how one vine can be vertical and the other at an angle. If there's any tension in the side vine, it will move the "vertical" vine to the side.
*October 11, 2015*

**algebra**

If we use x liters of 20% acid, then we use (300-x) liters of 35% acid. So, keeping track of the amount of acid in each part, we have .20x + .35(300-x) = .25(300)
*October 11, 2015*

**Math**

If there are d dark and m milk, d = 2m .25d + .15m = 10.00
*October 11, 2015*

**math**

recall that F = ma so, plug in your numbers. Watch the units.
*October 11, 2015*

**Algebra badly need help**

Just solve (x+1/x)^2=3 x^2+2+1/x^2 = 3 x^2 = (1±√3 i)/2 A little playing around will show you that that is the same as x = ±i^(1/6) So, since the polynomial is just a bunch of powers of x^6 = i, we have x^72+x^66+x^54+x^36+x^6+1 = i^12 + i^11 + i^9 + i^6 + ...
*October 11, 2015*

**Math**

Not sure what steps you took (I can never understand students who say they got something wrong and then don't show how they did it!), but just start off with your expression, and start substituting values, from the inside out: h(g(1.5)) = h(8-3*1.5) = h(8-4.5) = h(3.5) = 2...
*October 11, 2015*

**algebra**

Since we're dealing with the sides of a right triangle, I'd say the first tool to try is the Pythagorean Theorem. If the height is h, the the length of the leaning part of the bamboo is 10-h. So, you have a right triangle where 3^2+h^2 = (10-h)^2 Now just find h.
*October 11, 2015*

**math**

v(t) = -9.8t + 4.9 so, when is v=0? Use that value of t to find s(t)
*October 11, 2015*

**math**

26 = 5*5 + 1
*October 11, 2015*

**Maths**

clearly, that would be $25.00*8*6
*October 11, 2015*

**math**

c:b = 2:5 b:r = 1:3 = 5:15 So, c:b:r = 2:5:15 So, if r=30, c:30 = 2:15 c = 4
*October 11, 2015*

**Pre-calculus**

to double in 8 years, you need interest rate r where (1+r)^8 = 2 r = 0.0905 or 9.05% So, with initial amount P, you have P(1+.0905)^10 = 1700 Now just solve for P.
*October 11, 2015*

**Math**

(1) no (3) starting at (10,1.5), d decreases by .5 every 10 minutes d = 1.5 - 0.5((t-10)/10) = 3/2 - (t/10 - 1)/2 = 3/2 - t/20 + 1/2 = 2 - t/20 or, if you like decimals, d = 2.0 - 0.05t
*October 11, 2015*

**physics**

what's the trouble? Viewing from the side, the plane has moved from (350 cos31°,350 sin31°) to (770 sin154°,770cos154°) That is, from (300,180) to (-692,338) Now you can crank it out, right?
*October 11, 2015*

**Math (algebra)**

.235 = .2 + 0.35 You know that .35 = 35/99, so .035 = 35/990 .235 = 1/5 + 35/990 Now just add those fractions.
*October 11, 2015*

**PreCalc**

you have a phase shift only. The period is unchanged at 2π.
*October 11, 2015*

**calculus**

f(-1,y) = y^3+9y+9 f(1,y) = y^3-9y+9 f(x,-1) = 9x^2+9x-1 f(x,1) = 9x^2-9x+1 So, check for maxima of those on the interval [-1,1] Then, you have ∂f/∂x = 18x-9y ∂f/∂y = 3y^2-9x Extrema occur when both these are zero. Along the line ∂f/∂x=0 or ...
*October 10, 2015*

**ALGEBRA**

certainly velocity. And since acceleration is a change in velocity, I'd have to go with: Both.
*October 10, 2015*

**Algebra**

don't grok the ƒ{ part, but as for the rest, since complex roots occur in conjugate pairs, p(x) = (x-2)(x-(1+i))(x-(1-i)) = (x-2)((x-1)-i)((x-1)+i) = (x-2)(x^2-2x+1 + 1) ...
*October 10, 2015*

**Algebra**

You want t where 2^(t/6) = 20000/11000 = 1.818 t = 6log1.818/log2 = 5.174 hr Makes sense, since in 6 hrs it will double to 22,000
*October 10, 2015*

**Math**

If the tire has width w, then the inner radius is 2w, and the entire area has radius 3w. So, π(9w^2-4w^2)/π(9w^2) = 5/9 = .5555 or 56%
*October 10, 2015*

**MATH**

Not quite... The cost C is C = 8(2W+L)+15L since LW=400, = 8(2(400/L)+L) + 15L = 8(800/L + L) + 15L = 6400/L + 23L
*October 10, 2015*

**Calculus**

Looks like A&B to me. I mean, have you tried actually drawing a vector xi+yj, with its tail at (x,y)?
*October 10, 2015*

**Calculus**

using implicit differentiation, 2(x^2+y^2)(2x+2yy') = 16y' x(x^2+y^2) + y(x^2+y^2)y' - 4y' = 0 y' = -x(x^2+y^2)/(x^2y + y^3 - 4) So, the slope is vertical where x^2y-y^3 -4 = 0 y(x^2-y^2) = 4 Not easy to solve in general, but a little playing around gives x...
*October 10, 2015*

**Physics i need hlp i have exams next weeks that is why i post the question that which i cannot do**

c = av + b/v cv = av^2 + b av^2 - cv + b = 0 now just use the quadratic formula to find v. There will only be a maximum cost if a<0, since then the parabola will open downward.
*October 10, 2015*

**Geometery i need help**

well, let's assume the octagon is just a regular polygon. It consists of eight isosceles triangles, each with two sides of 12, and a vertex angle of θ=45°. So, if the base of each triangle (one side of the octagon) is s, (s/2) = 12 sin(θ/2) Now just plug in ...
*October 10, 2015*

**Geometery i need help**

clearly the band forms an octagon (with rounded corners) with a diameter of 3 oranges, or 24mm. Now play around a bit with the dimensions of an octagon.
*October 10, 2015*

**trig**

so, cotθ = (√7)/2, meaning x = √7 y = 2 since cotθ = x/y
*October 10, 2015*

**trig**

recall that sinθ = y/r cosθ = x/r tanθ = y/x r^2 = x^2+y^2 So, in QIII, where x and y are both negative, tanθ = √5 = √5/1, so y = -√5 x = -1 r = √6 cotθ = x/y = 1/√5 secθ = r/x = -√6 For #1, do you mean (1/2) &#...
*October 10, 2015*

**maths**

clearly, the bonuses are (2/3 * 105), 105, (4/3 * 105) So, the total is 3*105 = 315
*October 10, 2015*

**Calculus**

You are correct. I fear that the answer choices relate to a different problem, since with a simple power, no logs will arise.
*October 10, 2015*

**How many of these number maths??**

(a) 7!/(3!2!) = 420 (b) If 3,4 are used, that leaves 5!/3! = 20 (c) divisible by 5 means that 3,4,5 are used, leaving 4!/3! = 6 (2) divisible by 2 means that 2,3,4 are used, leaving 4!/2! = 12
*October 10, 2015*

**precalculus**

better review your trig functions sinθ = x/r cosθ = y/r r^2 = x^2+y^2 You have a 3-4-5 triangle. Draw the dang thing! x = -3, so cosθ = -3/5
*October 10, 2015*

**precalculus**

in QII, x is negative, so cosθ = -3/5 The rest is easy, right?
*October 10, 2015*

**maths**

as usual, draw a diagram. The tower's height is 20 + 20 cot30° * tan60° = 20(1+tan^2 60°) = 20 sec^2 60° = 20 * 4 = 80
*October 10, 2015*

**geometry**

Just knowing the four sides of a quadrilateral does not help find the area. As you know, if you build such a figure, you can squash it around, because the joints are not rigid. Unless you know some of the angles or something, you cannot determine the area.
*October 10, 2015*

**MATH!**

D is the best bet. It appears not to repeat.
*October 10, 2015*

**Calculus i need help**

x^2 y + sin(xy) + y^4 x^2 + 3xy = 5 Just use the product rule and the chain rule: d/dx(x^2 y) = 2xy + x^2 y' d/dx(sin(xy)) = cos(xy) (y + xy') d/dx(x^2 y^4) = 2xy^4 + 4x^2y^3 y' d/dx(3xy) = 3y + 3xy') So, add them all up, collect the y' stuff, and it's ...
*October 10, 2015*

**Maths**

well, how long does it take to go 200 miles, if you travel 100 miles each hour?
*October 10, 2015*

**Pre-Calculus**

you must have some ideas about some of these topics. At the very least, use a graphing utility to see what the graph looks like, and most of the answers should become clear immediately.
*October 10, 2015*

**Grrr.**

Oops. I thought we were doing serious math homework here. My bad :-( Like 16/64 = 1/4 because you can just cancel the 6's.
*October 10, 2015*

**Prove also need help**

this is nonsense. Of course 13/12 is not 2 You want to fix up your question a bit?
*October 10, 2015*

**Trigonometry prove need help**

tan 45-b = (tan45-tanb)/(1+tan45 tanb) = (1-tanb)/(1+tanb) square that, and you have (1-2tanb+tan^2b)/(1+2tanb+tan^2b) = (sec^2b-2tanb)/(sec^2b+2tanb) = (1/cos^2b - 2sinb/cosb)(1/cos^2b + 2sinb/cosb) = (1-2sinb cosb)/(1+2sinb cosb) = (1-sin2b)/(1+sin2b) I think you have a typo...
*October 10, 2015*

**pre calc**

Tough to describe the function, when it not given...
*October 10, 2015*

**pre calc**

first, put the words into algebra. If there is x at 5.5% and y at 11%, you have: x+y <= 130000 x >= 40000 y <= 60000 maximize p = .055x + .11y Now just use your favorite linear programming tool.
*October 10, 2015*

**pre calc**

You've posted a bunch of these. Some have been done for you. No ideas yet?
*October 10, 2015*

**calculus**

that is the definition of an inverse function. For any f(x), f(f^-1(x)) = x f^-1(f(x)) = x if f is single-valued, which sinh(x) is. If you take sinh^-1 of both sides, you have sinh^-1(sinh(sinh^-1(x))) = sinh^-1(x) but that only helps if you already know that sinh^-1(sinh(u...
*October 10, 2015*

**Math**

for the short diagonal x, x^2 = 5^2+10^2-2*5*10 cos50° for the long diagonal y, y^2 = 5^2+10^2-2*5*10 cos130°
*October 10, 2015*

**calculus**

y' = (x^2-10x+16)/(x-5)^2 y" = 18/(x-5)^3 So, max/min if y'=0 max if y" < 0, min if y" > 0 The rest is just algebra.
*October 10, 2015*

**calculus**

Assuming the two walls meet at a right angle, x+y = 30 a = xy = x(30-x) this is just a parabola, with vertex at x=15. As usual, the area is a square. You can take the derivative if you want, but no calculus needed here...
*October 9, 2015*

**math**

just multiply both elements by the same nonzero value. Think of fractions.
*October 9, 2015*

**Maths**

why take the derivative to find t? It won't help here. Just expand and solve: 0.2t^3 - 2.84t^2 + 9.282t = 2.4 + 0.75t 0.2t^3 - 2.84t^2 + 8.532t - 2.4 = 0 Now you can fall back on graphical methods, or Newton's method or something you have in your bag of tricks.
*October 9, 2015*

**math**

sin = y/r cos = x/r tan = y/x r is always positive In QIV, x is positive and y is negative In QII, x is negative and y is positive So, in QII, cos,tan,cot,sec are negative in QIV, sin,tan,cot,csc are negative only tan,cot are negative in both.
*October 9, 2015*

**pre calc**

S1: 2 is a factor of 1^2+7*1 and so on. Clearly, since if n is odd, n^2 and 7n are both odd, so their sum is even if n is even. n^2 and 7n are both even, so their sum is even. Or, if n is odd (2k+1)^2+7(2k+1) = 4k^2+4k+1+14k+7 = 2(2k^2+9k+4) and similarly for even n=2k
*October 9, 2015*

**math**

4x+5y=10 2x+3y=8 x = -5 y = 6 Now just find 3x+4y
*October 9, 2015*

**Pre-calculus**

9^(2x-6)=6^(8x-5) (2x-6) log9 = (8x-5)log6 (2log9)x - 6log9 = (8log6)x - 5log6 x = (6log9 - 5log6)/(2log9 - 8log6) you can massage that a few ways, but that's the idea
*October 9, 2015*

**Math**

see http://www.jiskha.com/display.cgi?id=1444227515
*October 9, 2015*

**Trig**

after t seconds, Michael has gone 2t ft and Alyssa has gone 3t feet. So, the distance z between them using the law of cosines, is found using z^2 = (2t)^2 + (3t)^2 - 2(2t)(3t)cos60° so, plug in your time (in seconds!) and find z.
*October 9, 2015*

**precalc**

looks like our calculator agree!
*October 9, 2015*

**precalc**

Draw triangle LBM where the ball is at B; angle B is 158°. Using the law of sines, if the ball's distance from M is x, then x/sin9° = 23/sin158°
*October 9, 2015*

**Algebra**

(x+3)/(x+8)-(x-8)/(x-3) =[(x+3)(x-3) - (x-8)(x+8)]/[(x+8)(x-3)] =[(x^2-9)-(x^2-64)]/[(x+8)(x-3)] = (-9+64)/[(x+8)(x-3)] = 55/[(x+8)(x-3)] You probably forgot to use parentheses to keep track of quantities. But aside from that, what did you do with your denominator?
*October 9, 2015*

**geometry**

since DB bisects the angle ABC, the two smaller angles are equal. So, 12x+24 = 20x+16 So, find x, and you can figure the angles.
*October 9, 2015*

**MATH HELP PLEASE**

4718(1+.05/4)^(4n) = 5323 (1.0125)^(4n) = 1.1282 4n = log(1.1282)/log(1.0125) n = 2.4 or, roughly 2 1/2 years
*October 9, 2015*

**Math**

-2+5x-7=3x-9+2x collect constants and x's on each side: 5x-9 = 5x-9 Since both sides are the same, it will be true for any value of x at all!
*October 9, 2015*

**Calculus**

using shells, v = ∫[0,1] 2πrh dy where r = 1-y and h = √y - y^2 v = 2π∫[0,1] (1-y)(√y - y^2) dy = 11π/30 using washers, v = ∫[0,1] π(R^2-r^2) dx where R = 1-x^2 and r = 1-√x v = π∫[0,1] ((1-x^2)^2-(1-√x)^...
*October 9, 2015*

**Maths**

|a|-3/4=-5/8 |a| = -5/8 + 3/4 = 1/8 clearly, a = 1/8 or -1/8
*October 9, 2015*

**math**

As always, draw a diagram. The parabola has vertex at (1,0) and opens to the right. The line x=2 intersects the parabola at (2,1) and (2,-1). So, you will be revolving the sort-of-semicircular chink around the line y=-2, which is below the curve. So, as you say, v = ∫2&#...
*October 9, 2015*

**Maths**

|a|-3/4=-5/8 |a| = -5/8 + 3/4 = 1/8 clearly, a = 1/8 or -1/8
*October 9, 2015*

**Maths**

L1 has intercepts at A:(-6,0) and B:(0,4) L2 has gradient -3/2 and passes through (0,4), so L2 is y-4 = (-3/2)(x-0) 3x+2y-8 = 0 L2 passes through B:(0,4) and C:(8/3,0) So, if you plot the points, you can see that you have a triangle with base along the x-axis, and its third ...
*October 9, 2015*

**Maths**

why plug in 3 for x? Why change the polynomial? x^2 + 2kx - 3k = 0 the discriminant is (2k)^2 - 4(1)(-3k) = 4k^2 + 12k For two equal roots, the discriminant must be zero. 4k^2+12k = 4k(k+3) = 0 k = 0 or -3 k=0: clearly x^2=0 has two equal roots k = -3: x^2-6x+9 = (x-3)^2 has ...
*October 9, 2015*

**Algebra**

(8+12)/(4*5)=1
*October 9, 2015*

**Algebra**

24000(1+.0685/12)^(12*18) = 82,067.98
*October 9, 2015*

**math**

It's a little easier to check if you mark your answers with the questions. Saves a lot of scrolling back and forth. However, the wily teacher opens a 2nd window to avoid that, too! #1: 20/2 – 32 – (-2)2 = 10 - 32 - (-4) = 10 - 32 + 4 = -22+4 = -18 Since that'...
*October 9, 2015*

**Algebra**

in 2 hours, Molly was 54 miles ahead. Mark was going 18 mi/hr faster, so it took him 54/18 = 3 hours to catch up. But 3 is not the answer to the question ...
*October 9, 2015*

**Physics/trig**

as you recall, the equation of motion is y = tanθ x - g/(2(v cosθ)^2) x^2 So, plugging in your numbers, y = 0.664x - 0.00806x^2 That's a parabola with vertex at x=41.2
*October 9, 2015*

**pre calculus**

This is just a standard linear programming problem. For x breakfast packages and y afternoon packages, you want to maximize p=1.50x+2.00y subject to x/3 + y/2 <= 45 2x/3 + y/2 <= 70 Now just use your favorite linear programming tools.
*October 9, 2015*

**pre calculos**

Just substitute k for n and you have Sk. Same for k+1. Sk+1: 1∙2+2∙3+...+k(k+1)+(k+1)(k+2) = [(k+1)(k+2)(k+3)]/3 Can't get much simpler than that.
*October 9, 2015*

**math**

shift 3 left scale by 2 vertically reflect in x-axis In this case, the order does not really matter. Try them and see.
*October 9, 2015*

**prasanna high school**

been done. http://www.jiskha.com/display.cgi?id=1444216542
*October 9, 2015*

**math - eh?**

Not sure what you are saying here. When you say the building and the pole are at equal angles, do you mean the top of the building, or the bottom? The top of the pole, or the bottom?
*October 9, 2015*