Wednesday

July 27, 2016
Total # Posts: 42,121

**Math**

the area of a circle with radius 15 is pi r^2 = 225pi
*May 31, 2016*

**math**

that would be 3(2y) + 1(y+3) = 6y+y+3 = 7y+3
*May 31, 2016*

**math plz help ;-;**

the equation of your line is y-8 = -3(x+4) Now massage that to see which choice it matches.
*May 31, 2016*

**math plz help ;-;**

plug in the points for your equation: (-4,8): -3(8) = -4-4 -24 = -8 BZZZT! Try again (1,-7): -3(-7) = 1-4 21 = -3 BZZZT! Not that one either.
*May 31, 2016*

**Physics**

start: a = v/t = 36km/hr/10s = 3.6 km/hr/s you can convert km/hr to m/s if you desire the usual m/s^2 value. to stop takes twice as long, so a = 1/2 as much.
*May 31, 2016*

**maths**

x+y = 14 1/x + 1/y = 7/20 Now just crank it out.
*May 31, 2016*

**maths**

3/7 * 3/7
*May 31, 2016*

**Math**

42
*May 31, 2016*

**math plz check ;-;**

none of those choices matches the equation. It can be written as -3y = 12-2x or y = 2/3 x - 4 And of course, no graphs are shown here...
*May 31, 2016*

**Algebra 1**

since distance = speed * time, 7(x+60) = 9(x-60)
*May 30, 2016*

**Math**

35000 * (1.05^40-1)/(1.05-1)
*May 30, 2016*

**math**

well, he has to make 9 cuts, so 9*10
*May 30, 2016*

**Math2**

I assume the bowl is circular, with a parabolic cross-section. If we let the vertex be at (0,0) then we have y = kx^2 where y(15) = 9 so, y = 1/25 x^2
*May 30, 2016*

**Math1**

it hits the ground when h=0. So, just solve -12(t+2)^2 + 120(t+2) = 0 or, if you prefer t^2-6t-16 = 0
*May 30, 2016*

**Math (Calculus)**

The slope at any point (x,y) is dy/dx = (dy/dt)/(dx/dt) = (2t+1)/(2t) We want to find t such that the tangent line is y=kx (that is, it passes through the origin). At that point y = kx, so k = (t^2+t)/(t^2+3) That means that we need the slope = k, or (2t+1)/(2t) = (t^2+t)/(t^2...
*May 30, 2016*

**Algebra**

we want x,y and k such that (x+k)(y+k) = 2xy xy+kx+ky+k^2 = 2xy y = k(x+k)/(x-k) So, there are lots of solutions, with varying dimensions. x y k xy (x+k)(y+k) 2 3 1 6 12 3 10 2 30 60 and so on
*May 30, 2016*

**Math - volume about y axis**

The region is bounded above by the x-axis. That is, it is the region below the axis. Since it is symmetric about the y-axis, we really only need to rotate the region from 0 to 1. So, using discs, v = ∫[0,1] πr^2 dy where r = x, so r^2 = x^2 = y+1 v = ∫[0,1] &#...
*May 30, 2016*

**Algebra**

arrange the scores in order. The 80th %ile is that score above 80% of the others. Your score is the 78th %ile, so it should be close to the score found above.
*May 30, 2016*

**math plz help ;-;**

-2(x-3)+4=8x-20 lose the parentheses: -2x+6+4=8x-20 move all the x stuff to the left, and the numbers to the right: -2x-8x = -20-6-4 simplify each sides: -10x = -30 divide by -10 to get x alone: x = 3 check that it fits the original equation: -2(3-3)+4=8*3-20 0+4 = 24-20 yep
*May 30, 2016*

**math @**

finally discovered the difference between () and ||, eh, Stefan?
*May 30, 2016*

**Math**

L+K=900 L=K+400
*May 30, 2016*

**algebra**

Since the tax is only on the dress, I'd say 10.50 + 12.00*1.09
*May 30, 2016*

**algebra 1**

Have you talked to "Ann"? http://www.jiskha.com/display.cgi?id=1464626459
*May 30, 2016*

**Math**

This is strange. D(0) = 21 D(8) = 9 D(14) = 0 I am having a hard time relating T values with travel time. Does he server the ball at T=0? Can't be, since then he is 21 ft from the ball. The distance is always decreasing, so clearly the equation describes the return trip ...
*May 30, 2016*

**Math**

Draw a diagram. The measures of the angles of triangle PQR are P:43 Q:83 r:54 Now just use the law of sines to get PR: PR/sin83 = 10/sin54
*May 30, 2016*

**Maths**

Draw a diagram. The measures of the angles of triangle PQR are P:43 Q:83 r:54 Now just use the law of sines to get PR: PR/sin83 = 10/sin54
*May 30, 2016*

**Physics**

I haven't even checked the math, but a water level of 3.26 micrometers? That's barely a film, let alone a depth. give me a break! When you get an answer, at least do a sanity check.
*May 30, 2016*

**Math**

well, if their ages now are x and y, then their ages in 7 years will be x+7 and y+7. Now just write your equations and solve them.
*May 30, 2016*

**Math**

I assume you meant e^(x-10) = 10 x-10 = ln10 x = 10+ln10
*May 30, 2016*

**algebra**

well, just substitute 3 for the b and evaluate the result. 12*(2/3)^3
*May 30, 2016*

**maths**

(8+15+x)/3 = 12 now just find x
*May 30, 2016*

**Maths**

each interior angle is the supplement of its exterior counterpart. So, the interior angles are 180-60, 180-130, 180-82 The sum of the interior angles is 360, so ...
*May 30, 2016*

**maths**

from the garbled language, I assume you want x where (8+15+x)/3 = 12
*May 30, 2016*

**Maths**

the other side of the triangle is √(7^2-5^2) = √24 now use that to evaluate the cosine
*May 30, 2016*

**Math**

h/20 = tan45°
*May 30, 2016*

**mathss**

well, clearly a,b,c are not zero at least one of d,e and f,h is zero.
*May 30, 2016*

**physics**

start: a = v/t = 30km/hr/10s = 3km/hr/s If you like, you can convert km/hr to m/s and your answer will be the usual m/s^2. since it takes twice as long to stop, the acceleration is half as much. And negative.
*May 30, 2016*

**Physics**

the question has been garbled.
*May 30, 2016*

**SCIENCE PLZZ HELP!!!**

#1 and 3 are ok, but the difference between speed and velocity is direction. The magnitude of the velocity is the speed, but the velocity is a vector quantity, which is a directed speed.
*May 29, 2016*

**Math - integrals**

x-1 = √5 tan(u) x^2-2x+6 = 5sec^2(u) dx = √5 sec^2(u) du Now plug it into the integral 5∫3 1/(x^2-2x+6)^3/2 dx 5∫3 1/(5sec^2(u))^3/2 √5 sec^2(u) du 5∫3 (√5 sec^2(u))/(5√5 sec^3(u)) du 5∫3 1/(5 sec(u)) du ∫3 cos(u) du ...
*May 29, 2016*

**Math - integrals**

x^2-2x+6 = (x-1)^2 + 5 if x-1 = √5 tanu then (x-1)^2+5 = 5tan^2u+5 = 5sec^2u and dx = √5 sec^2u du plug those into the integral and watch things simplify
*May 29, 2016*

**Math3**

namely, their average: (a+b)/2
*May 29, 2016*

**Math2**

(A) 3 colors, so 4 socks must have two the same (B) Depends on how many pairs of each color. Worst case is if there is only one red pair. Then he might need to take them all.
*May 29, 2016*

**Math (pre-celc 12)**

If it is supposed to be 2 * 5^x = 3^(x+1) then just take logs as usual: log(2) + xlog(5) = (x+1)log(3) log2 + xlog5 = xlog3 + log3 x(log5-log3) = log3-log2 x = (log3-log2)/(log5-log3) or, if you prefer, x = log(3/2)/log(5/3)
*May 29, 2016*

**Maths**

(5+2x)(12+2x)
*May 29, 2016*

**Integral calculus**

A good discussion is found at http://math.stackexchange.com/questions/1379341/how-to-find-the-surface-area-of-revolution-of-an-ellipsoid-from-ellipse-rotating google is your friend.
*May 29, 2016*

**Pre calculus**

you need to add/subtract rows and their multiples so the resulting matrix consists of a diagonal of 1's. To start, subtract R1 from R2: x + y + z = -5 0 - 2y + 2z = 4 4x + y + z = -2 Now subtract 4*R1 from R3: x + y + z = -5 0 - 2y + 2z = 4 0 - 3y - 3z = 18 Now work on the...
*May 29, 2016*

**Math**

tan(A) = (25-1.5)/23.5
*May 28, 2016*

**math**

100,000 * (1+.10/12)^(12*10) = 270,704.15
*May 28, 2016*

**math**

The sum-to-product formulas might be helpful here. I think you'll find that the imaginary part is zero.
*May 28, 2016*

**algebra 2**

If the downstream rate is x, then since time = distance/speed, 75.5/x + 75.5/(x-5) = 15
*May 28, 2016*

**maths**

add up the x- and y-components of the vectors, then convert the result back to distance and bearing.
*May 28, 2016*

**Algebra**

this does not factor. The discriminant is negative. b^2-4ac = 36-4*9*8 = -252 The quadratic has two complex roots.
*May 28, 2016*

**Algebra**

notice the coefficients are all 2 or 1. So, I'd try 2x^3y^3-x^2y^3 + 2x-y x^2y^3 (2x-1) + 2x-y (x^2y^3+1)(2x-y)
*May 28, 2016*

**Math**

That would be (6.3/9) * (10^6/10^2) = 0.7 * 10^4 = 7.0 * 10^3 = 7000
*May 28, 2016*

**Math**

5(4-y) = 5*4 - 5*y = 20 - 5y
*May 28, 2016*

**Physics**

=? = ↓ ß ≟ xx
*May 28, 2016*

**Algebra**

1/a + 1/m + 1/z = 1/2 so, 1/5 + 1/6 + 1/z = 1/2 1/z = 2/15 1/a + 1/z = 1/5 + 2/15 = 1/3 so, Ann and Zack take 3 hours to do the job together.
*May 28, 2016*

**math**

original fraction is n/d (n-5)/(d-3) = 1/2 (n/2)/(d+7) = 1/4 Solving these gives n = (d+7)/2, not a single unique solution. Naturally, d cannot be 3, but it is odd, and n must be even, so we can have original fractions of 6/5, 8/9, 10/13, ... or, in general, (2k+4)/(4k+1) for ...
*May 28, 2016*

**Calculus**

You should probably solve for y using the quadratic formula. It comes out y = √(2x^3+11) - 1
*May 27, 2016*

**Calculus**

http://www.wolframalpha.com/input/?i=%E2%88%AB[1,4]+%CF%80%2825-%285-1%2Fx%29^2%29+dx Did you use ln(x)? If you used base 10, that would be wrong. As you get along in math, you will find the serious folks use log(x) to mean ln(x). It's kind of confusing if you're not ...
*May 27, 2016*

**Calculus**

y = 1/x, the x-axis, and the lines x = 1 and x = 4 is rotated about: a) the x-axis: just treat the volume as a stack of thin discs, each of area πr^2, where r=y. So, v = ∫[1,4] π(1/x)^2 dx = 3π/4 = 2.35 rotating about the line y=5, you have to use washers...
*May 27, 2016*

**Pre-Algebra (Check my answers)**

Sorry. #1 was correct for 5*3^n
*May 27, 2016*

**Pre-Algebra (Check my answers)**

as shown, the table does indeed not represent a linear function. If the 8 was a typo and you meant 9, then y = 8x+9
*May 27, 2016*

**Pre-Algebra (Check my answers)**

try using 3^n instead of 3n, which looks like multiplication. |-1| = 1, so f(2) = 2*1*2 = 4 If you actually meant (-1)^x, then f(2) = 2*(-1)^2 = 2*1 = 2
*May 27, 2016*

**Math Help Please? (check my work?)**

#1 Why do you start adding 10 each time, when the sequence clearly did not? Try adding 9 instead. #2 ok
*May 27, 2016*

**Algebra - oops**

oops. I forgot to carry the denominator along. a/bc - 2b^2/ac - 4c^3/ab
*May 27, 2016*

**Algebra**

why all the words? That's what math symbols are for. 2a^2 - 4b^3 - 8c^4 ------------------------ = a-2b^2-4c^3 2abc
*May 27, 2016*

**Calculus**

What do you mean by "finding" x,y,z? (x-2)/-1 = (y-5)/5 = (z-7)/7 or x = 2-t y = 5+5t z = 7+7t When x = -3, t=5, so the line passes through (-3,30,42)
*May 27, 2016*

**algebra**

I understand csc(A) but have no idea what the Y6 is supposed to mean. If the Y is a typo, then I suppose you could have meant 3csc(A)-6=0 csc(A) = 2 or, if you prefer, sin(A) = 1/2 A = pi/6 or 5pi/6
*May 27, 2016*

**math**

C cannot be right, since it only has two terms. Take another stab at it and see whether you can come up with (A)
*May 27, 2016*

**math PsyDAG**

well, 1.1*3 = 3.3 so A and B are clearly wrong. when multiplying, add the powers.
*May 27, 2016*

**Geometry**

good work
*May 27, 2016*

**MATH - CALC**

f = x^3-3x^2 = x^2(x-3) f' = 3x^2-6x = 3x(x-2) f" = 6x-6 = 6(x-1) Now just apply the features you know: f=0: x-intercepts x=0: y-intercept f'=0 ==> critical point f"=0 ==> inflection f'>0 ==> increasing f">0 ==> concave up
*May 27, 2016*

**math plz help**

let's do the math in base 10. 325 = 1710 1314 = 2910 x+y+z = 3*17 = 51 x+y = 29 now you can get z, and convert to base 6.
*May 27, 2016*

**math**

well, the apothem is 1/2 as long as a side, right? doesn't matter whether it's inscribed in a circle.
*May 27, 2016*

**math**

75/.06
*May 27, 2016*

**math**

cost = 2400/12*15 revenue = 1350*4 + (2400-1350)/5*8 % gain or loss = ((revenue-cost)/cost - 1)*100
*May 27, 2016*

**maths**

whoa - too many = signs!
*May 26, 2016*

**math**

by now I'm sure you have figured out what it means to substitute a value, but just in case, that gives you 15+9d = 69 Now you can get d, and thus a+3d.
*May 26, 2016*

**Calculus**

dy/dx = √[(x^3)(y)] dy/dx = x^(3/2) y^(1/2) y^(-1/2) dy = x^(3/2) dx y^(1/2) = 1/5 x^(5/2) + c Now if you square both sides, you do not get x^5/25 + c, since (a+b) is not a^2+b^2 But proceeding to find c, √2 = 1/5 + c c = √2 - 1/5 Now square both sides and ...
*May 26, 2016*

**Math**

just see which pair of values does not work. replace x and y with the values given: -4(-9)-9 = 36-9 = 27 so it is a solution now try the others.
*May 26, 2016*

**Math**

f" < 0 indicates it is concave down. f' > 0 means that the graph is rising. So, an example would be a parabola which opens downward. So, the tangent line at x=0 will be above the curve. draw a graph and it should be clear now.
*May 26, 2016*

**Calculus**

right on
*May 26, 2016*

**calc**

I find it easier to leave the π in the answer. That way it's exact, instead of some decimal approximation. In any case, I got (a) 20π/9 (b) 125π/27 #2 27π/2
*May 26, 2016*

**calc**

#1 (a) discs ∫[0,5] π((5-y)/3)^2 dy (a) shells ∫[0,5/3] 2πx(5-3x) dx (b) shells ∫[0,5] 2π(6-y)(5-y)/3 dy #2 discs ∫[0,9/4] π(((3+√(9-4x))/2)^2-(((3-√(9-4x)))/2)^2) dx shells ∫[0,3] 2πy(3y-y^2) dy
*May 26, 2016*

**Math**

draw a side view. r^2 = 6^2 + 4^2 A = 4pi r^2
*May 26, 2016*

**Math**

Recall Euler's Formula: V+F-E=2 and check out https://en.wikipedia.org/wiki/Icosidodecahedron
*May 26, 2016*

**Math-Calc**

f = 6x(3-2x)^3 f' = 6(3-8x)(3-2x)^2 f(1) = 6 f'(1) = -30 so, the tangent line is y-6 = -30(x-1) see at http://www.wolframalpha.com/input/?i=plot+y%3D6x%283-2x%29^3,+y%3D-30x%2B36
*May 26, 2016*

**Math topic help**

well, your answer is the balance after about 15.6 years, so I suspect it's not right. What did you intend?
*May 26, 2016*

**Math (Vectors)**

looks good to me.
*May 26, 2016*

**Math (Vectors)**

if they are parallel, then the lengths will be proportional. 10/2 = 5, so the ratios of the other dimensions will also be 5:1
*May 26, 2016*

**Algebra II**

No, I didn't answer the question, but I gave you enough information so you could complete the solution. Sorry to be jerk-ish.
*May 26, 2016*

**Algebra II**

So, with the triangle drawn as you have said, with the right angle on the x-axis, you just have a scaled up 3-4-5 triangle, with sides 15,20,25. sinX = 15/25 = 3/5 cosX = 20/25 = 4/5 tanX = 15/20 = 3/4 Looks like time to review your basic trig functions and the Pythagorean ...
*May 26, 2016*

**College Algebra/Drug Calculations**

just multiply numerators and denominators 5/6 * 2/3 = (5*2)/(6*3) = 10/18 = 5/9
*May 26, 2016*

**Calc-Math**

just look up the values on the graph and multiply or divide the numbers.
*May 26, 2016*

**maths-word problem**

if the digits are t and u, then 10t+u = 6(t+u)-1 t-u = 1 Looks like 65 to me.
*May 26, 2016*

**maths**

the opposite side is √(y^2-x^2) so csc(Ø) = y/√(y^2-x^2)
*May 26, 2016*

**calculus-again frustrated-looking for steve asap**

sinh(y)=[(4sinh(x)-3)/(4+3sinh(x))] cosh(y) y' = (25cosh(x))/(4+3sinh(x))^2 cosh^2(y) = 1+sinh^2(y) = 25cosh^2(x)/(4+3sinh(x))^2 so, cosh(y) = 5cosh(x)/(4+3sinh(x)) y' = 5/(4+3sinh(x))
*May 26, 2016*