Thursday

July 30, 2015

July 30, 2015

Total # Posts: 32,558

**Geometry - incomplete**

depending on the labeling, any one of those could work.
*June 9, 2015*

**math**

100x + 200y <= 1400 0.6x + 0.8y <= 7.2 x/y >= 2/3 volume is .8x+1.2y Now just follow the usual optimization steps.
*June 9, 2015*

**gwalior glory school**

1st digit is 4 ones digit is 2 tens is 2+4 the number is 462 or, 451 if you only have one eye ...
*June 9, 2015*

**math**

3n+p = 70 n+3p = 50 add them together and you have 4n+4p = 120 so, 2n+2p = 60
*June 9, 2015*

**Algebra**

Just realize that a negative exponent switches the factor between numerator and denominator. (c^-3)^2d^4/d^-2 = c^-6 d^4/d^-2 = d^6/c^6 (e^4)^-3f^-5/f^-2 = e^-12 f^-5/f^-2 = e^-12 f^-3 = 1/(e^12f^3) (8^-7/8)^-2 = (8^-8)^-2 = 8^16
*June 9, 2015*

**MATH**

(ar)(ar^3) = 2ar^4 a^2r^4 = 2ar^4 ar^4 = 2r^4 a = 2 a(r^4-1)/(r-1) = 80 1+r+r^2+r^3 = 40 r = 3 So, the GP is 2,6,18,54,162,... (6)(54) = 2*162 2+6+18+54 = 80
*June 9, 2015*

**Algebra**

-5c^-3 = -5/c^3 So, (-5/c^3)^4 = 625/c^12
*June 9, 2015*

**Algebra**

here's one way. You want the concentration to be reduced by a factor of 15/25 = 3/5 So, the volume must be increased by a factor of 5/3. 5/3 * 120 = 200 ml So, if you add 80ml of water, the concentration is reduced from 25% to 15% Or, you can solve the equation .25*120 + ...
*June 9, 2015*

**maths**

factor out the x and you have (5/3 - 3/5)x = (25-9)/15 x = 16/15 x
*June 9, 2015*

**calculus1**

The graphs intersect at (0,-1) and (0,1). If this area is spun around the y-axis, you can think of the solid as a bunch of thin discs, of thickness dy, each with a hole in the middle. So, the area is a = ∫[-1,1] π(R^2-r^2) dy where R = 2-2y^2 and r = 1-y^2 a = &#...
*June 9, 2015*

**Calc 1**

true If the perimeter is p, then 2x+2y = p y = (p-2x)/2 = p/2 - x The area is a = xy = x(p/2-x) = p/2 x - x^2 This is just a parabola, with vertex at x = p/4 so, y = p/4 as well, and you have a square to get max area.
*June 9, 2015*

**science**

I believe it is C water can act as either a polar or nonpolar solvent.
*June 9, 2015*

**Calc 1**

what do you mean you "tried 0.5"? Tried for what? In my solution, the slope is 2, not 1/2.
*June 9, 2015*

**Calc 1**

clearly, s(0) = 0 s(1/2) = 1 So, the line is y=2t. since t is time and y is position, say, cm, then the slope of the line (cm/s) is the average velocity during the interval [0,0.5] See http://www.wolframalpha.com/input/?i=plot+y%3Dsin%28pi*x%29%2C+y%3D2x
*June 9, 2015*

**Kainantu sec..science(chemistry)**

PV=kT That is, PV/T is constant. (10)(1.93)/(273+100) = 1V/273 so just solve for V.
*June 9, 2015*

**Kainantu sec..science(chemistry)**

1 mole at STP is 22.4L 2.00g of CS2 is 2/76.14 = 0.0263 moles CS2 CS2+3O2 = CO2 + 2SO2 so, each mole of CS2 produces 2 moles of SO2. I expect you can take it from there...
*June 9, 2015*

**Maths**

assuming you measure the 12m along the ground, we have the horizontal distance from the base of the building is 12cos5° = 11.95 So, from eye level, the vertical distance to the top of the building is 11.95tan35°15' = 8.45 Add to that the 1.6m of your height, and ...
*June 8, 2015*

**math**

65*3 = 5x x = 39
*June 8, 2015*

**Physics**

since s = 1/2 at^2, the truck moved for t seconds, where 42 = 1/2 * 2.15 t^2 t = 6.25 seconds The car moved 1/2 * 3.47 * 6.25^2 = 67.77 meters so, it was 25.77 m behind the truck v = at, so you can figure out the speeds, I expect.
*June 8, 2015*

**Physics**

h(t) = 22.0 + 17.0t - 4.9t^2 v(t) = 17.0 - 9.8t So, find t when h=0, and use that to find v at that instant.
*June 8, 2015*

**Physics**

x(t) = 2.17+4.80t^2−0.100t^6 v(t) = 2.17+ 9.60t - 0.600t^5 a(t) = 9.6 - 3.0t^4 As you say, v(2) = 0, so a(2) = -38.4
*June 8, 2015*

**algebra**

x/(x-5)-3=5/(x-5) x/(x-5)-5/(x-5)=3 (x-5)/(x-5) = 3 1 = 3 not true for any value of x.
*June 8, 2015*

**Maths**

since the circumference is πd, just find d where πd*480 = 1200 m
*June 8, 2015*

**geometry**

clearly AD and BC are horizontal, and thus parallel. since B is 1 to the left of A, and C is one to the left of D, BA || CD. since AD and AB are not the same length, ABCD is a parallelogram and not a rhombus.
*June 8, 2015*

**Math**

can't see the diagram, but the area of a semi-circle is π/2 r^2 So, the area between two semi-circles of radii r and R is π/2 (R^2-r^2) You have circles of radii 14 and 16. So, plug in your numbers.
*June 8, 2015*

**statistics**

there were 15 cards, 5 of them diamonds. So, the chance of getting a diamond is 5/15 After that, having drawn a diamond, there are only 14 cards left, 4 of them diamonds. So, the chance that the 2nd draw is another diamond is 4/14.
*June 8, 2015*

**pre calc**

y = kx 9 = 12k, so k = 3/4 y = 3/4 x
*June 8, 2015*

**physics**

you have a right triangle, with the hypotenuse = 9 (his actual speed), and one leg=2 (the current) so, his resultant velocity must be upstream at an angle x such that sin(x) = 2/9 and the actual resultant speed s (due north) is √(9^2-2^2) Having gotten his speed s, the ...
*June 8, 2015*

**maths**

One subset is {1,3} I expect you meant to ask a little more . . .
*June 8, 2015*

**Calculus**

the revenue function is price * quantity, so R(x) = x(50-.01x) profit is revenue less cost, so P(x) = R(x)-C(x) = x(50-.01x) - (10+2x) = -0.01x^2 + 48x - 10 Now just set dP/dx=0 and evaluate P there. Or, just use your Algebra I skills and find the vertex of the parabola.
*June 8, 2015*

**math**

d/1400 = 8/450
*June 8, 2015*

**Math**

The digits are, as explained above, going right to left, 2 3 4*2=8 2+3=5
*June 8, 2015*

**math**

Grrrr! The Greek letter is pi The dessert is pie! Here is the graph. I expect you can use it to help with the other questions. http://www.wolframalpha.com/input/?i=-1800cos%289%CF%80%2F6+t%29%2B1800+for+0+%3C%3D+t+%3C%3D+10
*June 8, 2015*

**science**

at deceleration a, the velocity is v = 20-at so t = 20/a the distance traveled by the car is s = 20t + 1/2 at^2 = 20(20/a) - 1/2 a (20/a)^2 = 200/a So, to stop in 50 meters, 200/a = 50 a = 4 m/s^2 check: v = 20-4*5 s = 20*5 - 2*5^2 = 100-50 = 50
*June 8, 2015*

**speed time distance**

180 km/hr = 50 m/s the front of the train arrives at t=0. The front of the train leaves the platform at t = 365/50 = 7.3 The back of the train whose length is x leaves the platform at t = 7.3 + x/50 So, solve for x in 7.3 + x/50 = 21
*June 8, 2015*

**math**

Neither can I. You never asked a question.
*June 8, 2015*

**math**

The volume is x(30-2x)^2 The area is (30-2x)^2 + 4x(30-2x)
*June 8, 2015*

**Math**

also recall that cos(x) is negative in QII and QIII
*June 7, 2015*

**Pre-Algebra**

direct: y = kx, so k = y/x inverse: xy = k If y varies directly with x, find the constant of variation with x = 4 and y = -24 k = y/x = -24/4 = -6 If y varies inversely with x, find the constant of variation with x=5 and y = 10 k = xy = 5*10 = 50 and do the others in like wise.
*June 7, 2015*

**calc**

y is concave up if y" > 0 By the 2nd FT of Calculus, y' = -f(x) = -6/(x+1)^2 y" = 12(x+1)^3 so, where is y" positive?
*June 7, 2015*

**Inequalities**

Hmmm. I get (x-3)/(x-2) + 12/(x^2-x-2) > 8/(x+1) (x-3)(x+1) + 12 > 8(x-2) x^2-2x-3+12 > 8x-16 x^2-10x+25 > 0 (x-5)^2 > 0 So, all x except x = 2,5 and -1 Oddly enough, it appears that the real solution is x < -1 or x > 2 and x not 5. See http://www....
*June 7, 2015*

**Math (PLEASE HELP!)**

Evidently you do not understand the concept of a range. Your temperature is somewhere in 27±3 24 = 27-3 and 30 is 27+3 So, clearly the range is 24-30
*June 7, 2015*

**math**

as usual, maximum area is achieved in a square pen. To show this, note that if the dimensions are x and y, we have 2x+2y=200, so y = 100-x The area a = xy = x(100-x) = 100x-x^2 This is just a parabola, with its vertex at x=50. So, the pen is 50x50, or a square. Hmm. I see I ...
*June 7, 2015*

**Math**

you want m where 50 < 15+.10m
*June 7, 2015*

**Math**

total socks: 7 so, P(blue,red) = 2/7 * 2/7
*June 7, 2015*

**calc**

not sure what you mean by evaluate 4-x and 2x-8 Maybe a look at the graphs will help clarify your thinking: http://www.wolframalpha.com/input/?i=plot+y%3D4-x%2C+y%3D2x-8
*June 7, 2015*

**Walk two moons need help**

1st question: have you read it?
*June 7, 2015*

**statistics**

1569/23652 = 0.06633 You can do the rounding.
*June 7, 2015*

**statistics**

176/232 = 0.75862 You can round that as needed.
*June 7, 2015*

**Statistics**

since the yearly salary amounts to 2080*13.50 the benefits amount to 24% of that amount.
*June 7, 2015*

**algebra**

y = (10x+7)/(6x+3) now solve for x: (6x+3)y = 10x+7 6xy+3y = 10x+7 (6y-10)x = 7-3y x = (7-3y)/(6y-10) now x is y^-1, so to express f^-1(x), that is f^-1(x) = (7-3x)/(6x-10)
*June 7, 2015*

**Algebra**

for the survey, (b) looks better to me for the poems, you have (4/13)(2/12)
*June 7, 2015*

**Algebra**

(1.4*8)x10^(1+4) = 11.2x10^5 = 1.12x10^6
*June 7, 2015*

**Math**

if the base is x and the height is z, then x^2z = 4 The surface is the base and 4 sides, so a = x^2 + 4xz = x^2 + 4x(4/x^2) = x^2 + 16/x da/dx = 2x - 16/x^2 = (2x^3-16)/x^2 da/dx = 0 when x=2, so the container is 2x2x1
*June 6, 2015*

**Math (PLEASE HELP!)**

T = 27±3 |T-27| <= 3 Surely you can figure the range.
*June 6, 2015*

**Math**

1/7 = 3/(4x-1) 1 = 21/(4x-1) 4x-1 = 21 4x = 22 x = 11/2
*June 6, 2015*

**Math/Physics**

just given arbitrary vectors of some magnitudes, there is no way to compute their sum or difference. Unless you know the x- and y-components, you are stuck.
*June 6, 2015*

**Algebra**

x(x+18) <= 175 x^2+18x-175 <= 0 (x+25)(x-7) <= 0 So, -25 <= x <= 7 Or, for real-world tables, 0 <= x <= 7 There is probably a minimum positive width for a useful table.
*June 6, 2015*

**math**

The 4th row of Pascal's Triangle is 1 4 6 4 1 That is, (a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 So, (3+b)^4 is 3^4 + 4(3^3)b + 6(3^2)b^2 + 4(3^1)b^3 + b^4 = 81 + 108b + 54b^2 + 12b^3 + b^4 That's all very nice, but if your question is to solve for b in (3+b)^4 = 1...
*June 6, 2015*

**algebra**

when you raise to powers, you multiply exponents. Consider: 2^3 = 2*2*2 (2^3)^4 = (2*2*2)*(2*2*2)*(2*2*2)*(2*2*2) = 2^(3+3+3+3) 2^(3*4) = 2^12 So, (x^4 y^5)^5 = x^20 y^25
*June 6, 2015*

**Math**

1/4 + 1/s = 1/3 Now just solve for s.
*June 6, 2015*

**algebra**

6C3 = (6*5*4)/(3*2*1) = 20
*June 6, 2015*

**math**

Draw radii to the points of tangency. Draw a line connecting the centers. Draw a parallel line, 2 units closer to the tangent points. Now you have a right triangle with legs 10 and 6, with the tangent line forming the hypotenuse.
*June 6, 2015*

**math**

It won't seem hard after you've done a few dozen of them... I assume you meant 9^(x+2)=240+9^x That is the same as 9^2*9^x = 240 + 9^x 9^x(81-1) = 240 80*9^x = 240 9^x = 3 x = 1/2
*June 6, 2015*

**Math**

make that 2n/10 + 5n/7 + 303 = n
*June 6, 2015*

**Math**

"several" sevenths? What kind of nonsense is that? Anyway if that means 5/7, then 2/10 + 5n/7 + 303 = n n = 3535
*June 6, 2015*

**math**

angle ADB is an exterior angle to triangle ACD. So, it is equal to angles ACD+CAD. Thus angle C is 45 degrees.
*June 6, 2015*

**Math**

In the absence of any other information about how the amplitude changes, I'd just use a linear rate of change. For instance, y = (2+0.3x) sin(x) Or, if you insist on a constant amplitude, then averaging them seems like a good option.
*June 6, 2015*

**Math**

Assuming even change in height, the rate is 0.5 m/hr So, at what time t do you have 1.4 + 0.5t = 1.71 ??
*June 6, 2015*

**Math**

see your original post.
*June 6, 2015*

**math derivatives - eh?**

The given information says nothing about the height or width of the building. Without that, we do not know enough to work the problem.
*June 6, 2015*

**Math**

see your previous post.
*June 6, 2015*

**Math**

come on, man! The common difference is A2-A1 or A3-A2, which they gave you. The difference is shown above: -2p+10 or -3p-13 In either case, it is 56. Now do you think you can write the sequence?
*June 6, 2015*

**Math**

since the difference is constant, (5p+12) - (7p+2) = (2p-1) - (5p+12) -2p+10 = -3p-13 p = -23 Now you can find the difference, and write the terms of the sequence.
*June 6, 2015*

**Algebra II**

that it is.
*June 6, 2015*

**math**

you want, for m miles, 26+.12m = 24+.19m
*June 5, 2015*

**math**

a quadratic has x^2 as its highest power. So, A. Note that y decreases by 3 when x increases by 1. So, it will look like y = -3x But, when x=0, -3x=0, and y is -8. So, y = -3x-8 again, A
*June 5, 2015*

**Math- algebra**

the colony doubles 4 times every hour. So, after t hours, it has reached 2^(4t) cells. Clearly 4t = 13
*June 5, 2015*

**Geometry**

Since triangle DEC is a right triangle, with angle C = 30° and DC=10√3, DE = 10. No idea what the "length" of E is, since it's a point.
*June 5, 2015*

**Math**

24/8 = 3 x^4/x^-3 = x^7 z^-2/z = z^-3 Looks like C to me
*June 5, 2015*

**Basic Differential Equation**

You have (D^2+16)y = 0 so the solutions are e^(±4ix) or sin and cos (sin 4x)" = -16sin(4x) (cos 4x)" = -16cos(4x) If the answer is just sin(4x) then there must be some other conditions
*June 5, 2015*

**math**

you can choose each scoop 5 ways, so 5^3=125
*June 5, 2015*

**Math**

multiply by r and you have r^2 = -2rcosø-2rsinø x^2+y^2 = -2x-2y x^2+2x + y^2+2y = 0 (x+1)^2 + (y+1)^2 = 2
*June 5, 2015*

**geometry**

4'6" = 54" 9'7" = 115" so the scale is 115/54
*June 5, 2015*

**Algebra help**

you want the area of the deck-and-pool to be 832ft^2 greater than just the pool. So, solve for w in (w+8)(3w+8) - 3w^2 = 832 Having gotten w, you can now determine the size of the pool. Sounds to me like the problem is poorly stated. You already know the size of the pool. You ...
*June 5, 2015*

**math**

1 3/4 = 7/4 7/4 * 9 = ?
*June 5, 2015*

**math**

1 4/5 = 9/5 2/(9/5) * 7 and round up to a whole bottle
*June 5, 2015*

**Math**

"larger than" means you subtract, to find the difference. So, you have ((-1)^2 + (-2)^2 + 5^2) - (-1-2+5)
*June 5, 2015*

**Math**

Can you factor x^2+x-12? This will be very like that.
*June 5, 2015*

**Math of class-10**

∜3 is a root of the equation x^4 - 3 = 0 Using the Rational Root Theorem, any rational roots p/q of that equation must have denominator of 1, and numerator a divisor of 3. Since 3 and 1 are not roots, there are no rational roots.
*June 5, 2015*

**Calculus**

Looks like the curve y=e^x on the interval [2,7] rotated around the line x=1. I assume you meant to include dx in the integral. Since the volume of a shell of thickness dx is 2πrh dx, if r = x-1 and h=e^x that fits the bill.
*June 5, 2015*

**Math**

(1/2)(3/5) + (1/2)(2/6)
*June 5, 2015*

**algebra**

w(2w+2) = 312 Now solve for w.
*June 5, 2015*

**Math**

well, geez. Midnight to 10 am is not exactly a full day, now, is it? Most folks are sleeping during most of that period. When you ask "which" it implies that we be given a list to choose from. Maybe we were waiting for you to see that you had not supplied any choices...
*June 4, 2015*

**Math**

if sides a and b are equal, then a/sinA = b/sinB means that a/sinA = a/sinB so sinA = sinB Now, A and B need not be equal if their sines are equal, but in that case A = 180°-B and your triangle is just a flat line.
*June 4, 2015*

**math**

well, that would be when 20sin 2pi/40 (t-10) + 21 = 15 20sin 2pi/40 (t-10) = -6 sin 2pi/40 (t-10) = -.30 sin .304 = .3, so 2pi/40 (t-10) = pi+.304 or 2pi-.304 t = 31.935 or 48.065 or other values periods away.
*June 4, 2015*

**math**

domain is naturally all reals vertical shift is -1, so range is -1-2 to -1+2 amplitude is 2 period is 2pi/3 phase shift is -180
*June 4, 2015*

**Math**

First, disambiguate your expression. As it is written, it means (1/6^8)/3 - 2/3 = -3359231/5038848 or do you mean 1/6^(8/3) - 2/3 = 1/∛1679616 - 2/3 or something else entirely? Formatting is not the browser's forte.
*June 4, 2015*

**Math**

so she needs a circle of radius 3 (9π) and a rectangle which is 6πx12.
*June 4, 2015*

**Algebra II**

Your first step is wrong. Just subtract b². It should be 4b = 3b² 3b² - 4b = 0 b(3b-4) = 0 b=0 or b=4/3 b = -4/3 should immediately be seen not to work, since you have b² + 4b = 4b² If b is negative, there is no way you could subtract something from b&...
*June 4, 2015*