Steve@Math

name change

11 years ago

since there are only 4 colors, 5 socks must include a match

11 years ago

assuming the usual carelessness with parentheses, I see this as (x+2)/(x^3-16x) (x+2)/(x(x-4)(x+4)) so the domain is all reals except 0, -4, 4

11 years ago

If there were s student tickets and a adult tickets sold on the first day, then we have a = s-80 .85a + 1.30s = 1824 a = 800 s = 880 so, on day 1 they took in 65*800 + 32*880 = 80160

11 years ago

x = -3 is also excluded from the domain, eh? since x^2-9 = 0

11 years ago

or, just note that T20 = T11+9d = -13-18=-31

11 years ago

well, the terms are getting larger. If you carry on forever, the sun is 10/99.

11 years ago

a = 4b b = a-36 b=12, a=48 a+b=60

11 years ago

PRM and QSM are right triangles RM=MS PR=QS by SAS, the triangles are congruent, so PM=QM

11 years ago

that would be 32 x 794 = 25408 g

11 years ago

54 x 29 = ?

11 years ago

what's with all the words? ((14x^7 y^5 x^6)/(21x^12 y^2))^4 ((14x^13 y^5)/(21x^12 y^2))^4 ((2xy^3)/3)^4 16x^4 y^12 / 81

11 years ago

Dang. Gotta read more carefully.

11 years ago

8*6.5 - 5*1*1 = 47 m^2

11 years ago

2x^2 - 12x + 11 2(x^2 - 6x) + 11 2(x^2 - 6x + 9) + 11 - 2*9 2(x-3)^2 - 7

11 years ago

how can a single line be parallel to both perpendicular axes? I'll assume you mean it's parallel to 2x-4y=3. In that case, the slope is 1/2, so the line is y+5 = 1/2 (x-2)

11 years ago

the equation of a line parallel to the x-axis has the equation y = k. Since your line goes through (2,-5), you obviously want y = -5. Similarly for the line parallel to the y-axis.

11 years ago

just plug in n=1,2,3: a(1-2) = -a a(1-4) = -3a a(1-6) = -5a

11 years ago

Since T4 is 4 terms before T8, 13 - 4(9) = ?

11 years ago

There are obviously 70 multiples of 3 (210/3 = 70) So, how many multiples of 9 are there? Figure the difference.

11 years ago

assuming you just want confirmation of your efforts, (C): 18

11 years ago

30° = π/6 radians s = rθ = 12(π/6) = 2π

11 years ago

How far apart are the two points? That will be r. As you know, the circle with center at (h,k) with radius r is (x-h)^2 + (y-k)^2 = r^2 Now just plug in your numbers.

11 years ago

Draw your triangles. cosx = √8/3 cosy = 4/5, so siny = 3/5 Now just plug in your sum-of- angles formula.

11 years ago

You need to review your commonly used reference angles. sin π/6 = 1/2 sin π/3 = √3/2 Now you should have no trouble.

11 years ago

add up the value of the coins: 10*84 + 25q = 2215 840+25q = 2215 25q = 1375 q = 55

11 years ago

Since there are 5 red marbles and 11 others, then the chance is 11/16 * 10/15 * 9/14 = 33/112 that the first three draws are <b>not</b> red. So, that leaves 79/112 chance that at least one was red.

11 years ago

I assume you want to find a way to spend exactly £100 on some combination of animals. c/2 + 10p + 2s = 100 One solution is 9 pigs, 16 chickens, 1 sheep (90+8+2) Lots of other solutions exist by substituting a quantity of chickens or sheep for different nu...

11 years ago

that would be 2πr(r+c) = 2π(2)(2+1.5) = 14π before being licked... That's on odd shape for a piece of candy - like a hockey puck.

11 years ago

just complete the squares for x and y: x^2-6x + y^2+10y = -9 x^2-6x+9 + y^2+10y+25 = -9+9+25 (x-3)^2 + (y+5)^2 = 25 I assume you can take it from there, eh?

11 years ago

the primes are 2 3 5 7 11 ... The first time there is a gap of at least 3 is at n=7, so (A)

11 years ago

oops. less than, not less or equal, so (B) 2,3,5,7 are less than 8, and also less than 11.

11 years ago

(x-2)(x+1) = 0 now, if the product of two numbers is zero, one of the numbers must be zero. So, which value of x makes each factor zero?

11 years ago

the line with slope of -3 through (2,-5) is y+5 = -3(x-2) Learn the various forms of equations for a line; it will come in handy many times in the future.

11 years ago

(x+7)(4) = 2x that should get you started...

11 years ago

Can you not find the distance between two points? They have given you (h,k). All you need is r.

11 years ago

no, 42

11 years ago

F&arrowbold;

11 years ago

F1 = -14.5i + 25.1j F2 = 18.9i Now we want F3 such that (-14.5i+25.1j)+(18.9i+0j) + F3 = (0,0) That should help

11 years ago

zero exponent is always 1: 5^0 = 1 negative exponents switch between top and bottom. x^-4 = 1/x^4 1/x^-4 = x^4 4ab^0 = 4a(1) = 4a 5x^-4 = 5/x^4 3x^-2 / y = 3/(x^2 y)

11 years ago

express P and A as a function of L and W Then just use the chain rule to find the derivative (d/dt)

11 years ago

Hmmm. It appears your integrating factor is correct. Evaluating the right-hand side is intractable. a couple of online solvers both come up with y in terms of that integral, which is not evaluated. Beats me.

11 years ago

you are correct

11 years ago

A = L + 7 (A+1) = 2(L+1)+1 A = 12

11 years ago

figure the perimeter of the pool, then multiply it by 29

11 years ago

If there are 4x 20¢ coins, there are 3x 50¢ coins. 20 20¢ coins = $4.00, or 8 50¢ coins. So, we have (4x-20)/(3x+8) = 7/11 x = 12 So, now you can figure the value of the money.

11 years ago

the demand rises by 750 when the price drops by 30, so d = 250 + 25(180-x) the supply rises by 750 when the price rises by 10, so s = 750 + 75(x-80) Now just set d=s and solve for x, the equilibrium price. Plug that back into either equation to figure the...

11 years ago

you want to reduce the concentration by a factor of .1899/.2005 = 0.9471 So, the volume must increase by that factor, to 1580/.9471 = 1668 mL. Just subtract to figure the amount of water added.

11 years ago

how is typing in the question here any easier than typing it in on google? You will find many correct responses.

11 years ago

just plug in your numbers for F = GMm/r^2 expect a small answer!

11 years ago

balance the torques. 62(25) + 76(25-2) = 96(25) + x(25-2.5) Now just solve for x

11 years ago