Ask a New Question
Search
Questions and answers by
Steve@Math
Answers (51)
balance the torques. 62(25) + 76(25-2) = 96(25) + x(25-2.5) Now just solve for x
just plug in your numbers for F = GMm/r^2 expect a small answer!
how is typing in the question here any easier than typing it in on google? You will find many correct responses.
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.
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
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.
figure the perimeter of the pool, then multiply it by 29
A = L + 7 (A+1) = 2(L+1)+1 A = 12
you are correct
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.
express P and A as a function of L and W Then just use the chain rule to find the derivative (d/dt)
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)
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
F&arrowbold;
no, 42
Can you not find the distance between two points? They have given you (h,k). All you need is r.
(x+7)(4) = 2x that should get you started...
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.
(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?
oops. less than, not less or equal, so (B) 2,3,5,7 are less than 8, and also less than 11.
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)
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?
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.
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
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 not red. So, that leaves 79/112 chance that at least one was red.
add up the value of the coins: 10*84 + 25q = 2215 840+25q = 2215 25q = 1375 q = 55
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.
Draw your triangles. cosx = √8/3 cosy = 4/5, so siny = 3/5 Now just plug in your sum-of- angles formula.
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.
30° = π/6 radians s = rθ = 12(π/6) = 2π
assuming you just want confirmation of your efforts, (C): 18
There are obviously 70 multiples of 3 (210/3 = 70) So, how many multiples of 9 are there? Figure the difference.
Since T4 is 4 terms before T8, 13 - 4(9) = ?
just plug in n=1,2,3: a(1-2) = -a a(1-4) = -3a a(1-6) = -5a
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.
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)
2x^2 - 12x + 11 2(x^2 - 6x) + 11 2(x^2 - 6x + 9) + 11 - 2*9 2(x-3)^2 - 7
8*6.5 - 5*1*1 = 47 m^2
Dang. Gotta read more carefully.
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
54 x 29 = ?
that would be 32 x 794 = 25408 g
PRM and QSM are right triangles RM=MS PR=QS by SAS, the triangles are congruent, so PM=QM
a = 4b b = a-36 b=12, a=48 a+b=60
well, the terms are getting larger. If you carry on forever, the sun is 10/99.
or, just note that T20 = T11+9d = -13-18=-31
x = -3 is also excluded from the domain, eh? since x^2-9 = 0
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
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
since there are only 4 colors, 5 socks must include a match
