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FredR
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u = 2d or u - 2d = 0 u + d = 5 so the matrix would be 1 -2 0 1 1 5 add -1 times the 1st row to the 2nd row. Then you get 1 -2 0 0 3 5 multiply the 2nd row by 1/3: 1 -2 0 0 1 5/3 The 2nd row tells us that d = 5/3, so u = 10/3
A+B = 60 (1) A/3 = B/2 A = 3B/2 (2) combine (1) and (2) 3B/2+B = 60 3B/2+2B/2 = 60 5B/2 = 60 B = 24 A = 36
x = original # of pieces w = gave to wife = 10x/100 m = gave to meg = 3 (x-w-m)/4 = j = gave to joe 18 = x - w - m - j combine and solve for x
distance = velocity x time or time = distance / velocity total time=d1/v1+d2/v2=2.5 hrs (1) where d1 = car distance = 30 (2) v1 = car speed d2 = train distance = 120 - d1 (3) v2 = train velocity = v1+20 (4) combine (1), (2), (3), and (4) and solve for v1
It doesn't seem like there is enough information to determine the answer
sin = opposite / hypotenuse so, hypotenuse = 8 opposite leg = 7 adjacent leg = sqrt(8^2 - 7^2) = sqrt(15) 1/cot = tan tan = opposite / adjacent = 7/sqrt(15) in the 4th quadrant the tangent is negative, so add a minus sign: 1/cot(theta) = -7/sqrt(15)
cos(A) -1 = 0 or cos(A) = 1 arccos(1) = A = 0
a)centripetal acceleration is v^2/r. In this case, v is 30 mi/hr and r = 200 ft. The inertial force would be: F = ma = mv^2/r (1) This would be the friction force required to keep the car on its path. b) The static friction is: F = uN (2) where u is the
From 1920 to 1980 is 60 years. The record was shortened by 1.2 seconds. The slope of the line is the rise over the run, or: -1.2/60 for t=0, R(t) = 46.6 so the y intercept is 46.6. The format of a line is y = mx + b where m is the slope and b is the y
Assuming an ideal spring (linear) and using Hooke's law, the force generated by the spring is: F = -kx (1) where x is the distance the spring is stretched from the rest position. So, substituting the info into (1) you get: -10 = -k(2) or k = 5 lb/in work
Oops. Thanks for the correction.
It could be interpreted as the azimuth, which in surveying applications is typically measured in a clockwise direction from north.
if you bisect one of the angles, you have a right triangle with a hypotenuse of 1 cm and a short leg of 0.5 cm. Using the Pythagorean theorem, the height would be: height = square root (1^2 - 0.5^2)
let the 3 dimensions be a, b, and c where a and b are the bottom edges and c is the height. v is the volume s is the surface area a = 7b v = abc = 7b^2c or rearranging, c = v/7b^2 s = ab + 2ac + 2bc = 7b^2 + 14bc + 2bc = 7b^2 + 16bc = 7b^2 + 16bv/7b^2 =
rearranging the given line, y = (9/5)x - 4/5 (1) which has the form y = mx + b (2) where m is the slope and b is the y intercept. The new line would have the same slope to be parallel, so it would have the form: y = (9/5)x + b (3) to solve for b, input the
It doesn't look right to me sec Q - 1/sec Q = 1/cos - cos = 1/cos - cos^2/cos = (1-cos^2)/cos = sin^2/cos
using the Pythagorean theorem, opposite side = sqrt(13^2-12^2) sin x = opposite side / 13 tan x = opposite side / 12
Newton's law of gravitation: F = G m1 m2 /r^2 let m2 = mass of ice cube and s = G m1/r^2 so, F = s m2 rearranging, s = m2/F let V = orbital speed centripetal acceleration = V^2/r For an object to remain in orbit s must equal the centripetal acceleration
12.60 - 23.00 = A A / -1.30 = standard deviation
if you are traveling along the leg of the star, the deflection angle is the angle you have to turn to get to the next leg. d = deflection angle at a point = 180 - 12 = 168 since there are points, the total point deflection angle at the points is 1680. x =
180+arctan(12/-5) = 112.6
a = angular acceleration I = rotational inertia w = angular velocity v = linear velocity Torque = RxFT = Ia so, a = RxFT/I = (R/I)(4t-0.10t^2) w = integral(a) with respect to t v= wR
It looks right to me. The easiest way to check is to plug the number back into the formula: e^.6931 = 2 2*2-3 = 4-3 = 1
Permutations of 9 objects taken 5 at a time = 9!/(9-5)! where 4! = 4x3x2x1 and 9! = 9x8x7x6x5x4x3x2x1
yes, it could be. You just need to change the sign, so you would have 8-y over y-4 PLUS 7y-9 over y-4
dist from station to trains = D ta = time in hours from when A passes station to when trains A & B meet. D = 80ta D = 88(ta-15/60) solve for ta Time at catch-up = 1:20 AM + ta
NO. 500 = A - A x 0.28 = A (1-.28) = .72A so, A = 500/0.72 = 694
csc=1/sin cot=cos/sin so, csc^2-1/cotcsc = ((1/sin^2)-1)sin/(cos/sin) = sin((sin/sin^2)-sin)/(cos) = (sin^2/sin^2-sin^2)/cos = (1-sin^2)/cos = cos^2/cos = cos A ((sin^2/sin^2)-sin/(sin^2 cos)
1) it is not correct. The second term would be log3(-2) and you can't take the log of a negative number. 2) it is not correct. The base angle would be 45 degrees, so the cotangent would be 1.
9cos^2(t)+3cos(t)=0 subtract 3cos(t) from both sides 9cos^2(t)=-3cos(t) divide both sides by 3cos(t) 3cos(t)=-1 cos(t)=-1/3 t=arccos(-1/3)
# of ways of selecting winning quintuple = permutations of 5 objects = 5! = 5 x 4 x 3 x 2 x 1 probability of winning = n!/(r!(n-r)!) = 44!/(5!39!)
x = original # of tickets y = original price you have 2 equations: xy = 180 (x-2)(y+9)=180+36 substitute equation 1 into 2 and solve.
16-15.2 = 0.8 0.8/0.03 = n = 26 standard deviations so, erf(n/sqrt(2)) = approximately 1 so, 50% of the data would be between 15.2 and 16.
The formulas given look like they are for sinusoids and not circles.
the graph of y=sin(theta) is moved vertically along the y axis a distance of q to obtain y=sin(theta)+q.
= (18+4)r^2 +7s = 22r^2+7s
= (y/3)(y^2+5y+6) = (y/3)(y+2)(y+3)
for example, for x = 2, y = 0 x = 3, y = -1/3 x = 4, y = -2/3 x = 5, y = -1
A = amt invested at 8.5% B = amt invested at 10% A + B = 50000 or A = 50000 - B income = 0.085A + 0.10B = 4700 substitute A in terms of B and solve
c = # of chest freezers u = # of upright freezers c+u>=100 or u >= 100-c assume a total of 100 units cost = 250c + 400u = 250c +400(100-c) = 250c + 40000 - 400c = 40000 - 150c = 800 - 3c so obviously the more type c units, the lower the total cost. profit
cos(x)=15/17 so x = arccos(15/17) tan(76) = 13/x so x = 13/tan(76)
dmA/dt = cmB^2/mA where c is a constant mB=m-mA so dmA/dt = c (m-mA)^2/mA = (c/mA)(mA^2-2mmA+m^2) = cmA-2cm+cm^2/mA integrate with respect to t
sin(a+b)/cos(a)cos(b) = = (sin(a)cos(b)+cos(a)sin(b))/cos(a)cos(b) = sin(a)cos(b)/(cos(a)cos(b) + cos(a)sin(b)/(cos(a)cos(b) = sin(a)/cos(a) + sin(b)/cos(b) = tan(a) + tan(b)
It would take 10 minutes.
Moles of solute = 16 g / 60 g/M = = 0.267 M volume of H2O = 39g/1.3 g/ml = 30 ml = 0.03 l molarity = 0.267 M / 0.03 l = 8.89 M/l
the probability that the 1st marble is red is 10/13. So, the probability that both marbles are red is (10/13)^2 or (100/169)
(a) 3y = x - 10 substitute in the 2nd equation 3(x-2) = x - 10 3x - 6 = x - 10 simplifing... 2x = -4 or x = -2 and y = -4 this is a unique solution (b) this system has infinitely many solutions and cannot be solved without more information. 3x + 3y = 15
y = (x-2)^2 y = x^2-4x -4 dy = 2xdx - 4dx so, dy/dx = 2x-4 dx/dt = 2 so, dy/dt = dy/dx dx/dt = 2(2x-4) = 4x-8 for x=1 then dy/dt = 4(1)-8 = -4 units per second
dx/dt = 2 inches/min V = x^3 so, dV/dt = 2^3 = 8 cubic inches/min
Area = A x B Total length = 3000 = A + 2B B = (3000-A)/2 so, Area = A x (3000-A)/2 = = 3000A/2 - (1/2)A^2 take 1st derivative 0 = 1500 - A and solve for A A = 1500 and B = 750
