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
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)
grade 12 trig
cos(A) -1 = 0 or cos(A) = 1 arccos(1) = A = 0
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
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 static friction coeffic...
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 is the 1st integra...
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)
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