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R_scott
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Each case of soft drinks contains 24 cans. Syd bought 4 cases of drinks and Vangie bought 12 cases of drinks. How many total
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6:00 PM
a r + a r^2 = 12 ... a r (1 + r) = 12 a r^3 + a r^4 = 300 ... a r^3 (1 + r) = 300 dividing equations ... r^2 = 25 ... r = 5 substitute back to find a
combine like terms ... 6 x^2 ... the x-terms and numbers cancel each other
let B equal both (ii) 30 + 25 - B = 45 (iii) 25 - B (iv) 45 - B
k.e. = 1/2 m v^2 = 1/2 * 0.60 kg * (2.0 m/s)^2 = ? J
the sides are 2 cm high ... so the bottom of the box is a square , 10 cm on a side the original piece was 14 cm on a side ... 10 cm + 2 cm + 2 cm
sin(θ) * 12.1 N = sin(25º) * 6.40 N
number is twice 18
lift = π * 6 m
center of mass is at 50 cm .150 kg * g * 50 cm = f * 60 cm ... f is in Newtons
Moles = 10 cm^3 / 22.4 L = .01 L / 22.4 L average nucleons per atom = atomic mass fluorine is diatomic ... two atoms per molecule moles of nucleons = 2 * Moles * atomic mass
r^(6 - 3) = 80 / 10 ... r^3 = 8
exterior + interior = 180º exterior = 180º - interior 360º / exterior = sides
a mole of gas at STP occupies 22.4 L
a = p e^(r t) = $6000 * e^(.045 * 9)
y - 3 = [(3 - -1) / (-3 - -2)] (x - -3) y - 3 = -4 (x + 3)
L^2 = (2 m)^2 + (6 m)^2
total energy is ... 1/2 m v^2 + m g h ... (1/2 * 100 * 10^2) + (100 * 9.8 * 30) ... 5 kJ + 29.4 kJ = 34.4 kJ potential energy becomes kinetic energy at ground height ... 1/2 * 100 * v^2 = 34.4 kJ ... v^2 = 688 m^2/s^2
x = 1400 / 40
heat lost by copper is gained by water 25 kg * (100 - t) * (s.h. Cu) = 45 kg * (t - 45) * (s.h. H2O) s.h. Cu is ... 385 J/kg⋅ºC s.h. H2O is ... 4184 J/kg⋅ºC
3^x = 6.84 ... x log(3) = log(6.84) ... x = log(6.84) / log(3)
no 8/18 reduces to 4/9 ... which does not equal 1/2
24 L * .35 = ?
y-15=3x ... y = 3 x + 15 substituting ... -2 x + 5(3 x + 15) = -3 ... 13 x = - 78 solve for x , then substitute back to find y
increasing the denominator (bottom) makes the fraction smaller
cos(θ/2) = ± √{[1 + cos(θ)] / 2} 37.5 is half of 75 , which is half of 150 ... cos(150º) = - √3 / 2
m v^2 / r = 1.97E3 N - 650 N = 1320 N m v^2 = (1320 * 17.0) N⋅m = (1302 * 17.0) kg⋅m^2/s^2 v^2 = (1302 * 17.0 / 650) m^2/s^2
b = bio , c = chem , B = both b = c + 6 b + c + B = 60 ... b + c = 52 substituting ... c + 6 + c = 52 ... c = 23 substitute back to find b
(5 n) - 3
w = 5 d w + 4 = 3 (d + 4) substituting ... 5 d + 4 = 3 d + 12 solve for d , then substitute back to find w
2 k + 3 k = 30 ... k = 6 2 k : 3 k = 12 : 18
2 * {[6 / sin(48º)] + [6 / cos(48º)]}
the man's speed (relative to the ground) is less than the bus he must be walking toward the back
180 - (35.5 + 82.6) = ?
dropped a zero 2 p = 360
1 hr 36 min = 1.6 hr p - w = 320 mi / 2 hr = 160 mph p + w = 320 mi / 1.6 hr = 200 mph adding equations (to eliminate w) ... 2 p = 36 mph solve for p , then substitute back to find w
there is no compounding of the daily return... 1st response is not applicable
$100 * .01 * 250
[$100 * (1 + .01)^250] - $100
n = 5^s
π * (7 cm)^2 * (36º / 360º)
6/5 (10 t + u) = 10 u + t 60 t + 6 u = 50 u + 5 t 55 t = 44 u ... 5 t = 4 u 45 looks good
W = L + 5 W * L = 205 substituting ... (L + 5) * L = 205 ... L^2 + 5 L - 205 = 0 solve for L , then substitute back to find W
potential energy becomes kinetic energy m g h = 1/2 m v^2 v = √(2 g h) = √(9.8 m/s^2 * 270 m) = ? m/s
(s + 40 g) * k = (21.8 - 20) cm = 1.8 cm (s + 60 g) * k = (22.05 - 20) cm = 2.05 cm subtracting equations ... 20 g * k = .25 cm ... k = .0125 cm/g substituting ... (s + 40 g) * .0125 cm/g = 1.8 cm solve for s
1/5 , 1/6 , ... follow the pattern
(700 kph) * (1000 m/km) / (3600 s/h) = ? m/s
1) 5 m + 10 m = ? 2) √(5^2 + 10^2) = ? m
6 * 5 = 30 ... but ... A with B is the same as B with A ... so, 30/2
area = b h using 4 as b , then h is 3 smaller angle ... sine is 3/5 larger angle ... 180º minus smaller angle
