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henry2,
Questions and answers by visitors named
henry2,
Questions (1)
in the poem the tide rises the tide falls Longfellow uses figurative language in repetition in a paragraph identify the
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Answers (125)
$4.56/2.9L = $___ /L. $5.78/3.8L = $___ /L.
r = 84km/3.5h = 24km/h. d = r*T 1050 = 24T T =
Y = x^2+x-2/(x-1). -2 = -1*2, -1+2 = 1 = B. (x-1)(x+2)/(x-1) = 0 x+2 = 0 X = -2. solution: (-2, 0). We have only one solution, because the Eq was divided by (x-1). Test: Let X = -2.
d1 = 2 km. d2 = 13 * 1.5 = 19.5 kim. Tan A = d2/d1 = 19.5/2 = 9.75 A = 84.14 degrees = 1.47 radians.
As = 2(pi*r^2)+pi*2r * h. As = 2(3.14*25)+3.14*10*16 = ___ cm^2.
4pi/3 radians = 4*180/3 = 240 degrees. (240/360)C = 2C/3 = 1 meter. (2/3)3.14*2r = 1 r = 0.2387 m.
K = mg/d = 1*10/1.2 = 8.33 N/cm.
d = 100[60+180]+160[130o] d = (100*sin240+160*sin130)+(100*cos240+160*cos130)i d = 36-153i. d = sqrt(36^2+153^2) = 157 km.
All angles are measured CW from +y-axis. KA = 25km[90o]. AK = 25km[90+180]. KB = 15km[120o]. AB = AK+KB = 25[270o]+15[120o] AB = (25*sin270+15*sin120)+(25*cos270+15*cos120)i AB = -12-7.5i = 14.2km[58o]W. of S. = 14.2km[238o] = distance between the boats.
150-30 = 120 good eggs. Income = 120*0.17 + 30*0.1 = Gain = Income-22.50 = %Gain = (Gain/22.50)*100% =
or 3-14 = -11 = 11 floors downward.
1,2,3 2,3,4 3,4,5 4,5,6 To 11,1,2. 11 different ways.
conductivity = 1/P siemens/meter.
P = RA/L = 0.1*(2*10^-3)/(1*10^2) = 2*10^-6 ohm*cm. = resistivity. Conductivity = 1/P =
Given: M1 = 4kg, V1 = 2.5m/s. M2 = 0.2kg, V2 = 0. V3 = velocity of M1 and M2 after collision. Momentum before = Momentum after. M1*V1+M2*V2 = M 1*V3+M2*V3 4*2.5+0.2*0 = 4V3+0.2V3 4.2V3 = 10 V3 = 2.4 m/s.
A(x, y), M(-3, 3), B(-5, 1). -3-x = -5+3 X = -1. 3-y = 1-3 Y = 5.
0.8*40 = $32 = cost before tax. Total cost = 32 + 0.05*32 =
P = Po + Po*r*T = 7,200 + 7,200*0.052*5 =
I = Po*r*T = 20,000*0.045*(240/360) =
P = Po + Po*r*T = 10,100.14. 10,000+10,000*(0.035/360)T = 10,100.14 0.972T = 10,100.14-10,000 T = ___ days.
P = Po+Po*r*T = 20,000 Po+Po*0.05*60 = 20,000 Po+3Po = 20,000 Po =
P = Po + Po*r*T = 237,500 50,000+50,000*0.15*T = 237,500 7500T = 237,500-50,000 = 187500 T = 25 years.
I = P*r*T = 100 P*0.035*(150/360) = 100 P =
Mg = 9*10 = 90 N. = Wt. of crate. F = 45 N. = applied force. Fs = force of static friction. F-Fs = Ma 45-Fs = 9*0 = 0 Fs = 45 N. F*cos50 = 45 F = 70 N. = required force.
1. y+2 = 1/2(x-1). 2. Same procedure as #1. 3. (8, 5), (9, 6), (x, y). m = (6-5)/(9-8) = 1. y-5 = 1(x-8).
D = 150km[42o]+250km[90o]. D = (150*sin42+250*sin90)+(150*cos42+250*cos90)i D = 350+112i = 367km[72o].
Given: XY = 12km[180o], YZ = ___km[330o], XZ = 15km[??]. sinZ/12 = sin30/15. Z = 24o. 330-270 = 60o Y = 90-60 = 30o. X = 180-24-30 = 126o. YZ/sin126 = 15/sin30. YZ = 24 km. YZ = 24km[330o]. XZ = XY+YZ = 12[180o]+24[330o]. XZ =
All angles are measured CW from +y-axis. d1 = 500*2 = 1000km[0o]. d2 = 300*1.5 = 450km[53o]. D = d1+d2. D = (1000*sin0+450*sin53)+(1000*cos0+450*cos53)i D = 359.4 + 1270.8i = 1320.7km[16o].
25-20 = 5 cm increase. 50/5 = 100/x X = ___ cm increase. Length = 20+x.
Standard form: 4x-y = 7. 3x+4y = -9. Multiply first Eq by 4 and add the Eqs.: 16x-4y = 28 3x+4y = -9 sum: 19x = 19 X = 1. In Eq2, replace x with 1 and solve for y: 3*1+4y = -9
Jim has X cars. Alex has x+40 cars. (x+40)/x = 8/3. 3x+120 = 8x 8x-3x = 120 X = 24. x+40 = 24+40 = 64.
D = 20.5 m. = diameter. L = D + pi*D/2 = 20+(3.14*20.5/2) = 52.2 m.
Mg = 91*9.8 = 892 N. = Wt. of bike with load = normal force, Fn. 1. Fs = uFn = 0.33*892 = 294 N. = Force of static friction.
Note: The 2.25o given by the student should have been 225o.
Given: PA = ___ [225o], PB = ___ [116o], AB = 3.9km[258o], PA = ?. P = 180-45-26 = 109o. B = 90-64-12 = 14o. A = 180-109-14 = 57o. PA/sin14 = 3.9/sin109 PA = 1 km. PA = 1km[225o]. PB/sin57 = 3.9/sin109 PB = 3.5 km. PB = 3.5 km[116o].
F = 180-100-25 = 55o. Ext. angle = 180-55 = 125o.
a. (350/1000) * 100% = 35%. b. (400/1000) * 100% = c. (250/1000) * 100% =
4/16 = 1/4 pizza/student.
Gavin: $X. Emma:$2x/3. 2x/3 = 40 X = 60. 2x/3 + x = 40+60 = $100.
a. Yes, the original Eq should be used to calculate max. ht. : h max = 10x-5x^2 = 10*1-5*1^2 =
F(x) = 10x-5x^2. a. V^2 = Vo^2+2g*h = 0. 10^2+(-20)h = 0 h = 5 m. = max. ht. b. 10x-5x^2 = 0 when the ball hits gnd. Divide both sides by 5x: 2-x = 0 X = 2 s. to hit gnd.
A(-2, 3), C(x, y), B(6, 3). x+2 = 1/2(6+2) X = 2. y-3 = 1/2(3-3) Y = 3. Eq: (x+2)^2+(y-3)^2 = r^2.
P(-10, -2), C(x, y), Q(4, 6). a. x+10 = 1/2(4+10) X = -3. y+2 = 1/2(6+2) Y = 2. b. Radius = sqrt ((-3+10)^2+(2+2)^2) = sqrt 65. c. (x+10)^2+(y+2)^2 = 65.
a. V = L*W*h = 750*400*1.2 = 360,000 cm^3. V = 7.5*4*0.012 = ___ m^3. b. 1g/cm^3 * 360,000cm^3 = ___ grams = ___ kg.
PE1 = Mgh1 = 80*9.8*700 = ___ Joules. PE2 = Mgh2 = 80*9.8*400 = PE1-PE2 =
Vo = 95ft/s[25o] Yo = 95*sin25 = 40.1 Ft./s. Y = Yo+gTr = 0 40.1+(-32)Tr = 0 Tr = 1.25 s. = Rise time. Y^2 = Yo^2+2g*h = 0 40.1^2+(-64)h = 0 h = 25.1 Ft. h max = 6+25.1 = 31.5 h max = 0.5g*Tf^2 = 31.5 16Tf^2 = 31.5 Tf = 1.40 s. = Fall time. Tr+Tf =
90+60 = 150o CCW from +x-axis.
m-23 = 66.
(2, 9), Y < 4x+1. When X = 2, Y should be less than 9: Y = 4*2+1 = 9.
P = Po(1+r)^n. r = 0.08/4 = 0.02 = quarterly % rate. n = 6*4 = 24 compounding periods. P = 2412.66. Po = Int. = P-Po. po =