# Posts by Scott

Total # Posts: 2,496

**Math**

4 d 7 f 2 sides 4! * 7! * 2

**Physics 11**

f = m a ... 20 + x = 4 * 3 ... x = -8 friction = m g ? ... 8 = 4 * g * ?

**Math**

7x + 84 = - x^2 + 40 x - 6 x^2 - 33 x + 90 = 0 (x - 30)(x - 3) = 0

**algebra**

in 15 min the small hose fills 1/4 of the pool, while the big hose fills the other 3/4 3/4 in 15 min means 20 min to fill

**Physics**

corner to corner is one length and one breadth ... distance = 40 + 30 the diagonal is the hypotenuse of a 3-4-5 triangle ... displacement = ?(30^2 + 40^2)

**Lisa**

the 330 and 320 form a right triangle the hypotenuse is the horizontal displacement the horizontal and vertical displacements also form a right triangle since the three directions are orthogonal ... 330^2 + 320^2 + 10^2 = d^2

**Math**

4 m = 24 5 m = 40 N = 40 - 24

**physics**

use Pythagoras to find the resultant 1.13^2 + .96^2 = s^2

**science - physics**

60 kph = 50/3 m/s 50/3 m/s / 10 s = 5/3 m/s^2

**Math**

multiply by ... ?15 / ?15 3 ?15 / 15 cancel 3 ?15 / 5

**math**

1/4 ft * 1/2 ft * 25 ft = ? ft^3

**Science**

54 kph = 15 m/s the locomotive stops in 50 s ... (15 m/s) / (0.3 m/s^2) = 50 s the average velocity is 7.5 m/s ... (15 m/s + 0 m/s) / 2 = 7.5 m/s 7.5 m/s * 50 s = 375 m ... 25 m from the traffic light

**Pre-Calculus**

so, the radius of the wheel is 15 m ...(31 - 1) / 2 when the altitude is 25 m, the car is 9 m above the center ... 25 - 15 - 1 using Pythagoras, ... hypotenuse (radius) is 15 m ... vertical side is 9 m ... horizontal side is ? HINT (a big one) ... it's a Pythagorean triple...

**Science.**

missing data ... distance to child

**Physics 11**

yes

**Physics 11**

boat on crest ... half wavelength trough ... half wavelength crest between boats ... half wavelength boat in trough the crest between the boats is the max of the wave that the 2nd boat is in the min of (trough) that is a half wavelength from the crest between to the crest the ...

**Physics 11**

from reading the problem, it looks like the boats are one and a half waves apart crest and trough, with one crest in between

**Trigonometry**

a - 2 = 14 b + 5 = 19

**Math**

an angle inscribed in a circle is equal to half of the intercepted arc the opposite angles of a quadrilateral intercept the entire circle ... 1/2 * 360 = 180

**Physics**

the basic range equation is d = v^2 * sin(2?) / g v = launch velocity ? = launch angle g = gravitational acceleration level surface (same height) no air resistance

**Physics**

the horizontal (x) component of the launch speed is ... 284 m/s * cos(43º) the horizontal component is not affected by gravity ... so just multiply by the flight time

**Calculus**

looks good

**Physics 11**

velocity is usually a function of the transmission medium probably doesn't change

**Physics**

m g h

**Physics 11**

c is the speed of light (like E = m c^2) 3x10^8 m/s

**Physics 11**

frequency = speed / wavelength f = c / 3.2 m = 93.7 MHz

**Maths**

4 consecutive heads, then a tail .52^4 * .48

**Physics**

the net force will be toward the center of the square, along the diagonal it's symmetric, so you only have to do one corner use the gravitational formula ... the diagonal is direct ... the adjacent corners pull at a 45º angle along the diagonal (vector)

**Physics**

friction

**Physics**

time = work / power

**Physics**

yes ... this is a simple "trick" question

**Physics**

for objects in free fall ... time up equals time down decelerating on the way up ... accelerating on the way down

**Physics**

m * g * h / t 150 * 9.8 * 2.5 / 3 = ? Watts

**Math**

1500 = 300 e^(k * 5) ln(5) = 5 k k = 32.2% / hr y(t) = 300 e^(.322 t)

**trig**

? is in quad III the tangent of the reference angle is ... 8/15

**Chem**

Ka = [H+][A-] / [HA] 2.34 / 1.86 = 1.26 M 2 x + (.649 - x) = 1.26 Ka = x^2 / (.649 - x) pKa = -log(Ka)

**algebra**

(50 * 1.2) + (x * .8) = 1 (50 + x) 60 + .8 x = 50 + x

**physics**

a = f / m ... d = 1/2 a t^2 the acceleration is inversely proportional to the mass the distance is directly proportional to the acceleration

**Math**

P(t) = 750g (1/2)^(t/250) log(200/750) = (t/250) log(1/2)

**Math**

3500000 = 450 * 3^w log(70000/9) = w * log(3)

**math**

a = .2 (a + b + c) 4a = b + c b = .5 (a + c) ... 2b = a + c ... a = 2b - c ... b + c = 8b - 4c ... 5c = 7b ... 5b + 240 = 7b ... b = 120 c = b + 48 = 168

**Physics**

9.25 t = 1/2 * 2.75 * t^2 1.375 t^2 - 9.25 t = 0 t (1.375 t - 9.25) = 0 t = 0 ... when the black one passes t = 9.25 / 1.375 ... brown catches up

**physics**

force is in Newtons (not Joules) the force with which the truck is pulling is accelerating the truck and the trailer...you're on the right track...just use both masses ... f = (truck + trailer) * a the force pulling the trailer is the tension between them ... tension = ...

**Geometry**

(72 - 2 - 2)^2 + (64 - 2 - 2)^2 = D^2

**science**

fcc has 4 atoms per unit cell 6.02E23 amu per gram (1.2 * 6.02E23) / (63.5 * 4)

**Math**

12C5 (5x)^5 (2y)^7 12C5 * 5^5 * 2^7 792 * 3125 * 128 = 316800000

**Maths science English hindi s.st computer**

four parallel 1? resistors have an effective resistance of 1/4 ?

**Math**

cos(?) = 9/41 = x/r use Pythagoras to find y ... x^2 + y^2 = r^2 sin(?) = y/r ... remember Q IV tan(?) = y/r

**Math**

the common difference is -4 an = 1 - 4n

**Trigonometry**

? is 105º clockwise from the x-axis this gives a 75º reference angle with the negative x-axis the x and y coordinates are both negative x = - cos(75º) y = - sin(75º)

**maths**

18 cm ? 2b + 16 cm ? 50 cm subtracting 16 ... 2 cm ? 2b ? 34 cm dividing by 2 ... 1 cm ? b ? 17 cm

**algebra**

yes

**Maths**

sin(?) = 12/200

**Science**

add all the numbers in the formula if there is no number, a one is understood

**science**

(54000 m / 3600 s) / 20 s

**math**

divide the distance on each leg by the speed to find the time for each leg ave. speed = total distance/total time

**Math**

wow...missing parentheses sure caused a dustup

**Math, probability**

a fair cube is expected to come up an equal number of times on each side 6 sides, 60 rolls ... each side is expected 10 times a) 10 ... expected outcome b) 30 ... expected outcome c) 30 ... expected outcome d) 0 ... only 6 sides

**Algebra 2**

f(x) = (x-2)(x-2i)(x+2i)(x-4+?6)

**Algebra**

3 x = -146 - 4(x + 2) 7 x = - 154

**math help**

t = 100 (1 + .10)^8

**math**

9 / (3/4)

**maths**

the freight train has a 64 km head start ... 32 kph * 2 h the passenger train closes the distance at 20 kph ... 52 kph - 32 kph 64 / 20 = ?

**Physics**

1/2 m v^2 = m g h v = ?(2 g h) = ?(2 * 9.8 * .45) m/s

**Math**

the 5th term is 4 differences from the 1st term

**Physics 11**

the change in velocity (delta v) is ... 9 m/s 5 m/s in one direction, then 4 m/s in the opposite direction

**Physics 11**

work against friction is ... 2520 N * 130 m change in gravitational potential is ... m g h = 1200 kg * g * 27 m the sum is the total energy energy divided by time (5 s) is the power

**Physics 11**

yes

**math**

good answer

**Geometry**

Pythagoras w^2 + 13^2 = 59^2

**Physics 11**

the ball is in free fall with no air resistance gravity is the only force on the ball net force is ... m g

**MATH**

your answer is P(red) the question is P(not red) the two probabilities sum to 1 ... something is either red or not

**math**

14^2 = 196 31^2 = 961

**Algebra 2**

2t + 3r + f = 39 t = r + f 3r - 1 = f subs ... t = r + 3r - 1 = 4r - 1 subs ... 8r - 2 + 3r + 3r - 1 = 39 find r, then substitute back

**Math**

d = original dimension d plus 50% is 1.5 d 1.5 d minus 20% is 1.2 d 1.2^2 = 1.44 final is 44% larger than original

**Algebra**

yes

**physics**

drop speed relative to rest ... s = 90 kph / tan(40º) drop speed relative to vehicle ... s = 90 kph / sin(40º)

**Math**

6C3 combination implies that order is not a factor

**math**

Q IV ... x is positive, y is negative (b,-a)

**physics**

work = m g h in real life, terminal velocity becomes a factor

**physics**

60 mph = 88 ft/s a) the average speed while stopping is 30 mph or 44 ft/s ... the stopping time is ... 200 ft / 44 ft/s divide the initial speed by the stopping time to find the acceleration b) same process as a) but with a different initial speed

**Math**

D = s + 1 s = b + 3 ... b = s - 3 D + s + b = 31 s + 1 + s + s - 3 = 31 find s, then substitute back

**Math**

the area is found by multiplying the base by the height

**Physics**

is the lift moving up or down it makes a difference

**maths**

sin(?) = 18 / 30

**Algebra 2**

[(a1 + a23)/2] * 23 = s23 [(10 + a23) / 2] * 23 = 2760

**Math**

c) the pedal is at 32 cm twice per revolution ... going down and going up 4th time would be 2nd time during 2nd cycle

**physics**

the change in momentum of the car is equal to the change in momentum of the truck

**Math, identities and equalities**

expand the cube of the binomial ... the x^3 terms will cancel ... consolidate like terms match the term coefficients with the left hand side solve the resulting equations for a, b, and c

**Physics 11**

the work calculation is the total work the components are not separable ... not enough information

**Mathse**

use vector addition on the two components ... a^2 + b^2 = c^2

**Physics 11**

the friction/work hint is a good one 90 * g * .6 * d = 720

**Someone please help!!! Math**

OH for three... no such thing as an equilateral right triangle only one way each to make the other two

**Physics**

energy = work 2200 J = 1100 N * ? m

**Math**

looks good

**Math**

x^3 + 3x^2 + 2x = x^3 + 21 3x^2 + 2x - 21 = 0 (3x - 7)(x + 3) = 0 x + 3 = 0

**logs**

yes ... that would equal t

**physics**

the horizontal distance to the edge of the water is ... 10.0 m * tan(40º) the time to fall the 10 m is ... ?(2 * 10 / g) divide distance by time to find the speed

**Algebra 1 Please Help**

squaring a binomial ... use FOIL (2?5)^2 + 2 * 2?5 * 3?7 + (3?7)^2 20 + 12?35 + 63 = 83 + 12?35

**Math**

the 3rd side of a triangle must be long enough to connect the ends of the other two sides ... it must also be short enough that the other two sides can connect its ends difference of other two sides < ... 3rd side < sum of other two sides