# Posts by Scott

Total # Posts: 2,308

**chemistry**

100% * (1/2)^(120 / 4.55)

**ALGEBRA**

the "graphing form" is the vertex form f(x) = a (x - h)^2 + k ... a = 1 in this case vertex is (h,k)

**Physics**

momentum is conserved 65 kg * 2 m/s = 55 kg * v the canoe moves in the opposite direction

**Physics**

the index of the solution must be approximately the index of the material there is no refractive boundary between the material and the solution, so the material seems invisible

**Pre calc**

the height of the observer is ... h = 375 sin(25º) the height of the tower above the observer is ... h = 375 sin(39º) add the two heights to find the tower (remember to round)

**pre calc**

exact answer is given in "related questions" at bottom of page

**Algebra**

they each paint a fraction of the house ... the fractions add to one t / 10 + t / 15 = 1 3t / 30 + 2t / 30 = 1 5 t = 30

**precalc**

a) u - v b) u - 2 v c) 2 u + 4 v

**Maths**

c = 4 h 2 (c - 280) = h c = 8 (c - 280) solve for c, then substitute back to find h

**Preclc**

use the Law of Sines to find the distance from one of the people to the spot then use trig to find the height of the spot the angle at the spot is ... 180º - 75º - 45º = 60º 600 m / sin(60º) = x / sin(45º) ... x is the distance from the observer ...

**Physics**

"acquires" is an interesting term the muscle energy on the ground (like a spring), becomes gravitational potential energy during the jump it appears you have correctly calculated the height

**Physics**

looks actually correct to me

**pre calc**

because i made a typo...

**pre calc**

use the Law of Sines the angle at the ship is ... 180º - 46º - 63º 283 ft / sin(43º) = 350 ft / sin(63º) ... = x / sin(180º - 46º - 63º)

**algebra**

to save money with B ... A > B .2 m + 40 > .12 m + 56.32 .2 m > .12 m + 16.32 .08 m > 16.32 m > 204

**Precalculus**

the plane flew 360 mi east in the air but the air moved east and north ... this is the wind speed 15 mi E and 20 mi N forms a 3-4-5 triangle the wind is ... 25 mi / 2 hr ... on a bearing N of E the tangent of the bearing angle is ... 20 / 15

**math**

twice the teeth means half the revolutions teeth * speed = k = 48 * 3 = 144 96 * s = 144 ... s = 144 / 98

**physics**

nope E friction from the tires and normal force from the banking

**Math**

looks like a glass/class typo 2/3 of 30.9 kg 2 * 30.9/3 = ?

**Pre calc DUE TOMRROW**

i mis-transposed the 43º ... it should be 46º

**Pre calc DUE TOMRROW**

use the Law of Sines the angle at the ship is ... 180º - 46º - 63º 283 ft / sin(43º) = 350 ft / sin(63º) ... = x / sin(180º - 46º - 63º)

**geometry**

use ... a^2 + b^2 = c^2 ... twice

**Pre-cac**

the max is on the axis of symmetry x = 3 / 2a 15 = a (3 / 2a)^2 - 3 (3 / 2a) + 5 10 = (9 / 4a) - (9 / 2a) 10 a = 9/4 - 9/2 = - 9/4 a = -9/40

**physics**

google "uniform circular motion" look at the diagrams on wikipedia

**Math**

time = distance / velocity

**Math**

the peak of the jump lies on the axis of symmetry of the parabola ... x = -b / 2a = -21 / (2 * -16) time up equals time down time of flight (jump) = 2 * 21/32

**Chem: please help**

new conc = old conc * 15/50

**Chemistry help!!**

new conc = old conc * 15/50

**math**

? * (5/2)^2 * (8/2) * 10^6

**Chemistry**

i think you're right

**Maths**

2 (x^3 + 6^3) the sum of cubes is a standard factoring

**math**

the area for the measurements you give is not listed see "Heron's Formula"

**Math1**

100 (1 + .04)^12

**math**

.08 f + .05 s = 1330 .12 f + .02 s = 1500 solve the system using elimination or substitution

**Calculus**

integrate the derivatives Y"=6 ... y' = 6 x + c ... 4 = 6 * 1 + c ... c = -2 y' = 6 x - 2 ... y = 3 x^2 - 2 x + c ... 6 = 3 - 2 + c ... c = 5 y(x) = 3 x^2 - 2 x + 5

**physics**

a = v / t d = 1/2 a t^2

**physics/maths help damon or steve or scott**

6 V0 + 18 a = 246 V0 + 9 a = V1 3 V1 + 4.5 a = 69 subs ... 3 V0 + 27 a + 4.5 a = 69 ... times 2 ... 6 V0 + 63 a = 138 subtract to eliminate V0 ... 45 a = -108 ... a = -2.4 substitute back to find V0

**Physics**

the time of flight is the time it takes for the car to fall the difference in elevation of the two sides 21.3 - 1.8 = 1/2 * 9.8 * t^2 find t the car has to travel 61 m in time t the car's horizontal velocity is constant its vertical velocity increases due to falling ... Vv...

**Physics**

the energy is proportional to the SQUARE of the compression 4^2 = ?

**Math**

multiply the percentage (as a decimal) by the total number of students (500) to find out the numbers for the various sports then subtract the basketball number from the football number

**Math**

the built in Windows calculator (in scientific mode) works pretty well

**Math**

there are 365 compounding periods per year [1 + (.03 / 365)]^365 = 1 + APY

**chem**

the pressure is proportional to the absolute (Kelvin) temperature 1.89 / (24.0 + 273) = p / (38.0 + 273)

**SCIENCE**

for EQUAL branches total r = branch r / # of branches

**Math**

5.00 + (0.65 * x) = 10.00 + (0.45 * x) .20 x = 5.00

**Physicd**

7 N?s - 5 N?s = 2 N?s m v = 2 N?s m = 2/5 kg

**Algebra**

the slope means that the candle shortens by .4 cm per hour of burning how much taller would it be nine hours earlier? (how much did it shrink in nine hours?)

**math**

s + c = 72 / 3 s - c = 72 / 4 add the equations (to eliminate c) solve for s substitute back to find c

**math**

m?CAB = m?CBA = 30º m?BCM = 90º BM = 24 ... 30-60-90 TRIANGLE AM = CM

**science**

RL and LR are brown eyed but carry the blue recessive they represent half of the population

**math**

there are two "integer" multiples of 2 A 0.25 = 2^-2 B .05 = 2^-1

**Maths**

each component has an 80% chance of being good so the probability of a good (not defective) product is ... .8^4 probability of a defective product is ... 1 - .8^4

**Physical Science**

pH + pOH = 14 pOH = 12 [OH] = 10^-12

**Pre-cal**

s = 2 f - 5 s + f^2 = 115 f^2 + 2 f - 5 = 115 f^2 + 2 f - 120 = 0 (f + 12)(f - 10) = 0 f - 10 = 0 ... f = 10

**Math**

did you see the two sig fig part? 210000

**SCIENCE**

think you're right

**Maths**

the 8th term is 5 differences from the 3rd term, and 7 differences from the 1st term e = 2 t = 2 (f + 2 d) f + 7 d = 2 f + 4 d ... 3 d = f sum8 = 8 f + 28 d = 39 substituting ... 24 d + 28 d = 39 ... 52 d = 39 ... d = 3/4 substitute back to find f

**math**

there are 36 inches in a yard 580 / 36 = ?

**11 grade / physics**

Vv = V sin(30º) = ?(2 g h) Vh = V cos(30º) = Vv / tan(30º)

**Chemistry**

for the given volumes (22.5 + 22.5) * (30.17 - 23.50) * 1.00 ... * 4.184

**Chemistry 11**

looks fine

**Math**

the area is the average of the bases, multiplied by the height divide the area by 12 to find the number of boxes needed remember to round up to the next WHOLE number, if necessary

**Chemistry**

temp change times mass of H2O times specific heat of H2O (37.8 - 21.6) * 100 * 4.179 remember sig fig

**Physics**

momentum is conserved Mb * Vi = (Mb + Mw) * Vf Vi = (.005 + 8) * 0.50 / .005

**Physics**

frictional force is the normal force multiplied by the coefficient in this case ... Ff = Ma * g * 0.25 the frictional force keeps A moving with B ... so the max acceleration (without slip) is ... Ma / Ff the max force is the max acceleration multiplied by the sum of the masses...

**Physics**

1/2 m (v sin(60º)^2 = m g 10 km v = ?(g * 20 km) / sin(60º) this is the minimum velocity for the projectile to reach the height of the plane ... faster would be better

**Physics**

the weight change in the kerosene is the weight of kerosene displaced by the solid if the solid were kerosene, its weight would be 1.6 g ... (4.8 - 3.2) so the density ratio is ... 4.8 / 1.6

**Chemistry**

in (a) and (b), the hydroxides are probably in solution and the hydrogen is evolved as a gas

**pre calc**

the full solution is in the related questions at the bottom of the page

**pre calc**

the two given population numbers only have one significant digit either 670000 or 700000 is what the computer is looking for

**Chemistry**

the drops happen, but are not counted the fewer the drops to the endpoint, the lower the concentration of the unknown the concentration would appear to be lower than it actually is

**physics!**

velocity = circumference / period v = (2 * ? * 11.30 m) / 16.60 s plug it in and crank the crank

**pre calc**

1.7 = e^0.01t ln(1.7) = 0.01 t 100 ln(1.7) = t

**Math**

(300 * 200) - (280 * 180)

**Physics**

the centripetal acceleration is ... v^2 / r a) v^2 / r = g ... v^2 = 500 m * 9.8 m/s^2 b) his weight will be twice normal ... w = 2 * 80 * g

**physics**

95 km/h = 26.4 m/s 26.4 / 4.2 = stopping time (26.4 + 0) / 2 = ave stopping speed

**Math**

y varies inversely with x ... x y = k 25 * 5 = 125 ... x y = 125

**math**

the solution is the intersection of the lines

**Physics**

f = 1 / cycle time = 1 / (34.8 / 15)

**pre calc**

100 * 1/2 * 1/2 = 25 12 days = 2 half lives OR 1/4 = e^(12k) ln(1/4) = 12 k [ln(1/4)] / 12 = k ln(1/2) = k t t = ln(1/2) / {[ln(1/4)] / 12} = 6

**geometry**

it gives you the size of the interior angles of a regular polygon of n sides

**math**

90º is a quarter of a circle 1/4 * 1/4 = ?

**Algebra**

.9 x + 6.8 = .65 x + 7.3

**science**

the weight (force) of 50 kg is ... 50 kg * g ... g is the acceleration by gravity 200 * d = 50 * g * (2 - d)

**Pre calc**

pop = 900000 * k^t 800000 = 900000 * k^2 8/9 = k^2 ... k = 2?2 / 3 p = 900000 * (2?2 / 3)^5

**Math**

no ... is imaginary ... and irrational

**Math**

no ... is imaginary

**science**

a = f / m = 5224700 / 13500

**Algebra**

the corners are now three times farther from the origin just multiply each coordinate by 3

**Physics**

the beam weight is 50.3 N/m when Suki crosses a support to some distance (d), the equilibrium equation is ... (100.6 N * 1.00 m) + (595 N * d) ... = (251 N * 2.50 m) the distance from the end is ... 2.00 m - d

**Algebra2**

this is a circle, centered at the origin the domain and range are the same

**Math ASAP**

looks good

**Math**

180º - 65º - 65º = ?

**Calculus**

d = 2 r = 3 h ... r = 3/2 h v = 1/3 * ? * r^2 * h v = ?/3 * 9 h^3 / 4 dv = 9 ? h^2 dh / 4 dh = 4 dv / (9 ? h^2)

**Physics**

the incident angle is measured relative to the normal to the surface 90º - 60º = 30º

**Physics!**

.4 * 3000 = f (in Joules) 1) ma = 0utput / input = 3000N / 1000N 2) not enough info

**Math**

n, n+2, n+4 4(n+2) = n + n+4 + 2

**Physics**

that is the compression of the spring the more it's compressed, the greater the force

**Physics**

f = m a ... a = f / m a = (315 N * .15) / .00446 kg a is in m/s^2

**Physics**

ave speed = 1.4 rad/s stop time = 2 * ? * 60 / 1.4 acceleration = 2.8 rad/s / stop time ... negative ... wheel is stopping 30 * 2 * ? = 1/2 a t^2 + 2.8 t ... solve for t