# Posts by Damon

Total # Posts: 30,127

**word problem**

L = 4w+9 if L = 85 w = 76/4 = 19 if L = 49 w = 40/4 = 10 10</= w </= 19

**physics**

The mass has nothing to do with this problem since acceleration is not asked for F = k (2q)q/.1^2

**physics**

work = force * distance = q E d = change in potential energy = qV 48 = .1 V

**physics**

forces on q1 attractive k =10^-12* usual k to account for ? force in direction of Fq3=k(8)/.4^2 force in direction of Fq2=k(6)/.4^2 force perpendicular to opposite side Fy= Fq3 cos 30 + Fq2 cos 30 force parallel to opposite side Fx= Fq3 cos 60 - Fq2 cos 60 F = sqrt(Fx^2+Fy^2) ...

**math**

Does he pay at the beginning of the year or the end?

**Math**

y = k x 6 = 3 k k = 2 y = 2 x 2/3 = 2 x x = 2/6 = 1/3 ================= 25 = m (1.25) so m = 25/1.25 a = F/m = 35 /(25/1.25) = 1.25 * (35/25)=1.75 You could have written that down immediately by proportions but: 1. I am a physicist 2. You are doing algebra of direct variation.

**Chemistry**

2 * 6.49 * Avogadro's number which is about 6.02*10^23

**Math**

I can get it in powers of cos but not first power sin^8 = sin^2 * sin^2 *sin^2 *sin^2 but sin^2 = 1 - cos^2 sin^2*sin^2 = 1-2cos^2+cos^4 (1-2cos^2+cos^4)^2 = 1-4cos^2x+6cos^4x-4cos^6x+cos^8x

**Sorry, misread it**

assumed x with 3

**Pam**

If you do not do calculus you complete the square x^2 - 3 = y x^2 - 3 + (3/2)^2 = y + 9/4 (x-3/2)^2 = y+9/4 vertex at (3/2 , -9/4)

**Math**

6.848*10^9 people .22*57.9 *10^6 miles^2 so [ 6.848/(.22*57.9) ]10^3 =.5376 * 10^3 = 538

**Pam**

( 1,000,000 , 1,000,003 )

**Pam**

LOL - looks like = x to me, the 3 is tiny Perhaps she means where the slope is zero? dy/dx = [(x+8)(2x+11) -x^2-11x-16)]/bottom^2 where is numerator = 0? 0 =2x^2+27x+88-x^2-11x-16 0 = x^2 + 16 x + 72 complex roots :( none

**mATH..OHHHH NOOOOO PLEZ HELP I AM STRUGGLING**

In this link which MsSue gave you will find: http://www.mathsisfun.com/quadrilaterals.html That a square is also a rectangle and a rhombus.

**if a square, then a rhombus**

a square has 4 equal sides, so does a rhombus however a square has 90 degree corners a rhombus could have 90 degree corners, but not necessarily. therefore If a square, then a rhombus If a rhombus, maybe a square In other words, a square is a special case of a rhombus

**mATH..OHHHH NOOOOO PLEZ HELP I AM STRUGGLING**

Try to draw a square that is not a rhombus

**word problem, please help**

41 = 25 + 4 t 16 = 4 t t = 4 so t>/= 4

**physics**

Vi = 180,000 meters/3600 seconds average speed during deacceleration= Vi/2 distance = (Vi/2)* 20 = 10 Vi

**Algebra**

a (x-2)(x-3) = 0 a[ x^2 -5x + 6 ] = 0 f(x) = a[ x^2 -5x + 6 ] when x = 1 f(1) = 12 = a[ 1-5+6] 12 = 2 a a = 6 so f(x) = 6x^2 -30 x + 36

**math**

prob jeff not = 1/4 prob Justin will = 2/5 the two events are independent so (1/4)(2/5) = 1/10

**math**

letters a through z. There are 26 on my keyboard order matters so permutations of 26 taken 4 at a time = 26!/[22!]= 26*25*24*23 10 numbers (0 through 9) again order matters = 10!/8! = 10*9 so in the end 10*9*26*25*24*23

**Algebra**

new salary = x = 1s + .093 s so x = 1.093 s

**?hours?**

proofread how many hours?

**Science**

Normal force is indeed 810 mass = normal force /g = 810/9.81 = 82.57 kg but it depends what your class uses for g. I use 9.81 but some classes even use 10m/s^2 (I guess they live on a slightly more massive planet :)

**Math**

3x >/= -6 x>/= -2

**Math**

You are welcome.

**Math**

6 > 10 z - z^2 well if z = 0, the right hand side is 0 and 6 is > 0 so A works by the way, the others do not work ;)

**Science**

If the speed is constant and the cable is horizontal then the friction force equals the tension, 60 N The normal force is m g = 15.0 * 9.81

**Science**

Hard to say since I have no idea where point A is.

**Social Studies**

Then I would indeed pick A. However B was also true of many Germans and played a part in the rise of Adolph Hitler.

**Social Studies**

2. I think A AND B

**Social Studies**

1. NOT C, yes to the others

**Algebra**

x along river and y 600 = x + 2y so x = 600-2y A = x y A = (600-2y)y = 600y - 2y^2 I assume you do not know calculus so complete square to find vertex of parabola. y^2-300 y = -A/2 y^2 -300 y+(300/2)^2 = -A/2 + 150^2 (y-150)^2 = -(1/2) (A-45,000) vertex at y = 150 and A = 45,...

**College Algebra**

.05x +.06(10,000-x) = 550 .05 x + 600 -.06 x = 550 50 = .01 x x = 5000

**Need mass**

say m normal force m g Friction force = .5 mg so T - .5mg = m a T = m (1 + .5g) = 5.9 m

**Math**

first one wrong (1/3)base area*height second one wrong, same formula third one wrong, same formula again I have no idea what the bottom 2 are about.

**Math**

the volume of anything pointy with straight sides = (1/3) base area * height

**science**

Well, we will run out of fossil fuel first of all. It is by definition not "long term" because we are using it up faster than plants are turning into coal and oil and even gas. However there are more immediate obvious problems such as air pollution and land pollution...

**Physics help please**

If it is exactly the same distance from each speaker then there is no difference in path length between the two paths and the signals superimpose to give a maximum.

**do not do magic**

we do not see your figure or line of reflection.

**PHYSICS**

m a = 520*2 = 1040 N third law ---> -1040 N net force on car = car mass * car a = 1400 * 2 = 2800 N the tires exert 1040 N more on the road to accelerate the trailer, but that is not what you asked :)

**math**

try common factor 2 182/3 60 2/3

**PHYSICS**

If it moves with constant V the forces add to zero. to counter first Fx = -6.2 N to counter second Fy = +5 N |F| =sqrt(5^2 + 6.2^2 ) = 7.96 N quadrant II tan T = 5/-6.2 T = 180 - 38.9

**MAthematicas**

slope of x = 1 is anything/0 which is UNDEFINED

**Algebra 2**

easy way g = [(x-3)/2 ] ??? g(1) = [(1-3)/2 ] = -1 f(-1) = 1-3+6 = 4 --------- hard way f(g)=[(x-3)/2]^2+3[(x-3)/2 ]+6 put in x = 1 f(g)=[(1-3)/2]^2+3[(1-3)/2 ]+6 = 1 -3 +6 = 4

**Algebra I**

zero when x = 1, always + elsewhere 2 x^4 would be symmetric about y axis HOWEVER the -8x adds a sloping line to that which destroys that symmetry so not symmetric about any of them

**Math**

Why could he not be 877.375 meters high? Integer means 877 or 878

**MATH**

x = 3 has no y axis intersection. It is a vertical line through x = 3, no solution , no relation , no way :)

**Geometry**

I assume D is on AB? Is it an altitude hitting AB at 90deg ? if so A is 60 and B is 30 then CD = 8 sqrt 3 AC = 16 AB = 32 BC = 16 sqrt 3 so perimeter = 48 + 16sqrt3

**Algebra**

80 a +200 b = 160 *20 = 3200 and a + b = 20 so b=20-a 80 a + 200(20-a) = 3200 80 a + 4000 -200 a = 3200 120 a = 800 12 a = 80 a = 6 2/3 pounds so b = 20-a = 13 1/3 pounds

**math**

no, not distributive property of multiplication If the question were (5x+3)(8x+1) no plus sign, so TIMES then you could say 5x (8x+1) + 3 (8x+1) That would be distributive property.

**math**

You do not show any multiplications or divisions in your given problems. All you can do is collect like terms. (5x+3)+(8x+1) is the same as 5x+3+8x+1 which is 13 x + 4 =========================== similarly (2t^2-3t)-(7t^2-5t) = 2 t^2 - 3 t - 7 t^2 + 5 t Note that +5 t due to ...

**Algebra 2**

you do not have either + or - 3.57 people

**Math**

A. y = -2x/3 + 4 B. goes through (0,-2) and (3,0) (0+2)/(3-0) = 2/3 = (y+2)/(x-0) y+2 = 2x/3 2x -3y = 6 C. y = -3 x + b 2 = -3(1) = b b = 5 y = -3x + 5

**Trig identies, Calculus**

use x^2-y^2 = (x+y)(x-y) c^4-s^4 = (c^2+s^2)(c^2-s^2) = 1 (c^2-s^2) = c 2x

**matrix, linear transformations**

first h*g 4 12 0 6 then k * h*g 0 6 4 12 in detail h * g |1 2| | 4 0 | |4 12| |0 1| | 0 6 | |0 6 | then |0 1| |4 12| |0 6 | |1 0| |0 6 | |4 12|

**Calculus**

I agree because as x increases the integral decreases.

**you mean 3/4 I assume**

cos is - in quad two sin^2 T + cos^2 T = 1 9/16 + cos^2 T = 16/16 cos^2 T = 7/16 cos T = -sqrt 7/ 4 agree with you

**Math**

but what is 28/4 !!!! ???? LOL

**Chemistry**

do the silver one first Ag 108 g/mol Cl 35.5 g/mol AgCl = 143.5 g/mol so 5.57 g AgCl has (108/143.5)5.57 = 4.19 g of Ag so 8.12 g stuff has 4.19 g Ag so 4.72 g stuff has 2.44 g Ag so we have 4.72-2.44 = 2.28 g of C and H we got 7.95 g of CO2 C = 12 g/mol O2 = 32 g/mol CO2 = 44...

**Precal**

Actually I just used the trig identity tan = sin/cos same difference :)

**Precal**

sqrt(5) / 2

**Chemistry**

Oh, that Einstein guy again? how much energy in a photon of wavelength 902 nanometers ? approx 900*10^-9 = 9*10^-7 meter L f = c 9*10^-7 f = 3*10^8 f= (1/3) * 10^15 E = h f E = 6.6*10^-34 * (1/3)*10^15 E = 2.2 *10^-19 Nope not quite strong enough to spring an electron free.

**chemistry help, urgent!**

Google buffer

**Math**

B. does it have constant slope? (-1+5)/(0+2) = 4/2 = 2 ((1+1)/(1-0) = 2/1 = 2 (3-1)/(2-1) = 2/1 = 2 :) slope, m = 2) C. y = mx + b so y = 2x-1 D. y = 2(10)-1 = 19

**physics**

m * v = 8.0 m * 10 = 8.0 m = 0.8 kg (about a pound)

** NO**

4. It would be correct if you typed it correctly 5.1 * 10^-8 Your answer is nonsense. the ^ symbol means "to the power" or superscript 5. Nonsense again 10^12 (Please learn to type exponents) is smaller than ANYTHING * 10^13 or 14 or 21 or whatever bigger than 12 ...

**math**

1. -10 * -1 = 10 2. doubles to * 2 at 4 doubles to * 4 at 8 doubles to * 8 at 12 8*90 = 720 3. No because not a power of ten. Your answer is wrong 4.8 IS between one and ten 1007 is NOT a power of ten. Perhaps 10^7 ????

**Math**

Sketch a graph ! I notice a dip for example.

**Math**

multiply everything by 2 -6 + n = 18 so? add 6 to both sides

**Engineering Science N4**

at the point where it has acceleration = a and velocity = 0, I assume the static friction rules. normal force = m g - F sin 15 friction force = .4(mg-F sin 15) pull force horizontal = F cos 15 so Fcos15 -.4(mg-F sin 15)= m a

**math help asap**

Oh come on 15/10 = ? 24/16 = ? 30/20 = ?

**Educational Technology**

I am using library, school and other computers around town being very busy with local political issues. I google in to answer a question once in a while but usually do not remember to write my name in above.

**Hey, we did the one with the copies**

Your turn.

**Algebra**

machine 1 does n copies/170 min so n/170 copies/min machine 2 does n/119 copies/min so n = (n/170 + n/119) t 1/t = 1/170+1/119 = .014286 so t = 70 min

**math**

0 = -32 t + 28 so t = 28/32 google projectile motion in English units (acceleration of gravity = -32 ft/second^2)

**Math**

LOL, anywhere else !

**Algebra**

n and n+32 1/n + 5/(n+32) = 1/6 multiply all terms by 6n(n+32) 6(n+32) + 5(6n) = n(n+32) 6n + 192 + 30 n = n^2 + 32 n n^2 - 4n - 192 = 0 (n-16)(n+12) = 0 n = 16 n+32 = 48 ================ ================ check 1/16 + 5/48 = 1/6 ??? 3/48 + 5/48 8/48 1/6 sure enough

**math**

ratio is 1/2 so one of these: 8 4 6 3 4 2 2 1 difference is 2 so 42

**physics**

365.25*24 = h, number of hours distance = 2 pi r = 6.28* 1.5*10^8 km speed = distance/hours

**suggestion**

Maybe point (a,b,c) line like z = m x + n y + k find vector along line find vector from point to random spot on line do cross product, show independent of spot chosen on line. That cross product is perpendicular to the plane containing the two lines

**geomatry**

You are welcome :)

**Neil**

not area, perimeter

**geomatry**

sides a, b c perimeter = a+b+c perimeter 4 a + 4 b + 4 c =4(a+b+c)

**no way**

You say rectangle. I assume typo? Did you read what you wrote? All terms in the area are - ? -17.5 x^2 -138 x - 27 = 231? +17.5 x +138 x + 258 = 0 x = -3.05 or x = -4.84

**College Algebra**

f = 3(6x-5)^2 -4 = 3 (36 x^2 -60 x +25) - 4 = 108 x^2 -180 x +71

**Physics**

How much Ke should she have at 328 m? m g h = 71(9.81)(1370-328) Joules How much Ke does she actually have? (1/2) m v^2 = (1/2)(71)(68)^2 The difference is how much work is done by air friction. ========================== At 328 meters she has a total energy of m g 328 + (1/2...

**value**

log x^n = n log x a 2a 3a 4a .......

**Physics**

1/d^2

**algebra**

e^[2(3x^2-2)]

**Math**

small = s and b = big s = b-2 1/s + 2/b = 5/12 so 1/(b-2) + 2/b = 5/12 agreed multiply by 12 b (b-2) 12 b + 24(b-2) = 5 b(b-2) 12 b + 24 b - 48 = 5 b^2 - 10 b 0 = 5 b^2 - 46 b + 48 0 = (b-8)(5b-6) so use b = 8 s = b-2 = 6

**Physics**

normal force = m g max static = .75 m g so m a = .75 m g for slide to start or a = .75 g now the slide a = .75 g v = a t 33 = .75 g t t = 33/(.75*9.81)

**physics**

m g h = m g (3.4/2)

**Physics - Calculating work required**

This is a problem about POTENTIAL! The path does not matter ! That is the whole point. The voltage potential difference is the difference in potential energy of a unit charge at the two voltages. work done = charge * change in V

**Physics**

force = 55*9.81 Newtons pressure = force/area if area is in meters^2 then force is in Pascals

**physics**

Vi = 110,000 meters / 3600 seconds v = Vi + a t at end, v = 0 so 0 = Vi -49 t t = Vi/49 then d = Vi t + (1/2) a t^2 d = Vi (Vi/49) - (1/2)(49)(VI^2/49^2) d = (1/2)Vi^2/49

**Algebra**

I worked it out in the way you are probably used to. Scroll down

**Algebra**

44+47 = ? mi/h time = 69/speed

**Math**

divide by 1,000 kg/tonne

**arrgh upside down**

.3456/.84 = .4114