Thursday

November 26, 2015
Total # Posts: 22,915

**Algebra 2**

I did this below.. These are two straight lines with the same slope. They never intersect.
*October 12, 2015*

**math functions**

(3x-3) - (x^2-4) = 3 x - 3 - x^2 + 4 = -x^2 + 3x + 1
*October 12, 2015*

**Math**

n and n+2 (n+2) + 3 n = 234 4 n = 232 58 and 60
*October 12, 2015*

**Physics**

You do not say what directions are x and y gravity force down = m g = 1300 * 9.81 component of gravity force normal to road = m g cos 20 that is balanced by force up from road normal to road, no acceleration normal to road Component of gravity force down slope = m g sin 20 ...
*October 12, 2015*

**Math Decimals**

8 >& . <& & so & is middle of remaining 8 >& . <& 3 so 86.23
*October 12, 2015*

**Algebra 2**

LOL - no way second equation is 3 x - y = 2 or 6 x - 2 y = 4 two straight lines with the same slope. They never cross.
*October 12, 2015*

**Physics**

horizontal speed is constant so t = 40.5 / 7.9 so distance down = (1/2) g t^2 = 4.9 t^2
*October 12, 2015*

**Physics - hey, see one you see them all**

This problem is practically identical to the one I just showed you how to do with the two cliffs. horizontal: u = 44.1 cos 31.3 forever t = 43.7/u vertical h = 0 + Vi t - 4.9 t^2 where Vi = 44.1 sin 31.3
*October 12, 2015*

**Physical Science**

There is no horizontal force (ignoring air friction) so the horizontal velocity remains 40 m/s In the vertical direction there will be acceleration down of about 9.81 m/s If the original vertical speed was zero, after one second it will be 9.81 t = 9.81 m/s down
*October 12, 2015*

**Physics**

check some of the solutions listed below
*October 12, 2015*

**living skill**

Steve gave you the recipe. In series, voltage drops add, currents are the same. In parallel, currents add, voltage drops are the same.
*October 6, 2015*

**physics**

force from road id perpendicular to road surface, call it N N cos theta = m g N sin theta = m v^2/R so tan theta = m v^2/( R m g) = v^2/(R g) so theta = tan^-1 (v^2/Rg) N = m g /cos theta where theta = tan^-1(v^2/Rg)
*October 4, 2015*

**physics**

Oh my, one by one: 1. A 2 kg body moves with constant velocity when a force of 10 N is applied. What is the coefficient of kinetic friction between the table and the body? (Would I use F=m*g for this??) ================================== F = m a but a = 0 so there is no net ...
*October 4, 2015*

**physics sl**

F * 4.2 = work done by friction Ke at start = (1/2)(2.2)(5.5)^2 Ke at finsh = 0 chnge in potential energy = 0 because level so 4.2 F = 1.1* 5.5^2
*October 4, 2015*

**physics**

0 = h + Vi t - 4.9 t^2 or 0 = h + 20*6 - 4.9 * 36
*October 3, 2015*

**Algebra 2**

I get [(a-b)(a+b) / (a+b)^2 ] ----------------------- [(2a+3b)(a+b) / (2a+3b)] [ (a-b) / (a+b) ] = ------------------- [ (a+b) ] = ( a-b) / (a+b)^2
*October 2, 2015*

**E& M physics**

find distance from center of triangle to any corner, call it d then potential for each electron is Pe = k |Qe|^2 /d potentials add so multiply by 3
*October 2, 2015*

**Algebra 2**

I can not factor that are you sure it is not 4 x^2 - 9 that would be the difference of squares which is (2 x - 3)(2 x + 3)
*October 2, 2015*

**math**

It must be longer than 5 which is 8-3 It must be shorter than 11 which is 8+3
*October 2, 2015*

**Physics. Interesting group question**

Sure they apply. It is just that you need to include the rotation of the earth in your problem statement if doing a large scale problem. Google "coriolis acceleration" or "cyclone physics")
*October 2, 2015*

**Need help with projection map**

I do not agree. The important thing about the Mercator projection is that the direction of one point to another, say New York to Fastnet Rock south of Ireland is accurate on the chart even though the earth curves significantly between points. (angles on the sphere are as you ...
*October 2, 2015*

**math**

10/.6 = 100/6 = x^2 -5.8 x so x^2 - 5.8 x - 16.7 - 0 solve the quadratic
*September 30, 2015*

**Math**

line given is 4x − y = 4 rearranged that is y = 4 x - 4 so slope = 4 new slope = -1/old slope so m = -1/4 y = -x/4 + b to find b put the point in 2 = -(1/4)(-7) + b solve for b
*September 30, 2015*

**Physics**

energy at top = m g h + (1/2) m v^2 = 2 * 9.81 * 1.99 + .5*2 (2.77)^2 energy at bottom = (1/2) m v^2 = .5*2(2.34)^2 difference is loss to friction. If there were no loss to friction of course it would be going FASTER at the bottom :)
*September 30, 2015*

**physics**

h f where h = 6.626 * 10^-34 f = 1/T = c/lambda = 2.998*10^8/2.21 so E = 6.626*10^-34 * 2.998*10^8/2.21 Joules
*September 30, 2015*

**Physics 114**

7 = (1/2) g t^2 7 = .5 * g * 1.19^2 solve for g
*September 30, 2015*

**math**

b + (b-7) + (b-5) = 408 3 b - 12 = 408 3 b = 420 get it?
*September 30, 2015*

**Pre-Algebra**

(x, y) = ( (-2+1), (-5+3) ) = (-1 , -2 )
*September 28, 2015*

**Math**

tan theta = 25/15 so theta = tan^-1 (25/15)
*September 28, 2015*

**Science PLEASE help!**

A theory explains how something might work, for example the kinetic theory of gasses. A law states a relationship for a particular type of problem, such as F = m A , Newton's second law.
*September 28, 2015*

**Math**

the curve crosses the x axis at x = -3 the curve just grazes the x axis at x = -1, bringing TWO roots together.
*September 28, 2015*

**health**

hunger means your body NEEDS food. appetite means your body LIKES food. therefore I would pick b
*September 28, 2015*

**Algebra 1**

volume and time. put a measuring cup there. Find out how much comes out in ten seconds. Convert that to gallons per minute or whatever.
*September 28, 2015*

**mechanics**

A dot B = |A| |B| cos T 0 + 6 - 4 = sqrt(38) sqrt(8) cos T 2 = 17.44 cos T T = cos^-1 .1147 T = 83.4 degrees
*September 28, 2015*

**Math**

Now you are going to have to try. I already took this course and do not need the practice.
*September 28, 2015*

**Math**

3(x) = 2(120 - x) 3 x = 240 - 2 x 5 x = 240
*September 28, 2015*

**Science**

Looks reasonable to me.
*September 28, 2015*

**SS please check and help**

You are welcome.
*September 28, 2015*

**SS please check and help**

Iroquois - woods, long houses, fires inside for heat, woods animals, fish and nuts etc for food Navajo - desert environment, grew some crops, built brick type dwellings for protection from sun heat Cheyenne - plains, buffalo, tepees you could drag with horses to follow food. I...
*September 28, 2015*

**SS please check and help**

In other words woods, desert and plains
*September 28, 2015*

**SS please check and help**

Heavens, look at Iroquois and Navajo or Cheyenne :)
*September 28, 2015*

**your answer?**

I do not see your answer. Did you try to copy and paste? That does not work on this site. If I try to answer it would go on for pages. Are you asking about the Americas specifically or do you include Antarctica, the Arctic, Africa, South and SE Asia, China, Japan, Polynesia, ...
*September 28, 2015*

**Honors physics**

3.24 * cos 24 = east component 3.24 * sin 24 = north component By the way, look at the questions and answers below.
*September 28, 2015*

**Math**

1/4 + 5/8 = 2/8 + 5/8 = 7/8 = amount used for pepper and tomatoes 2 1/2 - 3/4 = 5/2-3/4 = 10/4 -3/4 = 7/4 = total amount used so 7/4 - 7/8 = 14/8 - 7/8 = 7/8 for lettuce
*September 28, 2015*

**math**

L + w = 41 so L = (41-w) and L-9 = w so (41-w) - 9 = w 2 w = 32 w = 16
*September 28, 2015*

**Math**

x is meters, y is inches Two points for line (1 , 39.37) and (2 , 78.74) we know right off that this line goes through the origin but anyway in our determination to follow the instructions we continue (y - 39.37)/(x-1) = (78.74-39.37)/(2-1) (y - 39.37)/(x-1) = 39.37 y - 39.37...
*September 28, 2015*

**math**

how about x-30 ?
*September 28, 2015*

**Math**

(7/9) total = 315 so total = 315 (9/5) = 63*9 = 567 so empty = 567 - 315 = 252
*September 28, 2015*

**math**

x and x+2 x^2 + (x+2)^2 = (2x+2)^2 - 48 x^2 + x^2 + 4 x + 4 = 4 x^2 + 8 x + 4 - 48 2 x^2 + 4 x - 48 = 0 x^2 + 2 x - 24 = 0 (x -4)(x+6) = 0 x = 4 and x+2 = 6
*September 28, 2015*

**Physics**

max use C = .38+.02 = .40 A = 1.77+.02 = 1.79 rho = 1.20 v = 10+ .5 = 10.5 for min subtract instead of add the uncertainties
*September 28, 2015*

**AP Physics**

method is here from previous questions, change numbers: http://www.jiskha.com/display.cgi?id=1358405742
*September 28, 2015*

**Calculus**

x^3+y^3 = 1 (x^3)'+(y^3)' = (1)' 3x^2+3y^2(y') = 0 3y^2(y') = -3x^2 y' = -3x^2/3y^2 y' = -x^2/y^2 GOOD y" = [(y^2)(-x^2)'-(-x^2)(y^2)']/y^4 then I disagree with next line y" = [(y^2)(-2x)-(-x^2)(2y DY/DX)]/y^4 WHERE dy/dx = -x^2/y^2
*September 27, 2015*

**science- physics**

90,000 m/3600 s = 25 m/s v = a t so 25 = 7.3 t t = 3.42 seconds distance car went = 25*3.42 distance cycle went = (25/2)* 3.42 (average speed is half max speed if a is constnt)
*September 27, 2015*

**Math**

linda --- 1 room/7 hours marvin ---- 1 room/x hours 5 hours (1/7 + 1/x) = 1 1/7 + 1/x = 1/5 multiply all terms by 35 x 5 x + 35 = 7x 2 x = 35 x = 17.5 hours
*September 27, 2015*

**precalc**

3 - |2x-4| </= 1 -|2x-4| </= -2 then |2x-4| >/= 2 so (2x-4) >/= 2 2 x >/= 6 x >/ = 3 OR -(2x-4) >/= 2 2x-4 </= -2 2 x </= 2 x </= 1
*September 27, 2015*

**Physics**

let speed of release = x horizontal speed =u = x cos 50 initial vertical speed Vi = x sin 50 4.6 = (x cos 50) t so t = 4.6/(x cos 50) 3 = 1.8 + (x sin 50) t - 4.9 t^2 1.2 = (x sin 50)(4.6/x cos 50) - 4.9 (4.6^2)/(x^2 cos^2 50) 1.2 = 4.6 tan 50 - 4.9 (4.6^2)/(x^2 cos^2 50)
*September 27, 2015*

**math please help**

20 and 30, scroll down
*September 27, 2015*

**Chemistry**

6*10^23 * 3*10^-23 = 18 I do not know what element that might be. It is between Oxygen and Fluorine
*September 27, 2015*

**Geometry**

see geography below
*September 27, 2015*

**Geography**

4 y + 2 = 6 y - 50 2 = 2 y - 50 52 = 2 y 26 = y seems to me y is positive without doing much.
*September 27, 2015*

**Physics**

horizontal problem: u = 40 cos 21 for the whole time t d = 33 so t = 33/(40 cos 21) vertical problem, same t initial vertical velocity Vi = 40 sin 21 h = Vi t - 4.9 t^2
*September 27, 2015*

**College algebra**

10% = 10/100 = .10 = .1
*September 27, 2015*

**College algebra**

yes
*September 27, 2015*

**College algebra**

900 + .1 x = 1200 + .15(x - 8,000) is breakeven point. Above that take the 15%
*September 27, 2015*

**college algebra**

h + w = 50 w = (2/3) h (5/3) h = 50 h = 30 w = 20
*September 27, 2015*

**Physics**

Since there is no friction, we do not need the normal force on the ramp. The component of gravitational force down the ramp is m g sin 30 The tension up the ramp is T so F = m a T - m g sin 30 = m a so T = m (9.81 * .5 + 0.9) = 146 (4.6 + 0.9)
*September 27, 2015*

**science**

F = m A when at rest. Force down on floor =weight = m g Force up from floor on shoes = m g Force up - Force down = 0 so A = 0 when accelerating upward the force up from floor on shoes must be greater thsn the force down, m g Fup - m g = m A so Fup = m g + m A
*September 26, 2015*

**science**

F = m A 19.6 = 2 A A = 9.8 m/s^2
*September 26, 2015*

**math**

distance = 362 = 4 s + 2(s-20)
*September 26, 2015*

**Simultaneous equation need help**

from second y = [(1 0) x - (1 0)] substitute (1 1) x + (1 0) [(1 0) x - (1 0)] =(1 0 0 0 1) now what is (1 0) (1 0) ? (1^2^1 +0*2^0)(1^2^1 +0*2^0) = 2^1*2^1 = 2^2 = (1 0 0) so we have (1 1) x + ((1 0 0) x - (1 0 0)] =(1 0 0 0 1) or (1 1 1)x = (1 0 1 0 1) x = [ 2^4 + 2^2 + 2^0...
*September 26, 2015*

**Z table**

check here: http://davidmlane.com/normal.html
*September 26, 2015*

**Z table**

http://davidmlane.com/normal.html
*September 26, 2015*

**Arithmetic**

2 miles (60 min/3 miles) = 40 min
*September 26, 2015*

**Arithmetic**

He bought 4 for $25 so 4 * 6.98 - 25
*September 26, 2015*

**physics**

well, for v = 10 for example F = .5 (.38)(1.77)(1.2)(100) for v = 15 then F = F at 10 (225/100) and for v = 20 F = F at 10 (400/100) etc graphs F vs v is parabola F vs v^2 is straight line
*September 26, 2015*

**Calculus**

y = 4 x (x+1)^-.5 y' = 4x[-.5(x+1)^-1.5] +(x+1)^-.5[4} = -2x /[(x+1)(x+1)^.5] + 4/(x+1)^.5 = -2x /[(x+1)(x+1)^.5] + 4(x+1)/[(x+1)](x+1)^.5 = (2x+4)(x+1)^-1.5 now do second y = cos^2 x y' = 2 cos x (-sin x) agree but then y" = 2 [ cos x (-cos x) -sin x(-sin x) ] =2...
*September 26, 2015*

**physics**

see http://www.jiskha.com/display.cgi?id=1443249092
*September 25, 2015*

**Discrete Math**

y = x^2-4x+2 parabola upright, holds water. Therefore it is single valued in its range but the inverse is not single valued (not a function) If x = 2, y = -6 now what does the inverse look like? x = y^2 -4y + 2 now if x = 5 then y^2 - 4 y - 3 = 0 y = [ 4 +/- sqrt(16+12) ]/2 y...
*September 25, 2015*

**Physics**

v = Vi + a t d = Vi t + (1/2) a t^2 so 1000 = Vi (23.2) + .007(23.2)^2 solve for Vi
*September 24, 2015*

**Calculus**

0^3 - 5 0^2 + 4 = 4 not 0 1 works so factor it out (x-1)(x^2-4x-4) = 0 solve that quadratric x = [ 4 +/- sqrt(16+16)]/2 = 2 +/- 2.83 = 4.83 and -.83 sure enough
*September 24, 2015*

**ignore what I did**

I misread the question
*September 24, 2015*

**Algebra**

(1 lawn/36 min)t + (1 lawn/90min)t = 1 lawn t/36 + t/90 = 1 t/6 + t/15 = 6 t/2 + t/5 = 18 5 t/10 + 2 t/10 = 18 7 t = 180 t = 180/7 = 25.7 minutes =================================== check (1/36 + 1/90)25.7 = ? .0388888... * 25.7 = .99999999.... sure enough
*September 24, 2015*

**Algebra**

I think you might mean y = A(1 – r)^t y = 25,000 (1-.11)^6 = 25,000(.497) = 12,424.53
*September 23, 2015*

**Math**

n , n+2 4(n+2) + 2 n = 194 4 n + 8 + 2 n = 194 6 n = 186 n = 31 n+2 = 33
*September 23, 2015*

**Calculus**

but AD + DF = AF so in the end AF
*September 22, 2015*

**Calculus**

Line them up on a number line. for 2) BC + EF + AB + DE + CD AB + BC + CD = AD DE + EF = DF so AD + DF
*September 22, 2015*

**Calculus**

if the function is undefined there for exsmple f(c) = 1/(x-1) now what is the limit as x ---> 1
*September 22, 2015*

**physics**

normal force = m g cos T force holding block = .4 m g cos T force of gravity down slope = m g sin T so .4 cos T = sin T so tan T = .4 T = 21.8 degrees
*September 22, 2015*

**Physics**

v= a t .12 * 3 % 10^8 = 9.8 t t = .0367 * 10^8 = 3.67 * 10^6 then d = (1/2) a t^2
*September 22, 2015*

**PHYSICS**

Hey, I answered this hours ago: http://www.jiskha.com/display.cgi?id=1442934045
*September 22, 2015*

**science**

mass = density * volume Fresh water density is about 1 gram/ milliliter so 250 grams or 0.250 Kg
*September 22, 2015*

**Physics**

sqrt (4^2 + 2^2) = sqrt (20) = 4.47
*September 22, 2015*

**Physics**

from the origin, 4 to the right and 2 up so tan (angle) = 2/4 = .5 so angle = 26.57 degrees
*September 22, 2015*

**Physics**

horizontal problem: u = 16 cos 60 = 8 m/s t = d/u = 8/8 = 1 second Vi = 16 sin 60 = 13.9 m/s h = Vi t - 4.9 t^2 h = 13.9 (1) - 4.9 (1)^2 h = 9 m
*September 22, 2015*

**Math**

F = q (V x B) we need to know the angle between V and B. I suspect it is 90 degrees q = 1.6 * 10^-19 V = 3,000 B = .05 then F = 1.6*10^-19 * 3 * 10^3 * 5 * 10^-2 N second one the same way but use V x B = |V| |B| sin theta
*September 22, 2015*

**Algebra**

half gone in 8 years? x = Xi e^-kt .5 = e^-8k ln .5 = -.693 = - 8 k so k = .08664 x = 18 e^-.08664 t when t = 20 x = 3.18 grams
*September 22, 2015*

**huh?**

maybe F = (9/5) C + 32 to convert to Fahrenheit from Centigrade? Are you trying to get there from the usual two points, freezing and boiling of water? 0 C = 32 F 100 C = 212 F F = m C + b 32 = m (0) + b b = 32 F = m C + 32 212 = m (100) + 32 180 = 100 m m = 180/100 = 9/5 so ...
*September 22, 2015*

**math**

( 0.78 + 3.9) * 10^6
*September 21, 2015*

**physics**

ma = m g sin 36.9 so a = g sin 36.9 d = (1/2) a t^2 17 = .5 (g sin 36.9) (8.6)^2 solve for g
*September 21, 2015*

**math urgent**

Reiny already showed you that this can not be true. http://www.jiskha.com/display.cgi?id=1442757482
*September 21, 2015*