Wednesday

September 2, 2015
Total # Posts: 11,307

**math**

one-to-one (injection): every element in the codomain is mapped by at most one element of the domain. onto (surjection): every element in the codomain is mapped by at least one element of the domain. We can see that f(x) is both one-to-one but not onto, since f(1)=4, f(2)=7, ...
*August 12, 2014*

**Algebra**

(E)--- 8x-14 ----(G)--- 4x+1 ----(F) Since EF=59, => 8x-14+4x+1 = 59 Solve for x.
*August 12, 2014*

**MATH help please**

(a) An equivalence relation must satisfy the following three requirements: 1. reflexivity (a,b)R(c,d) is reflexive if (a,b)R(a,b)is defined. Since a-2d=c-2b ≡ a+2b=c+2d so (a,b)R(a,b) is defined since a+2b=a+2b Therefore R is reflexive. 2. symmetry R is symmetric if (a,b...
*August 12, 2014*

**math**

In general, days of the week advance one day per year (on the same day/month). For leap years, days from first of January to the 28th of February will jump TWO days if it is a leap year. For example, January 1st 1900 was a Monday. January 1st 1901 was a Tuesday (note that 1900...
*August 12, 2014*

**math**

Break up 6^8 => 2^8*3^8 10^6 => 2^6 * 5^6 Multiply together the odd factor of each number to get the largest odd factor of the product.
*August 12, 2014*

**math**

Original square area = n*n u² New dimensions: (n+m), (n-m) New area = (n+m)(n-m)=n²-m² u² Change in area = difference of above areas.
*August 12, 2014*

**math**

Song 1: 1,2,3,4 song 2: 5,6,7,8 song 3: 9,10,11,12 ... song 9: 33,34,35,36 ... song ?: 73,74,75,76
*August 12, 2014*

**phy**

Given a=<-1,-4,2> b=<3,2,-2> c=<2,-3,1> bxc= i j k 3 2-2 2-3 1 =<-4,-7,-13> so a.(bxc) =<-1,-4,2>.<-4,-7,-13> =4+28-26 =6
*August 12, 2014*

**Chemistry**

Start with writing a balanced equation: 4Fe3O4+O2 => 6Fe2O3 Then use stoichiometric coefficients and molar masses to solve the remainder of the problem.
*August 12, 2014*

**phy111**

We assume that the wind is blowing east at x km/h, represented by the vector <x,0> The truck is moving along the vector <p,q>, where sqrt(p²+q²)=70 km/h, or <sqrt(70²-q²), q> We also know that the resultant of the two velocities equals <...
*August 12, 2014*

**phy111**

In what direction is the truck moving, due south, south-west? Answers will vary depending on the direction.
*August 12, 2014*

**phy111**

Displacement=0 velocity=displacement/time =0
*August 12, 2014*

**Maths**

The solution of this problem is based on finding numerical values of parameters a and b in the PDF f(x). Recall that the given PDF is zero over the domain R except that f(x) is non-zero over (0,1). Therefore all integration should be performed from -∞ to +∞ but for...
*August 12, 2014*

**Math**

1. Total area = 30*20 ft² Let x=width of walkway, final area = (30-2x)(20-2x)=400 Solve for x. 2. Lawn area = 60*80 m² Final lawn width = x Pool area = (60-2x)(80 2x)=(1/6)(60*80) Solve for x.
*August 11, 2014*

**math**

Solve by elimination: Eliminate y by adding equation (3) to equation 1: 4x+z=12 Similarly, add 2*(3) to equation (2): 5x+5z=30 You can continue the elimination process to find x and z. Back-substitute to find y.
*August 11, 2014*

**math**

Use the binomial distribution, p=P(chemistry)=0.4 n=12 (sample size) r=number taking chemistry P(n,r)=nCr*0.4^r*(1-0.4)^(n-r) where nCr=n!/((n-r)!r!) is binomial coefficient. For probability of getting AT MOST 4 taking chemistry, add up all possible cases, i.e. 0,1,2,3,4, or P...
*August 11, 2014*

**college physics**

360 degrees in 27.3 days. Angular speed = angle/time 360 degrees / 27.3 days = ? degrees/day? If necessary, use conversion 1 day = 86400 seconds to find degrees/s.
*August 5, 2014*

**college physics**

Assuming the mass of 5kg is suspended freely, i.e. under influence of gravity, then the force (tension) on the wire is equal to 5g, where g=acceleration due to gravity, 9.8 m/s².
*August 5, 2014*

**Geometry**

The longest side is the hypotenuse, 6a+1 So use Pythagoras: (a+1)^2+(6a)^2=(6a+1)^2 and solve for a, reject all values of a≤0.
*August 5, 2014*

**statistics**

Number of children = 3 Possible number of boys = {0,1,2,3} X=number of boys - number of girls ={-3,-1,1,3} P(X)= probability of event X. b g X P(X) 0 3 -3 1/8 1 2 -1 3/8 2 1 1 3/8 3 0 3 1/8
*August 4, 2014*

**science (physics)**

Initial velocity v0= <1,2> a(t)=<0,g> v=∫a dt = <0,gt>+C => v(t)=<1, gt+2> x(t)=∫v dt = <t,(1/2)gt²+2t>+C Assume x(t)=<0,0> at t=0, then C=0 x(t)=<t, 2t+(1/2)gt²)
*August 4, 2014*

**PHYSICS**

radius of earth = 6378.1 km Assuming g is not affected by other heavenly bodies, then Newton's law of gravitation applies: g=GMm/r^2=k/r^2 (k=a constant) => g'r'^2=gr^2 r'^2=r^2(g/g') r'=r sqrt(g/g') =6378.1 sqrt(9.80/0.980) = 2.02*10^4 km approx.
*August 4, 2014*

**operating system**

In simple terms, a deadlock is a situation where two or more (minimum 2) processes are stubbornly holding on to a resource while waiting for another, creating a situation where certain resources are never available, and hence no work gets done for either of the processes. ...
*August 4, 2014*

**physics**

1 kW = 1 kJ/s Work done = 22 kW * 60 s = 22*60 kJ = 1320 kJ If the efficiency changes to 75% from 100%, the work done is reduced in the same ratio.
*August 4, 2014*

**math**

Assuming you are referring to the normal distribution, and depending on the context, it could be sample value, observed value, sample limit, etc.
*August 3, 2014*

**phy113**

Degrees fahrenheit = deg.C*(9/5)+32
*August 3, 2014*

**Geometry**

First find area of base: hypotenuse = 13 one leg = 12 other leg = √(13²-12²)=5 Area of base = 5*12/2=30 unit² Volume of prism =(area of base)*height/3 =?
*July 14, 2014*

**maths needed fastly**

I'll solve one, and will let you try the others as exercise. x=(4/5)(x+10) multiply by (5/4) on both sides: 5x/4 = x+10 subtract x from each side 5x/4-x =10 x/4=10 x=40 Check: (4/5)(x+10)=(4/5)(40+10)=40=x ok.
*July 12, 2014*

**mathe**

Let the cone radius = R cone height = H Let cylinder height = h then cylinder radius, r = R(1-h/H) Volume of cylinder, V =πr²h =πR²(1-h/H)²h For maximum volume, dV/dh=0 π(1-h/H)^2R^2-(2πh(1-h/H)R^2)/H=0 which simplifies to: (H^2-4hH+3h^2)=0 which ...
*July 12, 2014*

**physics**

Initial speed = 98 km/h = 98/3.6 m/s Work done = KE(final)-KE(initial) = (1/2)1370(0²-(98/3.6)²) = -508 kJ (approx.)
*July 11, 2014*

**physics**

Work done = F.D = 1.57*0 = 0 J.
*July 11, 2014*

**statisticss**

Calculate Z=(x-μ)/σ and look up the normal distribution table for the probabilities.
*July 8, 2014*

**physics**

First 30km with a speed of 30km/h means it took 1 hour. For an average of 40km/h over 60 km, total time = 60/40 = 1.5 hours. Time left for 30 km = 1.5-1 = 0.5 h. Can you calculate the speed needed for the second half of the journey?
*July 8, 2014*

**Physics**

Both correct!
*July 7, 2014*

**Physics**

Just answered your question a moment ago. Please check your previous post.
*July 7, 2014*

**Physics**

Correct for both.
*July 7, 2014*

**Physics**

Let initial velocity = vi final velocity = vf (negative if opposite in direction) Change in momentum, ΔM = m(vf)-m(vi)=m(vf-vi) a) Case A: vf=0, ΔM=-m(vi) Case B: vf=-vi, ΔM=-2m(vi) The absolute value of the change is greater in case B. b) Force = ΔM/&...
*July 7, 2014*

**math**

x=blue, y=green x/(25+x+y)=1/4 Cross multiply and simplify 4x=25+x+y => 3x-y=25 y/(25+x+y)=1/3 Cross multiply and simplify 3y=25+x+y => -x+2y=25 Solving: 5x=75 => x=15 y=3*15-25=20 z+15/(60+z)=4/13 Cross multiply 13(z+15)=4(60+z) 9z=45 z=5 5 blue beads must be added.
*July 7, 2014*

**math**

If the roots of a quadratic equation are equal, that means the discriminant equals zero, or b²-4ac=0 (for ax²+bx+c=0) So expand the expression and see if you can factorize the result to suit the conditions.
*July 5, 2014*

**Physics**

For the velocity, equation PE and KE. PE=KE mgh = (1/2)mv² and solve for v. (both g and h are negative, y is positive upwards) For time, use y=vi*t+(1/2)at² for vi=0, equation simplifies to y=(1/2)at² (Notice that both y and g are negative)
*July 2, 2014*

**PHYSICS**

Resolve forces into x,y components. Let the x-components of forces be f1,f2 then f1+f2=332 f1-f2=122 Solving f1=(332+122)/2=227 f2=(332-122)/2=105 Assuming θ=0 (in direction of raft) so the forces are 227 and 105N
*July 1, 2014*

**not statistics**

It depends on the tax rate, hence on where the payment is done. If the tax rate is 10%, then the sales tax is 993.91*0.10=$99.39
*July 1, 2014*

**physics**

average speed = total distance travelled / total time Total distance travelled = 30km/h * 0.5 h + 25km/h * 1 2/3 h + 40km/h *2 h = 136.667 km Total time = 0.5+1 2/3 + 2 = 4.1667 h Average speed = 136.667/4.1667 = 32.8 km/h
*July 1, 2014*

**physics**

Given: θ=arctan(3/4)=36.87° approx. m=0.5 kg μ=0.25 weight of block can be resolved into force down plane, F = mg sin(θ) and normal force, N = mg cos(θ) Let P be applied force parallel to plane (upwards) to maintain equilibrium, then Consider ...
*July 1, 2014*

**Physics**

Don't know how much work "I" do during this motion because I don't have the equations to calculations human work done during physical activities. However, work done on the 10N block is zero, the way you calculated it. When you lift the block, you are doing ...
*July 1, 2014*

**computer system organization**

Have you reviewed the three given sites as suggested? What are your thoughts about the quality of information supplied?
*July 1, 2014*

**Physics**

Nice question! Use energy. KE=PE+W Let x=distance up the slope (1/2)mv²=mg(sin(&theta))+μmg(cos(&theta))x Solving for x: x=(v²/((2g(sin(θ)+μcos(&theta))
*June 30, 2014*

**Statistics**

Use binomial distribution. P(r)=nCr*(p^r)(q^(n-r)) n=7 p=1/6 q=1-1/6=5/6 we need to calculate P(5)+P(6)+P(7) P(5)=7C5*(1/6)^5*(5/6)^2 =0.001875 P(6)=1.25*10^-4 P(7)=3.57*10^-6
*June 30, 2014*

**Algebra**

If you plot the graph, it will be a parabola concave up, with a zero at x=0, or x=4. Therefore f(x)≤0 when x∈[0,4). The correct answer would be S={x|x∈(-∞,0]∪[4,+∞)}
*June 30, 2014*

**mathematics**

Assuming garden has uniform width w. 2*60*50=(50+w)(60+w) => w²+110w-3000=0 w=22.62 or -132.62 (rejected) So wall is 22.62 m from the house.
*June 30, 2014*

**physics**

In the case of uniform acceleration, distance = average speed * time distance = 0.04m average velocity = (350+210)/2 m/s = 280 m/s Time = distance / average speed = 0.04/280 s = 1.43*10-.4 s
*June 30, 2014*

**math**

Out of 100 questions, 48 correct answers 11 blank 41 wrong answers total 100 We don't get to see the scheme. Assume the scheme says: 1 point for each correct answer, -1 point for each incorrect answer, 0 for leaving it blank. So total points =48-41-0 =7
*June 30, 2014*

**physics**

Acceleration = (vf-vi)/t = -2.24/2.55 = -0.8784 m/s Δx = vi*t + (1/2)at² =2.24(2.55)+(1/2)(-0.8784)(2.55²) = 2.85 m
*June 30, 2014*

**Geometry**

Have you read the Wiki article on cyclic quadrilaterals? It gives a lot of information on the subject. If it still does not answer your question, you could post your question and we could all learn something. http://en.wikipedia.org/wiki/Cyclic_quadrilateral
*June 29, 2014*

**physics**

For maximum height, equate PE and KE mgh=(1/2)mv² h=(1/2)37.1²/9.8=70.2 m at 50m, KE is reduced by mgh => (1/2)mv²=(1/2)m(37.1)²-mg(50) v=sqrt((37.1²-2(g)(50)) =19.9 m/s
*June 28, 2014*

**physics**

Conversion factor: 1 m/s = 3.6 km/h Δv=(vf-vi)=vi+aΔt 163 km/h = 163/3.6 = 45.28 m/s 45.28=vi+5.79*3.6 vi = 24.4 m/s = 88.0 km/h
*June 28, 2014*

**Physics**

Since gravity on the moon is about 1/6th of that on earth, the mercury column would be 6 times higher for the same air pressure. However, the atmosphere on the moon is about 10^13 time rarer than that on earth, the height of the mercury column is practically nil on the moon.
*June 28, 2014*

**math**

If the shapes are similar, area increase as the square of the ratio of the lengths. In this case, the shape remains the same, so old area = d*n new area = (1.1d)(1.1n) = (1.1)² dn = (1.1)² old area Percentage increase in area = 1.1²-1 = 21%
*June 28, 2014*

**statistics**

Given: μ=0.5" σ=0.01" Rejection rate = 0.8% = 0.008 rejection criterion = z<Z X=μ+Zσ =0.5+0.01Z From normal distribution tables P(Z<-2.409)=0.008 so X=0.5-0.01*2.409 =0.4758 Ans: washers with diameters less than 0.476 are rejected.
*June 28, 2014*

**Physics**

Power = work done / time = mgh / time = 55kg * 9.8 m/s² * 11 m ÷ (25*60 s) = 4.0 J/s = 4.0 W
*June 27, 2014*

**Physics**

During the reaction time of 0.59 s, car is still moving at 11 m/s, so distance to be added is 11 m/s * 0.59 s = 6.49 m. The deceleration is a=-8 m/s² vi=11 m/s vf=0 m/s (trying to stop) vf²=vi²+2ax => x=(vf²-vi²)/2a =(0²-11²)/(2*(-8)) = 7....
*June 27, 2014*

**physics**

Archimedes principle states that the buoyant force on an immersed object equals the mass of liquid displaced by the object. 230 ml of water weighs 230g Buoyant force = mg = 0.230 kg * 9.8 m/s² = 2.25 N
*June 27, 2014*

**Physics**

First convert velocities to their resultant vector: v=sqrt(vx²+vy²) =6.03 m/s Momentum=mass*velocity = 2.73 kg * 6.03 m/s = 16.5 kg-m/s
*June 27, 2014*

**calculus**

Assuming choices are payment options, i.e. we will be paying the said amounts. Total amount of payment for each choice is $9600. Assuming we have $9600 in the bank, we calculate what would we have accumulated in interest at the end of two years. Choice 1: 3800 for 2 years + ...
*June 27, 2014*

**Physics**

See http://www.jiskha.com/display.cgi?id=1259306687 as an example. Post your answer for a check if you wish.
*June 27, 2014*

**physics**

Given v=30 m/s θ=33° h=15m g=9.8 m/s² Resolve velocity into horizontal (vx) ad vertical (vy) components, vy=v*sin(θ)= 16.34 m/s vx=v*cos(θ)= 25.16 m/s Find time t to reach ground h m below: y=vy*t+(1/2)at² -15=16.34t-4.9t² => 4.9t²-...
*June 27, 2014*

**physics**

Equate KE and PE to get mgh = (m/2)vy² => vy=√(2gh) Initial velocity v=vy/sin(θ) Substitute v in range equation range=v²sin(2θ)/g = (vy²/sin²(θ))*sin(2θ)/g = (vy²*2sin(θ)cos(θ)/(g*sin²(θ) = (2gh*...
*June 27, 2014*

**physics**

given: m1 = 75.00 g c1 = ? t1 = 52.00°C m2=225.0 g c2=2.045 J/g-°C t2=07.50°C tf=9.08°C Using ΣmcΔt=0 75.00*c1*(9.08-52.00)+225.0(2.045)(9.08-7.50)=0 Solve for c1 to get c1=0.2258 J/g-°C
*June 27, 2014*

**physics**

v1=√(2gh)=√(2*9.8*0.53)=3.223 m/s v1=√(2gh)=√(2*9.8*0.53)=3.162 m/s Let ω=angular velocity r=0.363 m ω1=v1/r=8.879 rad/s ω2=v2/r=8.710 rad/s Time for one turn between t1 & t2 Δt=2π/((8.879+8.710)/2) = 0.7145s Angular ...
*June 27, 2014*

**JEMS**

t=digit in the tens place the number is then (t+2)t(t-6) If (t+2) is odd, so is t. t must be ≥ 6 or else t-6 is negative. so t could be 7 or 9 But t cannot be greater than 7 (or else t+2 will have 2 digits) therefore the three-digit number must be 571.
*June 26, 2014*

**science..physics**

r=70 semi-circumference = πr = 70πm Displacement = 2r = 140m
*June 26, 2014*

**physics**

Distance = 15+20 = 30m Displacement = final position - initial position = √(15²+20²) = 25m (magnitude) Direction =atan(20/15) =atan(4/3) =53.13 °
*June 26, 2014*

**Physics**

Actual distance = sum of length of paths Distance = sum of northings and eastings = 125cos(35°)+125sin(35°) = 102.4+71.70 m = 174.09 m Time = Displacement / velocity = 125/2.5=50s Speed = Distance / time = 174.09/50 = 3.48 m/s
*June 26, 2014*

**physics**

Acceleration = (vf-vi)/t =(0-9)/4 = -2.25 m/s²
*June 26, 2014*

**physics**

Downward component of weight (along incline) = mg(sin(θ)) Normal component of weight = mg(cos(&theta)); Upward component of horizontal force (along incline) = F(cos(θ)) Normal component of horizintal force = F(sin(θ)) Coefficient of kinetic friction = μ ...
*June 25, 2014*

**Physics**

Force on the car in the direction of the rope is F = (489/2)÷sin(θ) where θ=angle formed by the rope and the car-tree line.
*June 25, 2014*

**math**

Not clear what balance and unpaid balance are. Usually statements give unpaid balance (from last month), minus the payments, add purchases to get the new balance. Is $253 the new balance? Also, charge accounts usually charge interest by the day. How does this account work if ...
*June 25, 2014*

**physics**

Total mechanical energy = KE + PE (assume PE at ground level = 0) KE=(1/2)mv² PE=mgh h=height above ground in m v=velocity in m/s
*June 25, 2014*

**lwtech**

Interpretation of given information: The location of the theodolite (surveyor) at point A. First stake (Point B) Second stake (Point C). Points B and C are on opposite sides of the river, and distance BC represents the width of the river. For point B: distance = 562 bearing = ...
*June 25, 2014*

**geometry**

Assuming you have two triangles where ΔABC~ΔDEF, then AC/DF = AB/DE or AC=(AB/DE)*DF =(35/14)*22 =55
*June 25, 2014*

**math**

Given three non-parallel lines. We are to find the vertices of a triangle formed by the intersection of the lines. The intersections are found by solving for the three systems of two equations at a time. For example: L1: x+y+6=0 L2: x+2y-4=0 Subtract L1 from L2 gives: y-10=0, ...
*June 25, 2014*

**Physics**

Use kinematics equation: Δx = vi(t) + (1/2)at² vi=0 g=-9.8 -500=(1/2)(-9.8)t² t=√(2(-500)/(-9.8)) =10.1 s (approximately)
*June 25, 2014*

**mathematics**

It is not too clear "current" means 2014 or 2015. Will assume baseline year to be 2015 when 20 more schools are built. Number of students in 2015=1200 Number of students in 2016=1200*1.2=1440 Teacher:student ratio in 2015=1:60 Number of teachers in 2015=1200/60=20 ...
*June 25, 2014*

**Math**

Let's summarize the situation. Company: Score Total number of employees: 6 Pay period: semi-monthly. Salary break down: 2 employees at $1100 per pay 4 employees at $850 per pay Proposed employee retirement fund: $75 per pay per employee SUTA contribution (extra expense for...
*June 25, 2014*

**Math**

SUTA tax (5.4% for first $7000) =7000*0.054 =$378 Equivalent FUTA tax rates are different from state to state because of possible credit reductions. Same goes for the FUTA tax base. Assuming the FUTA tax base is $7000, and that there is no credit reduction, the effective FUTA ...
*June 25, 2014*

**physical**

Using Newton's second law: F=ma m=9.39 kg F=Fa+Fb=9.39*1.4=13.146 F=Fa-Fb=9.39*0.950=8.9205 Fa=(13.146+8.9205)/2=11.03 N Fb=(13.146-8.9205)/2=2.11 N
*June 25, 2014*

**physics**

Using Newton's second law F=ma to find the mass, then again the same law to find the frictional force. Finally use Work done = F.D Units are in kg, m/s² and J. Mass m = 7750/9.8 = 790.82 kg Frictional force = ma = 790.82 kg * 10.6 m/s² = 8382.7 N Work done (...
*June 24, 2014*

**physics**

Using the equation for spring potential energy W=(1/2)kx² so ΔW=(1/2)k(x2²-x1²) =(1/2)(30 N/m)*(0.96²-0.19²) =13.3 J
*June 24, 2014*

**Physics**

Using Distance = speed * time (a) Total time of travel =800m/6s + 800m/8s = 233.3s (b) Total distance = 800+800=1600m Average speed = total distance/total time =1600m/233.3s = 6.86 m/s Try #2 which is similar to #1.
*June 24, 2014*

**Physics**

Magnitude of displacement (vector) =290m velocity=2.0m/s (a) Walking time = distance /speed = 290m / 2.0m/s = 145s (b) Northerly distance = 290cos(25°) = 262.8m Westerly distance = 290sin(25°) = 122.6m Total distance = 385.4m Average speed = 385.4m / 145s = 2.66 m/s
*June 24, 2014*

**geometry**

Isabelle, your earlier question has been answered: http://www.jiskha.com/display.cgi?id=1403640676
*June 24, 2014*

**geometry**

Isabelle, your earlier question has been answered: http://www.jiskha.com/display.cgi?id=1403640676
*June 24, 2014*

**Geometry**

If only special parallelograms are considered (i.e. not just any parallelogram), they would be - rectangle - rhombus - square Note: one of these choices will fit more than one situation. I will leave it to you to sort out which one is which, and feel free to post your answer ...
*June 24, 2014*

**Geometry**

Since x and b are supposed to be valid for any value, take x=b=2 (or any other number) and sketch the graph. If you recognize any shape, then use the properties of the shape to do the general proof. Example: for x=b=2, the points become: (0,0),(6,2),(36,2),(30,0) A sketch ...
*June 24, 2014*

**physics**

Red is the hardest to get, since there are only 20 of them. To make sure you get some red, the worst case is you picked all the black, blue and whites (totalling 120) and still not get a red. What is the minimum you need?
*June 24, 2014*

**MathMate Help!!!**

Thank you for the brilliant and elegant proof, Steve!
*June 24, 2014*

**pre cal**

Could you tell me how far have you gone with this one? Where do you get stuck?
*June 24, 2014*

**pre cal**

I see a lot of similar questions, with no sign of effort on them, nor feed back. Do you have a problem working out these questions or you are just looking for someone to do your take-home test?
*June 24, 2014*

**Physics**

Given a position function x(t), then x'(t)=dx(t)/dt=velocity, and x"(t)=d²x(t)/dt²=acceleration. where x' and x" are the first and second derivatives with respect to time. (a) solve for x'(t0)=0 (b) evaluate position = x(t0), and displacement = ...
*June 24, 2014*