Wednesday

December 7, 2016
Total # Posts: 11,499

**Calculus**

All limit problems are different. If only the limit is known without knowing the question, we don't know the answer. Please post the complete question in order and answer can be formulated.

*May 20, 2014*

**math**

There is a good chance you have missed out parentheses in the numerator and the denominator. All fractions have implicit (i.e. understood) parentheses in the numerator and denominator. I believe your question means (x²+6x+9)/(5x+15) even though the question does not ...

*May 20, 2014*

**Math**

Assuming you want the difference in days between dates. It's a messy counting problem. First note that the dates are written in MM/DD/YYYY format. Difference in Months: October-May=31+30+31+31+30=153 days Difference in days 28-17=11 days Difference in years 2002-1987=15 ...

*May 20, 2014*

**Proofreading required.**

Please proofread your question. Does your question use proper grammar, capitalization and spelling?

*May 20, 2014*

**Trig help**

I guess Steve's interpretation of the question is correct. There was no indication of power! Please go with Steve's solution.

*May 20, 2014*

**Trig help**

If you are looking at solving 3sin(θ)=sin-1(θ) then you would be solving f(x)=3sin(x)-asin(x)=0 in which case the trivial solution x=0 can be obtained by inspection. If you're looking for 3sin(θ)=1/sin(θ) then there are 4 solutions between 0 and 2&...

*May 20, 2014*

**math**

The best way to solve these problems is to follow the instructions, draw the figure with the coordinates of each point, and then apply the transformation (rotation). I will give you a kickstart by giving the coordinates of the figure (not a rectangle). A(0,0) B(2,0) C(4,0) D(4...

*May 19, 2014*

**math Question**

a=12 c=30/2=15 b=√(15^2-12^2)=9 For a hyperbola with horizontal axis, x²/a²-y²/b²=1 x²/12²-y²/9²=1

*May 19, 2014*

**math**

Did you read the post previous to yours? It should help you get the right answer.

*May 19, 2014*

**math**

First, find the rule for a rotation of 180°, which is R(O,180): (x,y)->(-x,-y) so (3,-5) -> (-(3),-(-5)) can you take it from here?

*May 19, 2014*

**Finite Math**

The 25% means that she pays up front $6000 so she only needs to finance $18000. Over a period of 36 months, n=36=periods of one month i, monthly interest = 0.1/12 P, principal = 18000 R=rate=1+i=1+1/120 Monthly payment, A =PRn(R-1)/(Rn-1) =18000*(1+1/120)36*(1/120)/((1+1/120)...

*May 19, 2014*

**Math Question**

Figure out the rule of the sequence in terms of n, the number of terms We have common difference =6 T(0)=-12 T(1)=-6 T(2)=0 T(3)=6 Let's try T(n)=-12+6n so T(1)=-12+6=-6 T(2)=-12+6*2=0 T(3)=-12+6*3=6,... It works. So T(50)=-12+50*6=288

*May 19, 2014*

**Algebra 2**

It's the same as calculating the distance between two points. ||A||= sqrt((-13-2)²+(9-1)²) =17

*May 19, 2014*

**Math**

P(Bull's eye)=(6/16)²=0.1406 approx.

*May 19, 2014*

**Math - need more data**

To calculate the probability, we need to know the size of the dartboard and the radius of the bull's eye. A six-inch radius bull's eye has to be bigger than the eye of any living animal on earth!

*May 19, 2014*

**physics**

90 km/hour = 90/3.6=25 m/s deceleration rate, a = 5 m/s² time= 25/5 = 5 seconds Must be a new braking system :)

*May 19, 2014*

**science**

Apply F=ma (Newton's second law) where F=force (in N) m=mass in kg a=acceleration in m/s² so F=ma=10kg * 5 m/s²=50 N If the spring balance is graduated in kilograms (mass), which is the usual case, then it will show ma/g=5 kg (approximately).

*May 19, 2014*

**Science**

Total force on 4 legs =5 N + 9*9.8 N = 93.2 N Area of contact of each leg = 2.5*10-3 m² Pressure on each leg = 93.2/(4*2.5*10-3 = 9320 N/m² = 9320 Pa (pascals)

*May 19, 2014*

**Chemistry**

Only volume and temperatures change, pressure remains constant, Charles's law applies. V1/T1=V2/T2 Temperatures must be converted to °K. V1=100 ml T1=27°C=300.15°K V2=80 ml. The above applies when the volume change is at constant pressure to avoid work done ...

*May 19, 2014*

**Physics**

If the string remains taut at all positions, the minimum tangential speed of the rock has to generate a centripetal force equal to the weight of the rock, or mg = mv²/r Here m=2 kg r=0.80 g=9.8 m/s² w=weight=mg=19.6 N solve for v: 2(9.8)=2v²/0.8 v=√(196/25...

*May 19, 2014*

**Physics**

Kinematics equation: Δx = vi*Δt + (1/2)aΔt² Here Δt = elapsed time, in seconds vi=initial velocity = +35 m/s a=acceleration due to gravity = -9.8 m/s² Δx=displacement = -100 m -100=35Δt + (-9.8/2)Δt² Solve by quadratic ...

*May 19, 2014*

**chemistry - typo**

Can't seem to get it right! the -4 should be in superscript.

*May 19, 2014*

**chemistry - typo**

(typo in format) Cross multiply to get 3.0*10-4(0.001-x)=x² Solve for x, reject negative root. and pH is -log10(x). It should be between 3 and 3.5.

*May 19, 2014*

**chemistry**

Denote the chemical formula for the acid to be HA, then Ka=[H+][A-]/[HA]=3.0*10-4 at a given temperature. ACE table for H+ A- HA A 0 0 0.001 C x x -x E x x 0.001-x At equilibrium, 3.0*10-4=x²/(0.001-x) Cross multiply to get 3.0*10-4(0.001-x)=x² Solve for x, reject ...

*May 19, 2014*

**physics**

Work done by force = work done by load (W) 60 N * 1.2 m = 0.2 m * W Solve for W to get W=60*1.2/0.2 N = 360 N

*May 19, 2014*

**physics**

Work done =mgh = 60 kg * 9.8 m/s² * 1.2 m = 705.6 j Power = work done / time = 705.6 j / 1.5 s. = 470 watts (approximately) The difference may be due to the assumed value of g in the question.

*May 19, 2014*

**strength of materials**

The deflection of a cantilever beam with a point load P at the end is given by δx=PL³/(3EI) P=point load L=length E=modulus of elasticity I=area moment of inertia all in compatible units. The area moment of inertia for the beam in question is equal to that of the ...

*May 18, 2014*

**algerba - mea culpa**

After reading your response, I rechecked and surely I found the culprit: me. I have the wrong sign! The correct solution should have read: This is how you could approach it: Start with: (0.052*310)/(310+g)=59.4 cross multiply to get (0.052*310)=59.4 (310+g) Expand and evaluate...

*May 18, 2014*

**algerba**

I looked over the possible answers, and 310 is among them. If you are in a chemistry or physics class, or if you have already worked on significant digits, 310 would be the correct answer to 3 significant digits, since the original data are all to that accuracy. Sorry that I ...

*May 18, 2014*

**algerba**

309.72862 is correct. This is how you could approach it: Start with: (0.052*310)/(310+g)=59.4 cross multiply to get (0.052*310)=59.4 (310+g) Expand and evaluate: 16.12=59.4g+18414 transpose 59.4g = 18414-16.12 g=(18397.88/59.4) =309.72862 as you had it.

*May 18, 2014*

**algerba - explain please**

Not sure what the question is asking. Explanations would definitely help.

*May 18, 2014*

**Geometry - erratum**

Add (4) and (5) 2CG²=a²++b²+3(x²+y²) =5(x²+y²)+3(x²+y*sup2;) (using (1) and (2))

*May 18, 2014*

**Geometry**

Nice problem involving 1. definition of sine 2. the cosine rule 3. the properties of medians, i.e. intersection of medians divides them in ratios of 2:1. First need to do a diagram in words. Given triangle ABC, with medians AX and BY meeting at G. AX and BY are orthogonal (i.e...

*May 18, 2014*

**algebra**

R(x) is the revenue as a function of the number of units sold, x, where R(x)=(100x-x²)x if p(x)=100x-x² then R(x)=0 whenever p=0. So the problem would be reduced to solve for p(x)=0, or 100x-x²=0 x(100-x)=0 which gives us x=100 would make p(x)=0 and R(x)=0. Put ...

*May 18, 2014*

**Physics**

Assume L for length of stick. Put the fulcrum at x from the 500-g mass. The centre of gravity of the stick is at the centre, L/2, where the mass is assumed to concentrate. Take moments about the fulcrum: 500*x=200*(L/2-x) Solve for x in terms of L: x=L/7

*May 18, 2014*

**algebra**

So the problem is : Pull out 2 letters out of 5 distinct letters {P, A, R, I, S} from a hat with replacement. Need P(R,vowel) P(R)=1/5 P(vowel)=2/5 With replacement, the two draws are independent, so multiplication rule applies. P(R,vowel)=(1/5)*(2/5)=2/25

*May 18, 2014*

**algebra**

Future value = 50000 period = semi-annual number of periods, n = 15*2=30 Annual interest rate, = 11% i = 0.11/2=0.055 Periodic payment (6 months) = P Let r=1+i=1.055 Future value, 50000 = P(1+r+r²+r³+...+rn-1) = P(rn-1)/(r-1) by factorizing therefore, using (r-1)=i, ...

*May 18, 2014*

**algerba**

No problem. Let us know any time if you need further help.

*May 18, 2014*

**algerba**

You will get two roots for x. Reject the negative one, and retain the positive root which is a little under 200.

*May 18, 2014*

**algerba**

Yes, the quadratic formula is: x=(-b±√(b²-4ac))/(2a) what you need inside the square-root sign is b²-4ac

*May 18, 2014*

**college algebra**

The cube root of 600 is the length of one side, s. If you use a calcultor, then you would type in: 600^(1/3) or 600 xy (1/3) depending on the type of calculator you have.

*May 18, 2014*

**physical**

C°C = (C+273.15)° K 20°C = (20+273.15)=293.15°K

*May 18, 2014*

**physical**

Use T=(m1T1+m2T2)/(m1+m2) =(50*20+50*40)/(50+50) =30°C Note: special case when the masses are equal, and the liquids are identical, the temperature is the mean of the two liquid temperatures.

*May 18, 2014*

**algerba**

Answered to your other post.

*May 18, 2014*

**algerba**

You're welcome!

*May 18, 2014*

**algerba**

Cross multiply to get: 8.44(3x²+100)=3900x+1000 Rearrange: 25.32x²-3900x-156=0 Solve for x using quadratic formula: x=-0.03999 or x=154.07 approx.

*May 18, 2014*

**bdu applied maths**

If x,y form the base, and z is the height, then Area of base = xy Lateral area = perimeter*height = 2(x+y)z Total surface area of open box =xy+2xz+2yz

*May 18, 2014*

**physisc**

From glass to air means from an optically denser medium to a lighter one, the light path will refract away from the normal.

*May 18, 2014*

**algebra**

Just a lucky guess, but you were right in the first place!

*May 18, 2014*

**algebra**

Fractions have implicit parenthese around the denominator and numerator. I have a feeling that the two log3 terms both belong to the denominator, in which case the expression should have been written, according to the BEDMAS rules, (2*log7 16)/((log3(√10+1)+log3(√...

*May 18, 2014*

**Chemistry**

When temperature remains constant, use Boyle's law: PV = constant, or P1V1=P2V2 Also, for conversion, use 1 atm = 101.325 kPa

*May 18, 2014*

**Chemistry**

Use ideal gas law: PV=nRT (T in °K) => P=nRT/V =1.3 mol *0.0821 L atm °K-1 mol-1 *(22+273.16)°K/13 L =2.423 atm =2.4 atm (approx.)

*May 18, 2014*

**chemistry**

Use c1v1=c2v2 c1=230 g/L c2=200 g/L v2=2 L Solve for v1. Water to be added equals v2 - v1=2L - v1

*May 18, 2014*

**Physics**

Use the kinematics equation: Δy = viΔt + (1/2)gΔt² Here Δy=-25' vi=0 (initial velocity) g=-32.2 ft/s² Δt=√(2Δy/g) =√(2*25/32.2) =1.246 s. (approx.) Velocity =1.246*(-32.2) fps =40.12 fps =40.12/5280*3600 mph =27.4 ...

*May 17, 2014*

**math**

In the Cartesian plane, +x-axis is to the right, -x is to the left. Similarly, +y is up, and -y is down. For example, a translation of 2 units to the left and 3 units up would be written as (x,y)->(x-2,y+3) so a point (5,3) would become (3,6). Apply the same logic to the ...

*May 17, 2014*

**c++**

If you could use further help, post what you've got so far, and we'll take it from there.

*May 17, 2014*

**Degree symbol**

The degree symbol (°) may be written (at this forum) as the following, with all spaces suppressed: & d e g ;

*May 17, 2014*

**physics**

Assuming question wants height of table, Δy. Use kinematics equation: Δy = viyΔt + (1/2)gΔt² Assuming 1.1m/s is horizontal velocity (slides off table), so viy=0 g=-9.8 m/s² Δy=0+(1/2)(-9.8)(0.35²) =-4.9(0.35²) =-0.060 m Floor ...

*May 17, 2014*

**Math -___-**

Detailed explanation on how to make a stem and leaf plot. Difficult to show without fixed format characters. http://www.purplemath.com/modules/stemleaf.htm

*May 17, 2014*

**physics**

Use Newton's second law: F=ma=100000kg * 2 m/s² = 200000 N

*May 17, 2014*

**Science Physics**

Similar to previous problem of crate. Use Newton's second law: F=ma where F is net force=applied - friction.

*May 17, 2014*

**Science Physics**

F=160 N m=20 kg Ffr=80 N Net force causing acceleration = 160-80 =80 N Use Newton's second law, F=ma a=F/m =80/20 =4 m/s²

*May 17, 2014*

**Algebra**

At vertex, x=6/10=0.6 Try f(-1)=-2 f(0)=9 f(2)=-1 So -1 and +2 could be approximate zeroes of f(x).

*May 17, 2014*

**UOS**

Pseudocode: read Capacity, MilesPerGallon MaximumRange:=Capacity*MilesPerGallon print "Maximum Range = ", MaximumRange

*May 17, 2014*

**CHemistry**

First calculate molecular masses of MH2 = (2*1.008) and MH2O = (2*1.008+15.999) Then mass of hydrogen produced =mass of water *(MH2/(2MH2O)) Note the factor 2 in the denominator comes from the stoichiometric coefficient of H2O.

*May 17, 2014*

**Math**

The answer to the question as posted is n=942!/(15), which is a large number approximately equal to: 1.45*10^2393. I suspect the factor 15n should read 15n. If that's the case, I offer the following. 942! is the product 942*941*940*....3*2*1 If n=1, then 942!/(15) should ...

*May 17, 2014*

**Math**

There are 20 yellow x green 80-x pink Total 100 P(YP)=(20/100)(80-x)/99 P(GY)=(x/100)(20/99) If P(YP)=P(GY) then 20(80-x)=20x after cancelling identical denominators Solve for x: x=1600/40=40 So there are 40 green and 40 pink mints. If P(YP)>P(GY) then 20(80-x)>20x 40x&...

*May 16, 2014*

**Physics**

1. calculate distance travelled before brakes have effect. 2. Once deceleration of 10 m/s² starts, car decelerates at that rate. Time taken to decelerate = initial velocity / 10 m/s² = 2 seconds. 3. Calculate distance travelled during those two seconds: vf²=vi&...

*May 16, 2014*

**Algebra**

I think the formula should read: h=-16t^2+vt+s Where s=initial height = 25' v=initial velocity = 96 fps t=time in seconds -16=half of acceleration due to gravity in ft/s² so the equation becomes h(t)=-16t²+96t+25 solve for t when h(t)=0.

*May 16, 2014*

**physics**

Time, t, to reach ground = Δx/vix = 2.329 s Use kinematics equation Δy=viy*t+(1/2)g(t)² = 10.92 m. (g=-9.81 m/s²), viy=0)

*May 16, 2014*

**math(urgent)**

Interest earned was simple interest, so in two years, the future value is 1000(1+2*0.1)=1200 With compound interest of rate i, the equivalent interest would be 1000(1+i)²=1200 or (1+i)=√(1200/1000)=1.095445 or 9.54% approx.

*May 16, 2014*

**math**

I think you mean "...from A to A'". Without the accompanying figure, every answer can be correct. You can figure it out as follows: Measure the horizontal distance between A and A', see if A' is to the left by 9 units of A. Similarly, check if A' is 3...

*May 15, 2014*

**physics**

Use conservation of momentum: m1u1+m2u2=m1v1+m2v2 m1=m2=0.5 kg u1=10 m/s u2=0 m/s v1=1.5 m/s Since m1=m2=0.5 kg, the masses can be cancelled out to give u1+u2 = v1+v2 10+0 = 1.5 + v2 v2=8.5 m/s

*May 15, 2014*

**math**

Agree! Please continue if you have a question.

*May 15, 2014*

**geomettry**

The first vertex is at 0° with the x-axis. The other vertices are at respectively θ= 60, 120, 180, 240 and 300° with the x-axis. The corresponding coordinates are at: ( 8cos(θ), 8sin(θ) ) Example: At θ=0° the vertex is (8cos(0°), 8sin(0&...

*May 14, 2014*

**NO SCHOOL**

Goods and Services Tax (applies to all Canada). Some provinces combine with the provincial sales tax to call it Harmonized Sales Tax (HST). Assume $100 includes GST, then the GST portion is $100*(10/110)=$100/11. If the GST rate is 15%, then the GST portion will be $100*(15/115).

*May 14, 2014*

**Math**

Increase per year = $900 Target=$65000 current amount = 3000 Number of years to generate exactly 65000 =(65000-3000)/900 = 68.89 years So minimum number of years to equal or exceed target = 69 years (a little more than "several" years!)

*May 14, 2014*

**Chemistry**

Heat, Q =mcΔT =1200*4.184*(86-24)° =... joules Divide by 1000 to convert to kj.

*May 14, 2014*

**maths**

I think you mean: In a triangle PQR. q=8cm , r=6cm . Cos(P)=1/12. Calculate the value of p. where capital letters mean vertices, lower-case letters mean lengths of sides opposite the corresponding vertex. Use cosine rule: p²=q²+r²-2qr(cos(P)) and solve for p.

*May 14, 2014*

**route surveying**

1 mph = 5280*.3048/3600=0.44704 m/s vi=50*.44704=22.352 m/s vf=0 a=-10 m/s² use kinematics equation vf²=vi²+2aΔx to solve for Δx. Subtract Δx from 80m to get the distance from the wall.

*May 14, 2014*

**Math**

z+7+5=4 or 14, so z=2 4+2+x+1 (from carry)=1 or 11, so x=4 0+0+1 (from carry) =y so y=1 x=4, y=1, z=2 What is xyz?

*May 13, 2014*

**french**

Jenna, In French, every accent counts, as well as capitalization. For example, "Ce concert est très ennuyeux."

*May 13, 2014*

**math/probability**

With replacement: P(BY)=P(B)*P(Y)=(6/21)*(7/21)=2/21

*May 13, 2014*

**Physics**

Without friction and air resistance, we equate potential (PE) and kinetic (KE)energies. Energies at bottom of hill = energies at top of hill. 0+(1/2)m*vi²=mgh+(1/2)m*vf^sup2; cancel m and substitute g=9.8 m/s² vi=20 m/s h=15 m (1/2)(20²)=9.8*15+(1/2)vf² ...

*May 13, 2014*

**Math**

Amount = Pert where P=principal, present value so 225500=Pe.0865*(8+125/365) solve for P to get: 109585 (approximately)

*May 13, 2014*

**physics**

Initial volume of liquid, V = 3.5 × 10-4 m³ Change in temperature, ΔT =(78-5)=73° Spillage, ΔV =3.6 × 10-6 m³ Coefficient of volume expansion over that of the can =(ΔV/V)/ΔT =(3.6*10^(-6)/(3.5*10^(-4)))/(78-5) =1.4090*10^(-4) ...

*May 13, 2014*

**Calculus**

This sounds like a physics problem: Work done = increase in potential energy (PE) Let PE at ground = 0 After lifting, the centre of gravity of the chain is at h=10 m above ground, so potential energy =mgh =(3 kg/m * 20m)*9.8 m/s²*10m = 5880 joules Using calculus, Let &rho...

*May 12, 2014*

**MATH 1 QUESTION PLS HELP ASAP!!!!!**

The given expression is a geometric progression, which means that successive terms are obtained by a constant multiplier. If you look at the quotient between successive terms, (a) does not even give a constant ratio, so (a) is out. For (c), there is a problem with the first ...

*May 12, 2014*

**pre cal**

Read from the following table: https://www.stat.tamu.edu/~lzhou/stat302/standardnormaltable.pdf You will see that 2.275% of data falls beyond μ+2σ and another 2.275% of data falls below μ-2σ.

*May 12, 2014*

**algorithms**

Algorithm of sorting exactly three object, a,b,c is like comparing masses of three objects on an ungraduated balance. if a>b then _if b>c then __a>b>c _else if a>c then __a>c>b __else __c>a>b else if a>c then __b>a>c _else if b>c then ...

*May 12, 2014*

**Physics HELP!!!!!!!!**

Oops, the reduction in volume is 41.1%, so ΔV=1.933 *10^-4 m³ and work done, W=22.4*0.021/1000*0.411*101300 =19.58 j =20 j (approx.)

*May 11, 2014*

**Physics HELP!!!!!!!!**

To get work done in joules, use metres, Pa, and m³ as units. Volume at 0° and 1 atm for n=0.021 V= 22.4*0.021 L = 4.704*10^-4 Pressure 1 atm = 101.3 kPa = 101300 Pa ΔV =0.41V =1.929 *10^-4 Using Work=PΔV, =101300*1.929*10^-4 = 19.54 J = 20 J (approx.)

*May 11, 2014*

**(2-8) Chemistry - Science (Dr. Bob222)**

For B, think of the ideal gas as a compressed spring. When the compressed spring expands, it does work on the surroundings. The external pressure has to stay constant to ensure work is done. If the external pressure decreases, the system can expand without doing any work.

*May 11, 2014*

**ALGEBRA**

I suspect the equation should read P = 14.7e-0.21a in order that the pressure decreases with altitude. For P=13.29 13.29=14.70.21a e0.21a = 13.29/14.7 take log on both sides -0.21a = ln(13.29/14.7) a=-ln(13.29/14.7)/0.21 = 0.4802 miles = 2535.3 feet (approx.) = 2535 feet (...

*May 11, 2014*

**Physics**

m1u1+m2u2=m1v1+m2v2 12*15+36*5=12*6+36*v2 Solve for v2.

*May 11, 2014*

**math**

effective rate =(1+i/n)^n - 1 where n is the number of compounding per year. For 7% compounded semi-annually, n=2, and i=0.07 effective rate =(1+0.07/2)²-1 = 7.1225%

*May 11, 2014*

**Chemistry help**

Think of some of the energy released has been used to transform water(l) to water(g) and adjust accordingly. Watch the stoichiometric coefficient of H2O.

*May 11, 2014*

**Calc**

Oops, I got the function wrong, got an extra t on the right hand side. Go with Steve's solution.

*May 11, 2014*

**Calc**

dy/dt=-0.15y dy/y = -0.15t ln(y)=-(0.15/2)t²+C f(t)=y=Ce-0.075t^sup2; At t=0, y=3 ml. f(0)=C=3 so, let t=time for second dose, f(t)=3e-0.075t^sup2; solve for f(t)=1 ml. -0.075t²=ln(1/3) t=sqrt(-ln(1/3)/-0.075) =3.8273 (approx.) Time for third dose (after second dose...

*May 11, 2014*

**chemistry**

Convert 811 cg to kilograms 811.0 cg = 811.0/100 g = 8.110/1000 kg =0.008110 kg Then use formula to calculate concentration in ppm. Keep the answer to 4 significant figures.

*May 11, 2014*