math question? what is constant rate of change?
926663
the slope of the graph. the formula is rise/run. take 2 points X1,Y1 and X2,Y2  plug them in the formula X2-X1/Y2-Y1. youll get the rate of change! Best of Luck!
415527
First of all, the sum is incompleted.  Is it for maximum height, or for range?  Secondly, let us assume you are asking for maximum height.  Then differentiating the given equation of motion, d = -5(t+1)^2 + 258 with respect to 't', we get dt/dt = -10(t+1).  At maximum height, the velocity is zero. \nTherefore, -10(t+1)=0, or t = -1.  This is absurd time can not become negative once the motion is initiated.  Check up there is something wrong in the problem.
980357
The answer has to be a speed!  It involves seeing that dy/dt = 1/2 x^(-1/2) dx/dt in this case.  Also, the distance from the origin is given by s^2 = x^2 + y^2; to get ds/dt (the required result), differentiate that wrt "t".  Put this all together yourself; it's the only way to learn rather than simply parrot.\n\nWithout a cast-iron guarantee, I get the value 27/(4 x sqrt 5) \ncm/s.\n\nLive long and prosper.
103569
Displacement.  If you mean its change in position *divided* by the time it took, that would be its average velocity.
1064629
find the surface area first (thats easy). Then given that the emissivity of th sphere is .851, and that the rate of change will be the change of temp/time. Since there is no time involved directly, we need to have the Surface area x emissivity rate / the temperature difference (in relation to the time, which is given within the emissivity rate). This should give u the net rate, depending on the emissivity/time relationship. You do the working out, otherwise you won't learn it.
735022
There will be a term proportional to distance added to a constant delay term, so the form will be t=kD + c. Solving for k and c:\nt1=.152=100k + c\nt2=.229=170k + c\nt2-t1=.077=70k\nThen k=.0011\nt1=.152=.11 + c --> c=.042\nt2=.229=.187 + c --> c=.042\nResult is\nt=.0011*D + .042
1129152
The answer is 10\nYou find the average rate of change of the function f(x) over a given interval, by computing the total change in the function f(x) over the interval, and then dividing this value by the size of the interval.\nYour interval size is 2- (-1) = 3. So you want:\n\n(f(2) - f(-1)) / 3  = (22-(-8))/3 = 30/3 =10\n\nf(2) = 2(2^3) - 2^2 +5*2 = 16-4+10 = 22\nf(-1) = 2(-1^3) - (-1^2) -5 = -2-1-5 = -8
372422
If each bug moves towards its target at constant rate r inches per second, then I think "infinitesimally speaking" because each target is moving at right angles to the direction of its seeker that the distance between seeker and target will lessen at the same constant rate of r inches per second.\n\nThus it will take 10 inches / (r inches per second) = 10/r seconds until all four meet in the center, during which time each will have traveled 10/r * r = 10 inches.
981618
I guess 1) just wants you to write: Want dA/dt, have da/dt and db/dt.\n(From now on I'll use A' to mean dA/dt etc).\n2)\nA=π*a*b, so using the product rule, A' = πa'b + πab'. So A' = 7πb - 11πa.\n3) Just substitute in: A' = 7*π*50 - 11*π*30 = 20π, so thats the rate it is increasing.\n4) Again substitute in, A' = 7*π*20 - 11*π*40 = -300π, so now its decreasing at that rate.
758038
omg i just had this test today too!!!!!\nu use hess's law to find the change--->write the equation, balance it \nand then use the moles (coefficients) to multiply the change is s for each element.\nhess's law says that the change in s (or h, or G, its usable for any of the thermo concepts) is equal to the sum of the entropy of the products minus the sum of the entropy of the reactants. its pretty easy after that--just plug in the values and there you go! \ngood luck! this was a hard test for me..so confusing.
846505
Well...\nIf Y/Y'=k/y then Y'/Y=y/k\nso (lnY)'=y/k\nIntegrating\nlnY=ln A+ (y^2)/2k      A- constant\nExponentiating\nY=A*exp( (y^2)/2k )\nAm I right?
778575
x^2=4500-5p^2\n-10p*dp/dt=2xdx/dt\nat p=20,x^2=4500-5(20)^2\n=4500-2000\n=2500\nx=rt2500\n=50\n-10*20*1=2(50)dx/dt\ndx/dt=-200/100=-2\nit is decreasingatthe rate of -2
601069
what is the current rate?
833158
1) increases\n2) decreases\n3) remains the same \n4) increases\n5) decreases\n6) increases
1047731
Assume dc/dz is constant, n(t=T)=-DA(dc/dz) T\nBut in reality dc/dz often changes.  Normally dc/dz should decrease with time.  If it is a close system, it will reach the equilibrium state.
11463
The changes propagate at the speed of light, so I guess the time taken is,\n\nt = r/c where c = speed of light
918850
If initial speed is V0 and the speed after a time t second is V,\n\nAcceleration is change in speed per unit time.\n\nAcceleration a = (V - VO) / t.\n\nThe average speed during this time t is \n\nV (average) = (V + V0) / 2.\n\nNote; if any quantity is changing uniformly, the average is found by adding the initial and final value and dividing the sum by 2.\n\nThe average speed can also be determined by dividing the total distance by total time of travel.\n \nV (average) = S / t.\n\nTherefore S/t = (V + V0) / 2.\n\nS = (V + V0)* t / 2.\n\nFrom a = (V - V0) / t we have V = V0 +at.\n \nTherefore S = {V0 +at + VO} * t /2 \n--------------- = {2 V0 t + at t } /2\n\n-------------- S = V0t + 0.5 a tt