Solve the following ode's numerically, using matlab.
for a given system of ode's,
dcc dt
\frac{dcs}{dt}= y_{sc}*(-r_{g})-r_{sm}
\frac{_{d_{cp}}}{dt}=r_{g}*y_{pc} given the parameters rd = 0.1 * cc; ysc = 1/0.08; ypc = 5.6; ks = 1.7; m = 0.03; umax = 0.33; rsm = m* cc; kobs = umax (1-cp/93)^0.52; rg = kobs (cc cs)/(ks+cs). initial conditions: cc(t=0) = 1; cs(t=0) = 250; cp(t=0) = 0.
use matlab to:
1. plot cc, cs, cp versus t
2. plot rg, rsm, rd versus t
3. plot kobs versus t

part a

x     y

0     8

1     7.5

2     7

3     6.5

4     6

5     5.5  

for every hour i deducted .5 inches from the length of the candle

part b

it is a function because for every x value there is only one y value

part   c

for every x value there is only one y value

x       y

0       8

1       7.6

2       7.2

3       6.8

step-by-step explanation:

solution:

the given function is ,

      =

the three roots of this quadratic function is , 0 ,8 and 10.all the roots are real and rational.

if a quadratic equation is of the form ,

sum of the roots = a + b +c= 0+8+10=18=  

product of roots taken two at a time = a b + b c+ c a = 0+80+0=80 =  

product of roots = a b c = 0=  

there are 11 doors in a hall starting from 0 to 11.the entry in the hall is permitted through gate number 0 and exit is allowed through only gate number 8 and 10.so, three roots(0,8,10) of quadratic function is given by entry and exit route of this hall.

x = -4

step-by-step explanation:

- 20 - 4(x + 3) = 4x

=> -20 - 4x - 12 = 4x

=> - 32 -4x = 4x

=> -32 = 4x + 4x

=> - 32 = 8x

=> -32/8 = x

=> -4 = x