2.) dT/dt = -t^2 + 3t
dT = (-t^2 + 3t)dt
T = -t^3/3 + 3t^2/2 + C
at 4:00 AM
80 = -4^3/3 + 3*4^2/2 + C
C = 80 + 4^3/3 - 3*4^2/2
C = 77.33
T = -t^3/3 + 3t^2/2 + 77.33
at 8:00 AM
T = -8^3/3 + 3*8^2/2 + 77.33
T = 2.67 deg
tis noteworthy that the segment contains endpoints of A and C and the point B is in between A and C cutting the segment in a 1:2 ratio,
![\bf \textit{internal division of a line segment using ratios} \\\\\\ A(-9,-7)\qquad C(x,y)\qquad \qquad \stackrel{\textit{ratio from A to C}}{1:2} \\\\\\ \cfrac{A\underline{B}}{\underline{B} C} = \cfrac{1}{2}\implies \cfrac{A}{C}=\cfrac{1}{2}\implies 2A=1C\implies 2(-9,-7)=1(x,y)\\\\[-0.35em] ~\dotfill\\\\ B=\left(\frac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \frac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)\\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Binternal%20division%20of%20a%20line%20segment%20using%20ratios%7D%20%5C%5C%5C%5C%5C%5C%20A%28-9%2C-7%29%5Cqquad%20C%28x%2Cy%29%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bratio%20from%20A%20to%20C%7D%7D%7B1%3A2%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7BA%5Cunderline%7BB%7D%7D%7B%5Cunderline%7BB%7D%20C%7D%20%3D%20%5Ccfrac%7B1%7D%7B2%7D%5Cimplies%20%5Ccfrac%7BA%7D%7BC%7D%3D%5Ccfrac%7B1%7D%7B2%7D%5Cimplies%202A%3D1C%5Cimplies%202%28-9%2C-7%29%3D1%28x%2Cy%29%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20B%3D%5Cleft%28%5Cfrac%7B%5Ctextit%7Bsum%20of%20%22x%22%20values%7D%7D%7B%5Ctextit%7Bsum%20of%20ratios%7D%7D%5Cquad%20%2C%5Cquad%20%5Cfrac%7B%5Ctextit%7Bsum%20of%20%22y%22%20values%7D%7D%7B%5Ctextit%7Bsum%20of%20ratios%7D%7D%5Cright%29%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf B=\left(\cfrac{(2\cdot -9)+(1\cdot x)}{1+2}\quad ,\quad \cfrac{(2\cdot -7)+(1\cdot y)}{1+2}\right)~~=~~(-4,-6) \\\\[-0.35em] ~\dotfill\\\\ \cfrac{(2\cdot -9)+(1\cdot x)}{1+2}=-4\implies \cfrac{-18+x}{3}=-4 \\\\\\ -18+x=-12\implies \boxed{x=6} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{(2\cdot -7)+(1\cdot y)}{1+2}=-6\implies \cfrac{-14+y}{3}=-6 \\\\\\ -14+y=-18\implies \boxed{y=-4}](https://tex.z-dn.net/?f=%5Cbf%20B%3D%5Cleft%28%5Ccfrac%7B%282%5Ccdot%20-9%29%2B%281%5Ccdot%20x%29%7D%7B1%2B2%7D%5Cquad%20%2C%5Cquad%20%5Ccfrac%7B%282%5Ccdot%20-7%29%2B%281%5Ccdot%20y%29%7D%7B1%2B2%7D%5Cright%29~~%3D~~%28-4%2C-6%29%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B%282%5Ccdot%20-9%29%2B%281%5Ccdot%20x%29%7D%7B1%2B2%7D%3D-4%5Cimplies%20%5Ccfrac%7B-18%2Bx%7D%7B3%7D%3D-4%20%5C%5C%5C%5C%5C%5C%20-18%2Bx%3D-12%5Cimplies%20%5Cboxed%7Bx%3D6%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B%282%5Ccdot%20-7%29%2B%281%5Ccdot%20y%29%7D%7B1%2B2%7D%3D-6%5Cimplies%20%5Ccfrac%7B-14%2By%7D%7B3%7D%3D-6%20%5C%5C%5C%5C%5C%5C%20-14%2By%3D-18%5Cimplies%20%5Cboxed%7By%3D-4%7D)
Answer:
1) The solve by graphing will the preferred choice when the equation is complex to be easily solved by the other means
Example;
y = x⁵ + 4·x⁴ + 3·x³ + 2·x² + x + 3
2) Solving by substitution is suitable where we have two or more variables in two or more (equal number) of equations
2x + 6y = 16
x + y = 6
We can substitute the value of x = 6 - y, into the first equation and solve from there
3) Solving an equation be Elimination, is suitable when there are two or more equations with coefficients of the form, 2·x + 6·y = 23 and x + y = 16
Multiplying the second equation by 2 and subtracting it from the first equation as follows
2·x + 6·y - 2×(x + y) = 23 - 2 × 16
2·x - 2·x + 6·y - 2·y = 23 - 32
0 + 4·y = -9
4) An example of a linear system that can be solved by all three methods is given as follows;
2·x + 6·y = 23
x + y = 16
Step-by-step explanation:
Answer:
Step-by-step explanation:
In logarithmic functions, if we have an exponent say the 
in 
, we can "pull it down" from the power and multiply it next to the log.
