A
minimum of algae is when we derive and equal 0
Derivation
[tex]\begin{gathered} (2)35t-360 \\ 70t-360 \end{gathered}[/tex]equal 0
[tex]\begin{gathered} 70t-360=0 \\ 70t=360 \\ t=\frac{360}{70} \\ \\ t=\frac{36}{7} \end{gathered}[/tex]the amout of algae reach minimum when t=36/7
B
to calculate the minumum number of algae we replace the days when algae reach minumum(previus exercise) on function
[tex]\begin{gathered} 35t^2-360t+1050 \\ 35(\frac{36}{7})^2-360(\frac{36}{7})+1050 \\ \\ 35(\frac{1296}{49})-\frac{12960}{7}+1050 \\ \\ \frac{6480}{7}-\frac{12960}{7}+1050 \\ \\ =\frac{870}{7}\approx124.3 \end{gathered}[/tex]the rounded minimum number of algae is 124
C
We replace the amount of algae to 900 on the original function
[tex]900=35t^2-360t+1050[/tex]simplify
[tex]\begin{gathered} 35t^2-360t+1050-900=0 \\ 35t^2-360t+150=0 \end{gathered}[/tex]now solve t factoring by
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]where x is the variable then t, a is 35, b is -360 and c 150
replacing
[tex]t=\frac{-(-360)\pm\sqrt[]{(-360)^2-4(35)(150)}}{2(35)}[/tex]simplify
[tex]\begin{gathered} t=\frac{360\pm\sqrt[]{129600-21000}}{70} \\ \\ t=\frac{360\pm\sqrt[]{108600}}{70} \\ \\ t=\frac{360\pm\sqrt[]{100\times1086}}{70} \\ \\ t=\frac{360\pm10\sqrt[]{1086}}{70} \\ \\ t=\frac{36\pm\sqrt[]{1086}}{7} \end{gathered}[/tex]then the days to have 900 algae is
[tex]\begin{gathered} t_1=\frac{36+\sqrt[]{1086}}{7}\approx9.85 \\ \\ t_2=\frac{36-\sqrt[]{1086}}{7}\approx0.46 \end{gathered}[/tex]we have 2 values for 900 algae 9.85 and 0.46 days
D Graph
we know the graph is a parable because maximum exponent is 2
we know points like,
minimum amount of algae A(5.14 , 124.3) ,
when the amount of algae is 900 B(9.85 , 900) C(0.46 , 900)
then we can palce the points of a graph and join them
