The value of h(t) when [tex]t=\frac{15}{32}[/tex] is 10.02.
Solution:
Given function [tex]h(t)=-16t^2+15t+6.5[/tex]
To find the value of h(t) when [tex]t=\frac{15}{32}[/tex]:
[tex]h(t)=-16t^2+15t+6.5[/tex]
Substitute [tex]t=\frac{15}{32}[/tex] in the given function.
[tex]$h\left(\frac{15}{32} \right)=-16\left(\frac{15}{32} \right)^2+15\left(\frac{15}{32} \right)+6.5[/tex]
[tex]$=-16\left(\frac{225}{1024} \right)+15\left(\frac{15}{32} \right)+6.5[/tex]
Now multiply the common terms into inside the bracket.
[tex]$=-\left(\frac{3600}{1024} \right)+\left(\frac{225}{32} \right)+6.5[/tex]
Now, in the first term, the numerator and denominator both have common factor 16. So reduce the first term into the lowest term.
[tex]$=-\left(\frac{225}{64} \right)+\left(\frac{225}{32} \right)+6.5[/tex]
To make the denominator same, take LCM of the denominators.
LCM of 64 and 32 = 64
[tex]$=-\left(\frac{225}{64} \right)+\left(\frac{225\times2}{32\times2} \right)+6.5\times\frac{64}{64}[/tex]
[tex]$=-\frac{225}{64} +\frac{450}{64}+\frac{416}{64}[/tex]
[tex]$=\frac{-225+450+416}{64}[/tex]
[tex]$=\frac{641}{64}[/tex]
= 10.02
[tex]$h\left(\frac{15}{32} \right)=10.02[/tex]
Hence the value of h(t) when [tex]t=\frac{15}{32}[/tex] is 10.02.
