Question

Q.

[tex]\boxed{\left(\footnotesize\lfloor\frac{40}{15}\rfloor+\lceil\frac{98}{23}\rceil \right)^{2}+\displaystyle\sum_{\theta=3}^{4}\left(2\theta\right)}[/tex]
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Answer 1

[tex]\displaystyle\left(\left\lfloor\frac{40}{15}\right\rfloor+\left\lceil\frac{98}{23}\right\rceil \right)^{2}+\sum_{\theta=3}^{4}\left(2\theta\right)=\boxed{\ \bf63\ }[/tex] Pembahasan[tex]\displaystyle\left(\left\lfloor\frac{40}{15}\right\rfloor+\left\lceil\frac{98}{23}\right\rceil \right)^{2}+\sum_{\theta=3}^{4}\left(2\theta\right)={\dots}[/tex]Notasi  [tex]\left\lfloor\dfrac{a}{b}\right\rfloor[/tex]  menghasilkan nilai pembulatan ke bawah dari  [tex]\dfrac{a}{b}[/tex] . Sedangkan  [tex]\left\lceil\dfrac{a}{b}\right\rceil[/tex]  menghasilkan nilai pembulatan ke atas dari  [tex]\dfrac{a}{b}[/tex] . Penyelesaiannya:[tex]\begin{aligned}&\left(\left\lfloor\frac{40}{15}\right\rfloor+\left\lceil\frac{98}{23}\right\rceil \right)^{2}+\sum_{\theta=3}^{4}\left(2\theta\right)\\{=\ }&\left(\left\lfloor{\bf2}\,\frac{10}{15}\right\rfloor+\left\lceil{\bf4}\,\frac{6}{23}\right\rceil \right)^2+2\sum_{\theta=3}^{4}\theta\\{=\ }&(2+5)^2+2(3+4)\\{=\ }&7^2+2(7)\\{=\ }&7\times9\\{=\ }&\boxed{\ \bf63\ }\end{aligned}[/tex]