13. Jika Meli ingin membeli 1 pak buku dan 2 pak pulpen maka jumlah uang yang harus dibayarkan adalah. a Rp24 600,00 c. Rp28.6900,00 b Rp27.600,00 d Rp29,600,00 (soal akm kelas 5 sd)​

Jawaban:

[tex]\frac{2}{5 \sqrt{3} } - \sqrt{5} = (\frac{2}{5 \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} }) - \sqrt{5} = (\frac{2 \sqrt{3} }{5 - 3}) - \sqrt{5} = (\frac{2 \sqrt{3} }{2}) - \sqrt{5} = (\frac{1 \sqrt{3} }{1}) - \sqrt{5} = (1 \sqrt{3}) - \sqrt{5} = \sqrt{3} - \sqrt{5} = - \sqrt{2} [/tex]

semoga membantu (:

[tex]\frac{2}{5 \sqrt{3} - \sqrt{5}} = \frac{2}{5 \sqrt{3} - \sqrt{5}} \times \frac{5 \sqrt{3} + \sqrt{5}}{5 \sqrt{3} + \sqrt{5}} = \frac{(2)(5 \sqrt{3} + \sqrt{5})}{(5 \sqrt{3} - \sqrt{5})(5 \sqrt{3} + \sqrt{5})} [/tex]

[tex]= \frac{10 \sqrt{3} + 2 \sqrt{5}}{25 \sqrt{9} + 5 \sqrt{15} - 5 \sqrt{15} - \sqrt{25}} [/tex]

[tex]= \frac{10 \sqrt{3} + 2 \sqrt{5}}{25 \sqrt{9} - \sqrt{25}} [/tex]

[tex]= \frac{10 \sqrt{3} + 2 \sqrt{5}}{25 \times 3 - 5} = \frac{10 \sqrt{3} + 2 \sqrt{5}}{75 - 5} = \frac{10 \sqrt{3} + 2 \sqrt{5}}{70} = \frac{5 \sqrt{3} + \sqrt{5}}{35} [/tex]

Jawaban:

[tex] \frac{2}{5 \sqrt{3} } = \frac{2}{5 \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } = \frac{2 \sqrt{3} }{5 - 3} = \frac{2 \sqrt{3} }{2} = \frac{1 \sqrt{3} }{1} = 1 \sqrt{3} = \sqrt{3} [/tex]

semoga membantu (:

[tex]\frac{2}{5 \sqrt{3}} = \frac{2}{5 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2 \sqrt{3}}{5 \sqrt{9}} = \frac{2 \sqrt{3}}{5 \times 3} = \frac{2 \sqrt{3}}{15} [/tex]

Terdapat suatu bentuk limit trigonometri tak hingga sebagai berikut:

[tex]\lim_{x \to \infty} \frac{\frac{2}{x}\text{tan}\frac{3}{x}}{\text{sin}^2\frac{6}{x}}[/tex]

Nilai limit tersebut adalah [tex]\bf\frac{1}{6}[/tex] (C).

Diketahui:

[tex]\lim_{x \to \infty} \frac{\frac{2}{x}\text{tan}\frac{3}{x}}{\text{sin}^2\frac{6}{x}}[/tex]

Ditanya: nilai limit

Jawab:

• Pemisalan

Misalkan y = [tex]\frac{1}{x}[/tex]. Saat x → ∞, y → 0.

• Nilai limit

[tex]\lim_{x \to \infty} \frac{\frac{2}{x}\text{tan}\frac{3}{x}}{\text{sin}^2\frac{6}{x}}\\= \lim_{x \to \infty} \frac{2\cdot\frac{1}{x}\text{tan}(3\cdot\frac{1}{x})}{\text{sin}^26\cdot\frac{1}{x}}\\= \lim_{y \to 0} \frac{2y\text{tan}3y}{\text{sin}^26y}\\= \lim_{y \to 0} \frac{2y\text{tan}3y}{\text{sin}6y\cdot\text{sin}6y}\\= \lim_{y \to 0} \frac{2y}{\text{sin}6y}\cdot\frac{\text{tan}3y}{\text{sin}6y}\\= \lim_{y \to 0} \frac{2y}{\text{sin}6y}\cdot\lim_{y \to 0}\frac{\text{tan}3y}{\text{sin}6y}[/tex]

[tex]=\frac{2}{6}\cdot\frac{3}{6}\\=\frac{1}{3}\cdot\frac{1}{2}\\=\frac{1}{6}[/tex]

Jadi, limit ini bernilai [tex]\bf\frac{1}{6}[/tex] (C).

Materi tentang Menhgitung Nilai Limit Trigonometri Tak Hingga pada https://brainly.co.id/tugas/22137058

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[tex]{(8)}^{\frac{1}{3}} + {(27)}^{\frac{2}{3}} - {(125)}^{\frac{1}{3}} [/tex]

[tex]= {({2}^{3})}^{\frac{1}{3}} + {({3}^{3})}^{\frac{2}{3}} - {({5}^{3})}^{\frac{1}{3}} [/tex]

[tex]= {(2)}^{3 \times \frac{1}{3}} + {(3)}^{3 \times \frac{2}{3}} - {(5)}^{3 \times \frac{1}{3}}[/tex]

[tex]= {(2)}^{1} + {(3)}^{2} - {(5)}^{1} [/tex]

[tex]= 2 + 9 - 5 = 11 - 5 = 6[/tex]