Question

bantu lah minimal,yang ngemis mati aja​

Answer 1

Verified answer

Jawab:

Penjelasan dengan langkah-langkah:

16. [tex]p(5\sqrt{2}+\sqrt{6})=28\\p=\frac{28}{5\sqrt{2}+\sqrt{6}}\\p=\frac{28}{5\sqrt{2}+\sqrt{6}}.\frac{5\sqrt{2}-\sqrt{6}}{5\sqrt{2}-\sqrt{6}}\\p=\frac{28(5\sqrt{2}-\sqrt{6})}{(5\sqrt{2}+\sqrt{6})(5\sqrt{2}-\sqrt{6})}\\p=\frac{28(5\sqrt{2}-\sqrt{6})}{25.2-6}\\p=\frac{28(5\sqrt{2}-\sqrt{6})}{44} \\p=\frac{7(5\sqrt{2}-\sqrt{6})}{11}\\p=\frac{35\sqrt{2}-7\sqrt{6}}{11}[/tex]

17. Dik = s = [tex](4\sqrt{3}+2)[/tex] cm

    Dit = Luas = ..........[tex]cm^{2}[/tex]?

Rumus Luas Persegi=

= s . s

= [tex](4\sqrt{3}+2)(4\sqrt{3}+2)=(4\sqrt{3})^{2}+8\sqrt{3}+8\sqrt{3}+2^{2}=16.3+16\sqrt{3}+4\\48+16\sqrt{3}+4=48+4+16\sqrt{3}=52+16\sqrt{3}[/tex]

Maka,luas Perseginya adalah [tex]52+16\sqrt{3}[/tex] [tex]cm^{2}[/tex]

NOTE:

JIKA---->(a+b)(a-b)= [tex]a^{2}-b^{2}[/tex]

& [tex](a+b)^{2}[/tex]  atau (a+b)(a+b)= [tex]a^{2}+2ab+b^{2}[/tex]