sertakan cara dan rumus yaa
terimakasih:)
terimakasih:)
aplikasi integrAL
dijawab:
y = y
x² + 4 = 6x - x²
2x² - 6x + 4 = 0
a = 2
b = -6
c = 4
D = b² - 4ac
D = (-6)² - 4(2)(4)
D = 36 - 32
D = 4
[tex] \sf \: L= \frac{d \sqrt{d} }{6 {a}^{2} } [/tex]
[tex] \sf \: L= \frac{4 \sqrt{4} }{6(2) ^{2} } [/tex]
[tex] \sf \: L= \frac{4(2)}{6(4)} [/tex]
[tex] \sf \: L= \frac{8}{24} [/tex]
[tex] \sf \: L= \frac{1}{3} \: satuan \: luas[/tex]
