tolong di bantu y kak
Matematika
okta111001
Pertanyaan
tolong di bantu y kak
2 Jawaban
-
1. Jawaban AnugerahRamot
Eksponensial
[tex] a. \ \frac{1}{4t^6} \\
= \frac14t^{-6} \\
\\
b. \ \frac{5}{(a+b)^3} \\
= 5(a+b)^{-3} \\
\\
c. \ \frac{2}{(b^2+ c^3)^3} [/tex]
= 2(b² + c³)⁻³ -
2. Jawaban Rushinom
a
1
___
4t^6
1/4t ^ -6
b.
5
____
( a + b ) ^ 3
5 ( a + b ) ^ -3
c.
2
_____
( b^2 + c^3 ) ^3
2 ( b^2 + c^3 ) ^ - 3