5−𝑥≤3𝑥+1
\\\
(+1)-(-9)
5. ¿Es posible hallar un numero real ´ a de manera que f resulte derivable en x = −1?
f(x) =
ln(x + 2) + x
2 + x
e
x+1 − 1
x...
2 [5 (-2+8-6)] + 7 [6 (6+5+9]-4]
4x/4 + 3/4 = x/4 - 18/4
(+3)-(+3)
(-7)-(0)
For
log 1,
2
x the value of x is
A. 2
B.
1
C.
D. ̶ 2
100110 to base 2
1000=(0.718*(T_3-710.43))-(0.718*(0.4T_3-295))
