Answer:

![\sqrt[3]{0.95} \approx 0.9833](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B0.95%7D%20%5Capprox%200.9833)
![\sqrt[3]{1.1} \approx 1.0333](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B1.1%7D%20%5Capprox%201.0333)
Step-by-step explanation:
Given the function: ![g(x)=\sqrt[3]{1+x}](https://tex.z-dn.net/?f=g%28x%29%3D%5Csqrt%5B3%5D%7B1%2Bx%7D)
We are to determine the linear approximation of the function g(x) at a = 0.
Linear Approximating Polynomial,
a=0
![g(0)=\sqrt[3]{1+0}=1](https://tex.z-dn.net/?f=g%280%29%3D%5Csqrt%5B3%5D%7B1%2B0%7D%3D1)

Therefore:

(b)![\sqrt[3]{0.95}= \sqrt[3]{1-0.05}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B0.95%7D%3D%20%5Csqrt%5B3%5D%7B1-0.05%7D)
When x = - 0.05

![\sqrt[3]{0.95} \approx 0.9833](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B0.95%7D%20%5Capprox%200.9833)
(c)
(b)![\sqrt[3]{1.1}= \sqrt[3]{1+0.1}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B1.1%7D%3D%20%5Csqrt%5B3%5D%7B1%2B0.1%7D)
When x = 0.1

![\sqrt[3]{1.1} \approx 1.0333](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B1.1%7D%20%5Capprox%201.0333)
Answer:
hhhci
Step-by-step explanation:
Answer:
x > 5.5
Step-by-step explanation:
4(x - 3) - (2x - 1) > 0
4x - 12 - 2x + 1 > 0
4x - 2x - 11 > 0
2x - 11 > 0
2x - 11 + 11 > 0 + 11
2x > 11
2x ÷ 2 > 11 ÷ 2
x > 5.5
Answer: 41.5 min
Step-by-step explanation:
This problem can be solved with the Radioactive Half Life Formula:
(1)
Where:
is the final amount of the radioactive element
is the initial amount of the radioactive element
is the time elapsed
is the half life of the radioactive element
So, we need to substitute the given values and find
from (1):
(2)
(3)
(4)
Applying natural logarithm in both sides:
(5)
(6)
Clearing
:
This is the time elapsed
Answer:
I seen this on a test before and brainly said that we cant answer test questions i think so sorry dude
Step-by-step explanation: