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:
The solution is (0, 3/4)
Step-by-step explanation:
Please copy and share the instructions. Here they are: Solve the following system of linear equations.
Both of the equations can be reduced (simplified):
2x+8y = 6 => x + 4y = 3
15x + 20y = 15 => 3x + 4y = 3
Let's use the elimination by addition and subtraction method. Multiply the first equation by -1, obtaining
-x - 4y = -3
Add the second 3x + 4y = 3
equation to the
first.
We get: 2x = 0.
Thus, x = 0. Substituting 0 for x in the 1st original equation yields:
2(0) + 8y = 6. Then y = 6/8, or y = 3/4.
The solution is (0, 3/4).
1 is x=13.5
2 is x=-260
3 is x=38
4 is x=-37^38
i think this is correct
Answer:
32
Step-by-step explanation:
32 is the GCF of the whole numbers and up to x2 can go into both
The answer should be 32x2
Hope this helps
Answer:
This is actually a reduction, and the scale factor of this dilation is 0.5.