These are 3 questions and 3 answers.
Question 1.
Answer: bot equations have the same potential solutions, but equation A may have extraneous solutions
Explanation:
1) Equation A

Replace x = -4 in the right side of the equation:

Which is not defined, so this is a extraneous solution.
2) Equation B.
Since it is a cube root, it is defined for any (negative, zero, and positive) values of x.
Question 2.
Answer: cubing both sides once.
Explanation:
![\sqrt[3]{5x-2} -\sqrt[3]{4x} =0\\ \\ \sqrt[3]{5x-2} =\sqrt[3]{4x}\\ \\ (\sqrt[3]{5x-2} )^3=(\sqrt[3]{4x})^3\\ \\ 5x-2=4x\\ \\ 5x-4x=2\\ \\ x=2](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B5x-2%7D%20-%5Csqrt%5B3%5D%7B4x%7D%20%3D0%5C%5C%20%5C%5C%20%20%5Csqrt%5B3%5D%7B5x-2%7D%20%3D%5Csqrt%5B3%5D%7B4x%7D%5C%5C%20%20%5C%5C%20%20%28%5Csqrt%5B3%5D%7B5x-2%7D%20%29%5E3%3D%28%5Csqrt%5B3%5D%7B4x%7D%29%5E3%5C%5C%20%20%5C%5C%20%205x-2%3D4x%5C%5C%20%5C%5C%20%205x-4x%3D2%5C%5C%20%5C%5C%20%20x%3D2%20)
3) Question 3.
Answer: c - 2 = 25 + c + 10√c
Explanation:

Answer:
99 038.52 USD
Step-by-step explanation:
In order to solve this question we have to use the compound interest formula
A = P*(1+i)ˆn
A: final amount including principal
P: principal amount
i: interest per year
n: total number of years
A = 40000 (1+0.12)ˆ8
A = 99 038.52 USD
It’s False, you’d have to divide the number of minutes by the miles to get how many minutes it took PER mile to ride your bike. 30/5 is 6 therefore it takes 6 minutes per mile. if you were to multiply 6 times 12 miles it would equal 72 minutes.
<h3>Explanation:</h3>
Any techniques that you're familiar with can be applied to polynomials of any degree. These might include ...
- use of the rational root theorem
- use of Descartes' rule of signs
- use of any algorithms you're aware of for finding bounds on roots
- graphing
- factoring by grouping
- use of "special forms" (for example, difference of squares, sum and difference of cubes, square of binomials, expansion of n-th powers of binomials)
- guess and check
- making use of turning points
Each root you find can be factored out to reduce the degree of the remaining polynomial factor(s).