By the knowledge and application of <em>algebraic</em> definitions and theorems, we find that the expression - 10 · x + 1 + 7 · x = 37 has a solution of x = 12. (Correct choice: C)
<h3>How to solve an algebraic equation</h3>
In this question we have an equation that can be solved by <em>algebraic</em> definitions and theorems, whose objective consists in clearing the variable x. Now we proceed to solve the equation for x:
- - 10 · x + 1 + 7 · x = 37 Given
- (- 10 · x + 7 · x) + 1 = 37 Associative property
- -3 · x + 1 = 37 Distributive property/Definition of subtraction
- - 3 · x = 36 Compatibility with addition/Definition of subtraction
- x = 12 Compatibility with multiplication/a/(-b) = -a/b/Definition of division/Result
By the knowledge and application of <em>algebraic</em> definitions and theorems, we find that the expression - 10 · x + 1 + 7 · x = 37 has a solution of x = 12. (Correct choice: C)
To learn more on linear equations: brainly.com/question/2263981
#SPJ1
Answer:
Step-by-step explanation:
1). Step 4:
[Since,
]
![x=\sqrt[3]{5\times 5\times 5\times 5}](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B5%5Ctimes%205%5Ctimes%205%5Ctimes%205%7D)
Step 5:
![x=\sqrt[3]{(5)^3\times 5}](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B%285%29%5E3%5Ctimes%205%7D)
![x=\sqrt[3]{5^3}\times \sqrt[3]{5}](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B5%5E3%7D%5Ctimes%20%5Csqrt%5B3%5D%7B5%7D)
2). He simplified the expression by removing exponents from the given expression.
3). Let the radical equation is,

Step 1:

Step 2:

Step 3:

Step 4:

4). By substituting
in the original equation.



There is no extraneous solution.
C is same thing as 1c
1c + 1/4c
(In simple words: Adding a whole to 1/4.)
1 + 1/4 = 1 1/4 or 5/4
same thing when having c:
1c + 1/4c = 1 1/4c or 5/4 c
Answer: 1 1/4c or 5/4c