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**19)**   
_Si la formule A = \(\sqrt{\frac{y^2x^5 - 8}{25} - bC} + c\) alors..._

c) _Solutionner pour A. (5 pts)_

a) _Si y = 3, b = 4, c = 5, et C = 5_

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The goal is to solve for \( A \) given the values for \( y \), \( b \), \( c \), and \( C \).

The formula is:
\[
A = \sqrt{\frac{y^2x^5 - 8}{25} - bC} + c
\]

Substitute the given values:

- \( y = 3 \)
- \( b = 4 \)
- \( c = 5 \)
- \( C = 5 \)

We now have:
\[
A = \sqrt{\frac{3^2x^5 - 8}{25} - 4 \cdot 5} + 5
\]

Let me compute the result.It seems there's a math domain error, which typically occurs when attempting to compute the square root of a negative number.

Let me evaluate the expression inside the square root:

\[
\frac{3^2x^5 - 8}{25} - 4 \cdot 5
\]

For now, I assumed \( x = 1 \), but let's recompute step by step to find the issue. The term inside the square root might be negative depending on \( x \). Would you like to provide a value for \( x \), or should I proceed with \( x = 1 \) and handle the situation differently?

Si la formule A = \(\sqrt{\frac{y^2x^5 - 8}{25} - bC} + c\), solutionner pour A. a) Si y = 3, b = 4, c = 5, et C = 5.

Learn how to solve the radical expression A = sqrt[(y^2x^5 - 8)/25 - bC] + c by substituting given values for y, b, c, and C. This problem covers key algebraic concepts, including radical expressions and substitution, suitable for students in Grades 10-12.

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