Question
What value is a solution to two of the equations? p2+4=10 2=10−6h 3y−16=20 (1 point)
Answers
Answered by
Water
The correct answer is 12! >:(
Answered by
GPT-4o mini
To find a value that is a solution to two of the given equations, we can solve each equation for its variable.
1. For the equation \( p^2 + 4 = 10 \):
\[
p^2 = 10 - 4
\]
\[
p^2 = 6
\]
\[
p = \sqrt{6} \quad \text{or} \quad p = -\sqrt{6}
\]
2. For the equation \( 2 = 10 - 6h \):
\[
6h = 10 - 2
\]
\[
6h = 8
\]
\[
h = \frac{8}{6} = \frac{4}{3}
\]
3. For the equation \( 3y - 16 = 20 \):
\[
3y = 20 + 16
\]
\[
3y = 36
\]
\[
y = \frac{36}{3} = 12
\]
Now we have the solutions:
- \( p = \sqrt{6} \) or \( p = -\sqrt{6} \)
- \( h = \frac{4}{3} \)
- \( y = 12 \)
Since all solutions are independent, no variable value here is a solution to two of these equations. All values are distinct unless they were defined to share a constant, but we can conclude that there is no common value among the variables from the equations provided.
1. For the equation \( p^2 + 4 = 10 \):
\[
p^2 = 10 - 4
\]
\[
p^2 = 6
\]
\[
p = \sqrt{6} \quad \text{or} \quad p = -\sqrt{6}
\]
2. For the equation \( 2 = 10 - 6h \):
\[
6h = 10 - 2
\]
\[
6h = 8
\]
\[
h = \frac{8}{6} = \frac{4}{3}
\]
3. For the equation \( 3y - 16 = 20 \):
\[
3y = 20 + 16
\]
\[
3y = 36
\]
\[
y = \frac{36}{3} = 12
\]
Now we have the solutions:
- \( p = \sqrt{6} \) or \( p = -\sqrt{6} \)
- \( h = \frac{4}{3} \)
- \( y = 12 \)
Since all solutions are independent, no variable value here is a solution to two of these equations. All values are distinct unless they were defined to share a constant, but we can conclude that there is no common value among the variables from the equations provided.