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Sagot :
To find the values of [tex]\( y \)[/tex] corresponding to the given [tex]\( x \)[/tex] values based on the equation [tex]\( y = x^2 - 10 \)[/tex]:
### Step-by-Step Solution:
1. For [tex]\( x = -6 \)[/tex]:
Substitute [tex]\( x = -6 \)[/tex] into the equation:
[tex]\[ y = (-6)^2 - 10 \][/tex]
Calculating the value:
[tex]\[ y = 36 - 10 = 26 \][/tex]
So, [tex]\( y = 26 \)[/tex] when [tex]\( x = -6 \)[/tex].
2. For [tex]\( x = -1 \)[/tex]:
Substitute [tex]\( x = -1 \)[/tex] into the equation:
[tex]\[ y = (-1)^2 - 10 \][/tex]
Calculating the value:
[tex]\[ y = 1 - 10 = -9 \][/tex]
So, [tex]\( y = -9 \)[/tex] when [tex]\( x = -1 \)[/tex].
3. For [tex]\( x = 0 \)[/tex]:
Substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = 0^2 - 10 \][/tex]
Calculating the value:
[tex]\[ y = 0 - 10 = -10 \][/tex]
So, [tex]\( y = -10 \)[/tex] when [tex]\( x = 0 \)[/tex].
4. Given [tex]\( y = 6 \)[/tex]:
Substitute the known [tex]\( y \)[/tex] value [tex]\( y = 6 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[ 6 = x^2 - 10 \][/tex]
Rearrange the equation to solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 = 6 + 10 \][/tex]
[tex]\[ x^2 = 16 \][/tex]
Taking the square root of both sides:
[tex]\[ x = \pm 4 \][/tex]
Since [tex]\( x \)[/tex] was not specified, one possible value for [tex]\( x \)[/tex] could be [tex]\( 4 \)[/tex] or [tex]\(-4\)[/tex].
5. Given [tex]\( y = 90 \)[/tex]:
Substitute the known [tex]\( y \)[/tex] value [tex]\( y = 90 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[ 90 = x^2 - 10 \][/tex]
Rearrange the equation to solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 = 90 + 10 \][/tex]
[tex]\[ x^2 = 100 \][/tex]
Taking the square root of both sides:
[tex]\[ x = \pm 10 \][/tex]
Since [tex]\( x \)[/tex] was not specified, one possible value for [tex]\( x \)[/tex] could be [tex]\( 10 \)[/tex] or [tex]\(-10\)[/tex].
Combining all these, the completed table with the missing [tex]\( y \)[/tex] values is:
[tex]\[ \begin{tabular}{|l|l|l|l|l|l|} \hline $x$ & -6 & -1 & 0 & 4? & 10? \\ \hline $y$ & 26 & -9 & -10 & 6 & 90 \\ \hline \end{tabular} \][/tex]
The calculated missing [tex]\( y \)[/tex] values for the given [tex]\( x \)[/tex] values are:
### Detailed [tex]\(\ y\)[/tex] Values:
- When [tex]\( x = -6 \)[/tex], [tex]\( y = 26 \)[/tex].
- When [tex]\( x = -1 \)[/tex], [tex]\( y = -9 \)[/tex].
- When [tex]\( x = 0 \)[/tex], [tex]\( y = -10 \)[/tex].
These values match the obtained results when solving the problem step by step.
### Step-by-Step Solution:
1. For [tex]\( x = -6 \)[/tex]:
Substitute [tex]\( x = -6 \)[/tex] into the equation:
[tex]\[ y = (-6)^2 - 10 \][/tex]
Calculating the value:
[tex]\[ y = 36 - 10 = 26 \][/tex]
So, [tex]\( y = 26 \)[/tex] when [tex]\( x = -6 \)[/tex].
2. For [tex]\( x = -1 \)[/tex]:
Substitute [tex]\( x = -1 \)[/tex] into the equation:
[tex]\[ y = (-1)^2 - 10 \][/tex]
Calculating the value:
[tex]\[ y = 1 - 10 = -9 \][/tex]
So, [tex]\( y = -9 \)[/tex] when [tex]\( x = -1 \)[/tex].
3. For [tex]\( x = 0 \)[/tex]:
Substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = 0^2 - 10 \][/tex]
Calculating the value:
[tex]\[ y = 0 - 10 = -10 \][/tex]
So, [tex]\( y = -10 \)[/tex] when [tex]\( x = 0 \)[/tex].
4. Given [tex]\( y = 6 \)[/tex]:
Substitute the known [tex]\( y \)[/tex] value [tex]\( y = 6 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[ 6 = x^2 - 10 \][/tex]
Rearrange the equation to solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 = 6 + 10 \][/tex]
[tex]\[ x^2 = 16 \][/tex]
Taking the square root of both sides:
[tex]\[ x = \pm 4 \][/tex]
Since [tex]\( x \)[/tex] was not specified, one possible value for [tex]\( x \)[/tex] could be [tex]\( 4 \)[/tex] or [tex]\(-4\)[/tex].
5. Given [tex]\( y = 90 \)[/tex]:
Substitute the known [tex]\( y \)[/tex] value [tex]\( y = 90 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[ 90 = x^2 - 10 \][/tex]
Rearrange the equation to solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 = 90 + 10 \][/tex]
[tex]\[ x^2 = 100 \][/tex]
Taking the square root of both sides:
[tex]\[ x = \pm 10 \][/tex]
Since [tex]\( x \)[/tex] was not specified, one possible value for [tex]\( x \)[/tex] could be [tex]\( 10 \)[/tex] or [tex]\(-10\)[/tex].
Combining all these, the completed table with the missing [tex]\( y \)[/tex] values is:
[tex]\[ \begin{tabular}{|l|l|l|l|l|l|} \hline $x$ & -6 & -1 & 0 & 4? & 10? \\ \hline $y$ & 26 & -9 & -10 & 6 & 90 \\ \hline \end{tabular} \][/tex]
The calculated missing [tex]\( y \)[/tex] values for the given [tex]\( x \)[/tex] values are:
### Detailed [tex]\(\ y\)[/tex] Values:
- When [tex]\( x = -6 \)[/tex], [tex]\( y = 26 \)[/tex].
- When [tex]\( x = -1 \)[/tex], [tex]\( y = -9 \)[/tex].
- When [tex]\( x = 0 \)[/tex], [tex]\( y = -10 \)[/tex].
These values match the obtained results when solving the problem step by step.
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Acceleration refers to
a.increasing speed
b.decreasing speed
c.changing direction
d.all of the above
Acceleration refers to
a.increasing speed
b.decreasing speed
c.changing direction
d.all of the above
