Answer :
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12. Analysis of Simplification:
We need to determine if the student's simplification of the expression [tex]\(\frac{x^{-1} y^3}{x y^{-2}}\)[/tex] is correct.
Let's follow the steps given by the student and analyze each one:
[tex]\[ \frac{x^{-1} y^3}{x y^{-2}} \][/tex]
Step 1: Combine the exponents of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] in the numerator and denominator separately.
[tex]\[ = x^{-1-1} \times y^{3 - (-2)} \][/tex]
This step simplifies to:
[tex]\[ = x^{-2} \times y^{3+2} \][/tex]
[tex]\[ = x^{-2} \times y^5 \][/tex]
Step 2: Simplify the final expression.
[tex]\[ = \frac{1}{x^2} \times y^5 \][/tex]
So, the correct simplification should be:
[tex]\[ = y^5 x^{-2} \][/tex]
The student's solution was:
[tex]\[ x^{-1+1} \times y^{3-2} = x^0 y \Rightarrow y \][/tex]
From our correct simplification, we can see that the student made an error by incorrectly adding the exponents of [tex]\(x\)[/tex] in the numerator and denominator. Instead of combining the terms correctly, the student incorrectly simplified [tex]\(-1+1\)[/tex] and added the powers of [tex]\(y\)[/tex] as [tex]\(3-2\)[/tex] instead of [tex]\(3-(-2)\)[/tex].
Therefore, the correct final result is [tex]\( y^5 x^{-2} \)[/tex] not [tex]\( y \)[/tex], and the student was incorrect in this simplification.
13. Calculation of Orbital Period:
We are given Kepler’s formula:
[tex]\[ T = 0.2 R^{\frac{3}{2}} \][/tex]
where [tex]\(T\)[/tex] is the period in Earth days and [tex]\(R\)[/tex] is the distance from the planet to the sun in millions of kilometers.
Jupiter:
- [tex]\(R_{\text{Jupiter}} = 778.5\)[/tex] million kilometers
Calculating [tex]\(T_{\text{Jupiter}}\)[/tex]:
[tex]\[ T_{\text{Jupiter}} = 0.2 \times (778.5)^{\frac{3}{2}} \][/tex]
This gives us:
[tex]\[ T_{\text{Jupiter}} = 4344.281121773774 \text{ Earth days} \][/tex]
Venus:
- [tex]\(R_{\text{Venus}} = 108.2\)[/tex] million kilometers
Calculating [tex]\(T_{\text{Venus}}\)[/tex]:
[tex]\[ T_{\text{Venus}} = 0.2 \times (108.2)^{\frac{3}{2}} \][/tex]
This gives us:
[tex]\[ T_{\text{Venus}} = 225.09761153775045 \text{ Earth days} \][/tex]
Comparison:
To find which planet has the longer period, compare the periods of Jupiter and Venus.
[tex]\[ T_{\text{Jupiter}} = 4344.281121773774 \text{ days} \][/tex]
[tex]\[ T_{\text{Venus}} = 225.09761153775045 \text{ days} \][/tex]
Jupiter has the longer period.
Difference in Periods:
Calculate the difference in periods:
[tex]\[ \Delta T = T_{\text{Jupiter}} - T_{\text{Venus}} \][/tex]
[tex]\[ \Delta T = 4344.281121773774 - 225.09761153775045 \][/tex]
[tex]\[ \Delta T = 4119.183510236024 \text{ days} \][/tex]
Conclusion:
Jupiter has the longer period, and it is longer by 4119.183510236024 Earth days compared to Venus.
Date:
12. Analysis of Simplification:
We need to determine if the student's simplification of the expression [tex]\(\frac{x^{-1} y^3}{x y^{-2}}\)[/tex] is correct.
Let's follow the steps given by the student and analyze each one:
[tex]\[ \frac{x^{-1} y^3}{x y^{-2}} \][/tex]
Step 1: Combine the exponents of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] in the numerator and denominator separately.
[tex]\[ = x^{-1-1} \times y^{3 - (-2)} \][/tex]
This step simplifies to:
[tex]\[ = x^{-2} \times y^{3+2} \][/tex]
[tex]\[ = x^{-2} \times y^5 \][/tex]
Step 2: Simplify the final expression.
[tex]\[ = \frac{1}{x^2} \times y^5 \][/tex]
So, the correct simplification should be:
[tex]\[ = y^5 x^{-2} \][/tex]
The student's solution was:
[tex]\[ x^{-1+1} \times y^{3-2} = x^0 y \Rightarrow y \][/tex]
From our correct simplification, we can see that the student made an error by incorrectly adding the exponents of [tex]\(x\)[/tex] in the numerator and denominator. Instead of combining the terms correctly, the student incorrectly simplified [tex]\(-1+1\)[/tex] and added the powers of [tex]\(y\)[/tex] as [tex]\(3-2\)[/tex] instead of [tex]\(3-(-2)\)[/tex].
Therefore, the correct final result is [tex]\( y^5 x^{-2} \)[/tex] not [tex]\( y \)[/tex], and the student was incorrect in this simplification.
13. Calculation of Orbital Period:
We are given Kepler’s formula:
[tex]\[ T = 0.2 R^{\frac{3}{2}} \][/tex]
where [tex]\(T\)[/tex] is the period in Earth days and [tex]\(R\)[/tex] is the distance from the planet to the sun in millions of kilometers.
Jupiter:
- [tex]\(R_{\text{Jupiter}} = 778.5\)[/tex] million kilometers
Calculating [tex]\(T_{\text{Jupiter}}\)[/tex]:
[tex]\[ T_{\text{Jupiter}} = 0.2 \times (778.5)^{\frac{3}{2}} \][/tex]
This gives us:
[tex]\[ T_{\text{Jupiter}} = 4344.281121773774 \text{ Earth days} \][/tex]
Venus:
- [tex]\(R_{\text{Venus}} = 108.2\)[/tex] million kilometers
Calculating [tex]\(T_{\text{Venus}}\)[/tex]:
[tex]\[ T_{\text{Venus}} = 0.2 \times (108.2)^{\frac{3}{2}} \][/tex]
This gives us:
[tex]\[ T_{\text{Venus}} = 225.09761153775045 \text{ Earth days} \][/tex]
Comparison:
To find which planet has the longer period, compare the periods of Jupiter and Venus.
[tex]\[ T_{\text{Jupiter}} = 4344.281121773774 \text{ days} \][/tex]
[tex]\[ T_{\text{Venus}} = 225.09761153775045 \text{ days} \][/tex]
Jupiter has the longer period.
Difference in Periods:
Calculate the difference in periods:
[tex]\[ \Delta T = T_{\text{Jupiter}} - T_{\text{Venus}} \][/tex]
[tex]\[ \Delta T = 4344.281121773774 - 225.09761153775045 \][/tex]
[tex]\[ \Delta T = 4119.183510236024 \text{ days} \][/tex]
Conclusion:
Jupiter has the longer period, and it is longer by 4119.183510236024 Earth days compared to Venus.
