At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Join our platform to get reliable answers to your questions from a knowledgeable community of experts. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
Sure, let's work through the problems step-by-step.
### Part a
Given:
[tex]\[ x = 7^{a - b} \][/tex]
[tex]\[ y = 7^{b - a} \][/tex]
To show that [tex]\(x \cdot y = 1\)[/tex]:
[tex]\[ x \cdot y = 7^{a-b} \cdot 7^{b-a} \][/tex]
Using properties of exponents:
[tex]\[ 7^{a-b} \cdot 7^{b-a} = 7^{(a-b) + (b-a)} \][/tex]
Simplifying the exponent:
[tex]\[ 7^{(a-b) + (b-a)} = 7^0 \][/tex]
Since any non-zero number raised to the power of 0 is 1:
[tex]\[ 7^0 = 1 \][/tex]
Thus:
[tex]\[ x \cdot y = 1 \][/tex]
### Part b
Given:
[tex]\[ x + y = x \][/tex]
We need to show that [tex]\( x^y = 1 \)[/tex].
Since [tex]\( x + y = x \)[/tex], subtract [tex]\( x \)[/tex] from both sides:
[tex]\[ y = 0 \][/tex]
Any non-zero number raised to the power of 0 is 1:
[tex]\[ x^y = x^0 = 1 \][/tex]
Thus:
[tex]\[ x^y = 1 \][/tex]
### Part c
Given:
[tex]\[ a = 2 \][/tex]
[tex]\[ b = 1 \][/tex]
We need to show that:
[tex]\[ (a+b)^2 = a^2 + 2ab + b^2 \][/tex]
First, calculate [tex]\( (a+b)^2 \)[/tex]:
[tex]\[ (2 + 1)^2 = 3^2 = 9 \][/tex]
Now calculate the right-hand side:
[tex]\[ a^2 + 2ab + b^2 \][/tex]
[tex]\[ 2^2 + 2(2)(1) + 1^2 \][/tex]
[tex]\[ 4 + 4 + 1 = 9 \][/tex]
Both sides are equal:
[tex]\[ 9 = 9 \][/tex]
Thus:
[tex]\[ (a+b)^2 = a^2 + 2ab + b^2 \][/tex]
### Part d
Given:
[tex]\[ a = 5 \][/tex]
[tex]\[ b = 3 \][/tex]
We need to show that:
[tex]\[ (a-b)^2 = a^2 - 2ab + b^2 \][/tex]
First, calculate [tex]\( (a-b)^2 \)[/tex]:
[tex]\[ (5 - 3)^2 = 2^2 = 4 \][/tex]
Now calculate the right-hand side:
[tex]\[ a^2 - 2ab + b^2 \][/tex]
[tex]\[ 5^2 - 2(5)(3) + 3^2 \][/tex]
[tex]\[ 25 - 30 + 9 = 4 \][/tex]
Both sides are equal:
[tex]\[ 4 = 4 \][/tex]
Thus:
[tex]\[ (a-b)^2 = a^2 - 2ab + b^2 \][/tex]
### Part e
Given:
[tex]\[ p + q + r = 0 \][/tex]
We need to show that the expression [tex]\[ x^{p+q} \times x^{p-q} \times x^{q+r} \times x^{q-r} \times x^{r+p} \][/tex]
simplifies to 1.
Using properties of exponents:
[tex]\[ x^{p+q} \times x^{p-q} \times x^{q+r} \times x^{q-r} \times x^{r+p} = x^{(p+q) + (p-q) + (q+r) + (q-r) + (r+p)} \][/tex]
Simplifying the exponent:
[tex]\[ (p+q) + (p-q) + (q+r) + (q-r) + (r+p) \][/tex]
Combining like terms:
[tex]\[ p + q + p - q + q + r + q - r + r + p \][/tex]
Grouping all similar terms together:
[tex]\[ (p + p + p) + (q + q - q + q) + (r + r - r) \][/tex]
[tex]\[ 3p + 2q - q + q + 2r - r \][/tex]
[tex]\[ 3p + q + r \][/tex]
Since [tex]\( p + q + r = 0 \)[/tex]:
[tex]\[ 3p + q + r = 3(p + q + r) \][/tex]
[tex]\[ 3 \times 0 = 0 \][/tex]
Thus:
[tex]\[ x^{0} = 1 \][/tex]
Therefore, we have shown that:
[tex]\[ x^{p+q} \times x^{p-q} \times x^{q+r} \times x^{q-r} \times x^{r+p} = 1 \][/tex]
This covers all parts of the given problem statement.
### Part a
Given:
[tex]\[ x = 7^{a - b} \][/tex]
[tex]\[ y = 7^{b - a} \][/tex]
To show that [tex]\(x \cdot y = 1\)[/tex]:
[tex]\[ x \cdot y = 7^{a-b} \cdot 7^{b-a} \][/tex]
Using properties of exponents:
[tex]\[ 7^{a-b} \cdot 7^{b-a} = 7^{(a-b) + (b-a)} \][/tex]
Simplifying the exponent:
[tex]\[ 7^{(a-b) + (b-a)} = 7^0 \][/tex]
Since any non-zero number raised to the power of 0 is 1:
[tex]\[ 7^0 = 1 \][/tex]
Thus:
[tex]\[ x \cdot y = 1 \][/tex]
### Part b
Given:
[tex]\[ x + y = x \][/tex]
We need to show that [tex]\( x^y = 1 \)[/tex].
Since [tex]\( x + y = x \)[/tex], subtract [tex]\( x \)[/tex] from both sides:
[tex]\[ y = 0 \][/tex]
Any non-zero number raised to the power of 0 is 1:
[tex]\[ x^y = x^0 = 1 \][/tex]
Thus:
[tex]\[ x^y = 1 \][/tex]
### Part c
Given:
[tex]\[ a = 2 \][/tex]
[tex]\[ b = 1 \][/tex]
We need to show that:
[tex]\[ (a+b)^2 = a^2 + 2ab + b^2 \][/tex]
First, calculate [tex]\( (a+b)^2 \)[/tex]:
[tex]\[ (2 + 1)^2 = 3^2 = 9 \][/tex]
Now calculate the right-hand side:
[tex]\[ a^2 + 2ab + b^2 \][/tex]
[tex]\[ 2^2 + 2(2)(1) + 1^2 \][/tex]
[tex]\[ 4 + 4 + 1 = 9 \][/tex]
Both sides are equal:
[tex]\[ 9 = 9 \][/tex]
Thus:
[tex]\[ (a+b)^2 = a^2 + 2ab + b^2 \][/tex]
### Part d
Given:
[tex]\[ a = 5 \][/tex]
[tex]\[ b = 3 \][/tex]
We need to show that:
[tex]\[ (a-b)^2 = a^2 - 2ab + b^2 \][/tex]
First, calculate [tex]\( (a-b)^2 \)[/tex]:
[tex]\[ (5 - 3)^2 = 2^2 = 4 \][/tex]
Now calculate the right-hand side:
[tex]\[ a^2 - 2ab + b^2 \][/tex]
[tex]\[ 5^2 - 2(5)(3) + 3^2 \][/tex]
[tex]\[ 25 - 30 + 9 = 4 \][/tex]
Both sides are equal:
[tex]\[ 4 = 4 \][/tex]
Thus:
[tex]\[ (a-b)^2 = a^2 - 2ab + b^2 \][/tex]
### Part e
Given:
[tex]\[ p + q + r = 0 \][/tex]
We need to show that the expression [tex]\[ x^{p+q} \times x^{p-q} \times x^{q+r} \times x^{q-r} \times x^{r+p} \][/tex]
simplifies to 1.
Using properties of exponents:
[tex]\[ x^{p+q} \times x^{p-q} \times x^{q+r} \times x^{q-r} \times x^{r+p} = x^{(p+q) + (p-q) + (q+r) + (q-r) + (r+p)} \][/tex]
Simplifying the exponent:
[tex]\[ (p+q) + (p-q) + (q+r) + (q-r) + (r+p) \][/tex]
Combining like terms:
[tex]\[ p + q + p - q + q + r + q - r + r + p \][/tex]
Grouping all similar terms together:
[tex]\[ (p + p + p) + (q + q - q + q) + (r + r - r) \][/tex]
[tex]\[ 3p + 2q - q + q + 2r - r \][/tex]
[tex]\[ 3p + q + r \][/tex]
Since [tex]\( p + q + r = 0 \)[/tex]:
[tex]\[ 3p + q + r = 3(p + q + r) \][/tex]
[tex]\[ 3 \times 0 = 0 \][/tex]
Thus:
[tex]\[ x^{0} = 1 \][/tex]
Therefore, we have shown that:
[tex]\[ x^{p+q} \times x^{p-q} \times x^{q+r} \times x^{q-r} \times x^{r+p} = 1 \][/tex]
This covers all parts of the given problem statement.
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
