Answer :
Let's solve each part of the problem step by step.
Expression: [tex](16)^{0.16} \times (16)^{0.04} \times (2)^{0.2}[/tex]
We can simplify this expression using the laws of exponents:
The law of exponents states that [tex]a^m \times a^n = a^{m+n}[/tex]. So, we can combine the first two terms:
[tex](16)^{0.16} \times (16)^{0.04} = (16)^{0.16 + 0.04} = (16)^{0.20}[/tex]
Since $16 = 2^4[tex], we can rewrite[/tex](16)^{0.20}$ as:
[tex](2^4)^{0.20} = 2^{4 \times 0.20} = 2^{0.80}[/tex]
Now we multiply by [tex](2)^{0.2}[/tex]:
[tex]2^{0.80} \times 2^{0.2} = 2^{0.80 + 0.2} = 2^{1.0} = 2[/tex]
Thus, the expression evaluates to 2.
Answer for Part 1: The expression is equal to (b) 2.
Equation: [tex]\left(\frac{4}{11}\right)^{x-1} = \left(\frac{11}{4}\right)^{x-5}[/tex]
Let's solve for [tex]x[/tex].
We can rewrite the right-hand side of the equation as:
[tex]\left(\frac{4}{11}\right)^{-(x-5)} = \left(\frac{4}{11}\right)^{-x+5}[/tex]
Now the equation becomes:
[tex]\left(\frac{4}{11}\right)^{x-1} = \left(\frac{4}{11}\right)^{-x+5}[/tex]
Since the bases are the same, we equate the exponents:
[tex]x - 1 = -x + 5[/tex]
Solving for [tex]x[/tex], we add [tex]x[/tex] to both sides:
[tex]x + x - 1 = 5[/tex]
[tex]2x - 1 = 5[/tex]
Now, add 1 to both sides:
[tex]2x = 6[/tex]
Finally, divide both sides by 2:
[tex]x = 3[/tex]
Answer for Part 2: [tex]x[/tex] is equal to (a) 3.
Area of the Rectangle:
The length of the rectangle is $3^3[tex]metres, and the width is $9^{-2}[/tex] metres.
Calculate the dimensions:
- Length: $3^3 = 27$ metres.
- Width: $9 = 3^2[tex], so $9^{-2} = (3^2)^{-2} = 3^{-4}[/tex].
Now calculate the area using the formula for the area of a rectangle:
[tex]\text{Area} = \text{Length} \times \text{Width} = 27 \times 3^{-4}[/tex]
Convert $3^{-4}$ to a positive exponent:
[tex]3^{-4} = \frac{1}{3^4} = \frac{1}{81}[/tex]
Thus, the area is:
[tex]27 \times \frac{1}{81} = \frac{27}{81} = \frac{1}{3}[/tex]
Answer for Part 3: The area of the rectangle is (c) [tex]\frac{1}{3}[/tex] metres.
