Question

1. enter an expression equivalent to \\(\frac{d^8}{d^5}\\) in the form \\(d^m\\)
2. a student reasons that the product of \\(5^3\\) and \\(5^{-4}\\) is equal to 5 raised to the power of the product of the exponents: \\(5^3(5^{-4}) = 5^{-12}\\). enter an expression equivalent to \\(5^3(5^{-4})\\) to show that the student’s statement is false.
3. enter an expression equivalent to \\((-3xy)(3x^3y^2)\\) in the form \\(ax^my^n\\)
4. solve this inequality for y \\(-3y + 12 \geq 6\\)

Explanation:

Question 1

Step1: Recall exponent rule for division

When dividing exponents with the same base, we subtract the exponents: $\frac{a^m}{a^n}=a^{m - n}$. Here, the base is $d$, $m = 8$, $n = 3$.
$\frac{d^8}{d^3}=d^{8 - 3}$

Step2: Calculate the exponent

$8-3 = 5$, so $\frac{d^8}{d^3}=d^5$

Step1: Recall exponent rule for multiplication

When multiplying exponents with the same base, we add the exponents: $a^m\times a^n=a^{m + n}$. Here, the base is $5$, $m = 3$, $n=-4$.
$5^3\times5^{-4}=5^{3+(-4)}$

Step2: Calculate the exponent

$3+(-4)=3 - 4=-1$, so $5^3\times5^{-4}=5^{-1}$ (or $\frac{1}{5}$)

Step1: Multiply the coefficients and the variables separately

First, multiply the coefficients: $-3\times3=-9$. Then, for the $x$ terms: $x\times x^3=x^{1 + 3}$ (using $a^m\times a^n=a^{m + n}$), and for the $y$ terms: $y\times y^2=y^{1+2}$.

Step2: Calculate the exponents for variables

For $x$: $1 + 3 = 4$, so $x\times x^3=x^4$. For $y$: $1+2 = 3$, so $y\times y^2=y^3$. Then combine them with the coefficient: $-9x^4y^3$

Answer:

$d^5$

Question 2