Question

question 9: josue discovered a pattern when working with exponents:
$(4^3)(4^6)=4^9$
$(2^6)(2^7)=2^{13}$
$(7^3)(7^6)(7^8)=7^{17}$
based on this pattern, what should josue select as an answer for $(q)(q^6)(q^x)$
a) $q^{7 + x}$
b) $q^{6 + x}$
c) $q^{6x}$
d) $q^{7x + 1}$
question 10: if $wy\
eq0$, which is equivalent to $\frac{-8w^3y^2}{2wy^2}$?
a) $-4w^2y$
b) $-4w^2$
c) $-6w^2y$
d) $-6w^4y^4$
question 11: which expression is equivalent to $(-2a^{-2}b^3)(5a^5b^4)$?
a) $-3a^3b^7$
b) $\frac{-3b^{12}}{a^{10}}$
c) $-10a^3b^7$
d) $\frac{-10b^{12}}{a^{10}}$
question 12: if $km\
eq0$, which expression is equivalent to $\frac{kp^4m^5}{(k^2m^3)^4}$?
a) $k^5p^4m^2$
b) $k^9p^4m^{17}$
c) $\frac{p^4}{k^7m^7}$
d) $\frac{p^4}{k^5m^2}$
notes:

Explanation:

Question 9

Step1: Identify exponent addition rule

When multiplying same bases, add exponents: $a^m \cdot a^n = a^{m+n}$

Step2: Rewrite $q$ as $q^1$

$q = q^1$

Step3: Add all exponents

$q^1 \cdot q^6 \cdot q^x = q^{1+6+x} = q^{7+x}$

Question 10

Step1: Split coefficients and variables

Separate constants, $w$-terms, $y$-terms: $\frac{-8}{2} \cdot \frac{w^3}{w} \cdot \frac{y^2}{y^2}$

Step2: Simplify each part

$\frac{-8}{2}=-4$; $\frac{w^3}{w}=w^{3-1}=w^2$; $\frac{y^2}{y^2}=1$

Step3: Multiply simplified parts

$-4 \cdot w^2 \cdot 1 = -4w^2$

Question 11

Step1: Multiply coefficients first

$-2 \times 5 = -10$

Step2: Combine $a$-terms (add exponents)

$a^{-2} \cdot a^5 = a^{-2+5}=a^3$

Step3: Combine $b$-terms (add exponents)

$b^3 \cdot b^4 = b^{3+4}=b^7$

Step4: Multiply all results

$-10 \cdot a^3 \cdot b^7 = -10a^3b^7$

Question 12

Step1: Expand denominator exponents

$(k^2m^3)^4 = k^{2 \times 4}m^{3 \times 4}=k^8m^{12}$

Step2: Rewrite fraction with expanded denominator

$\frac{kp^4m^5}{k^8m^{12}}$

Step3: Simplify $k$-terms (subtract exponents)

$\frac{k}{k^8}=k^{1-8}=k^{-7}=\frac{1}{k^7}$

Step4: Simplify $m$-terms (subtract exponents)

$\frac{m^5}{m^{12}}=m^{5-12}=m^{-7}=\frac{1}{m^7}$

Step5: Combine all simplified parts

$p^4 \cdot \frac{1}{k^7} \cdot \frac{1}{m^7} = \frac{p^4}{k^7m^7}$

Answer:

Question 9: a) $q^{7+x}$
Question 10: b) $-4w^2$
Question 11: c) $-10a^3b^7$
Question 12: c) $\frac{p^4}{k^7m^7}$