zero and negative practice b
simplify the following expressions.
any base to the zero power equals one.
1) $(7y)^0$
2) $-6(x^3y^5)^0$
3) $-8^0x^3z^9$
4) $-(-r^8y^{-1})^0$

write the following so that all exponents are positive.
$x^{-a}=1/x^a$ and $1/x^{-a}=x^a$
5) $5^{-2}x^4y^9$
6) $\frac{3y^4}{12x^{-7}}$
7) $\frac{2(a^7b^{-4})^2}{c^1}$
8) $3a^{-4}b^3$

simplify using the rules of exponents.
simplified solutions will not have negative exponents.
9) $\frac{a^{16}b^{13}}{a^6b^{13}}$
10) $\frac{3x^{-10}y^{12}}{8y^{10}}$
11) $\frac{-(d^4b^3)^2}{d^9b^6}$
12) $25(g^{-11}h^2)^{-3} \cdot g^4h^{-6}$

Question

zero and negative practice b
simplify the following expressions.
any base to the zero power equals one.
1) $(7y)^0$
2) $-6(x^3y^5)^0$
3) $-8^0x^3z^9$
4) $-(-r^8y^{-1})^0$

write the following so that all exponents are positive.
$x^{-a}=1/x^a$ and $1/x^{-a}=x^a$
5) $5^{-2}x^4y^9$
6) $\frac{3y^4}{12x^{-7}}$
7) $\frac{2(a^7b^{-4})^2}{c^1}$
8) $3a^{-4}b^3$

simplify using the rules of exponents.
simplified solutions will not have negative exponents.
9) $\frac{a^{16}b^{13}}{a^6b^{13}}$
10) $\frac{3x^{-10}y^{12}}{8y^{10}}$
11) $\frac{-(d^4b^3)^2}{d^9b^6}$
12) $25(g^{-11}h^2)^{-3} \cdot g^4h^{-6}$

Accepted Answer

### Problem 1: $(7y)^0$
# Explanation:
## Step1: Apply zero - exponent rule
The zero - exponent rule states that for any non - zero base $a$, $a^0 = 1$. Here, the base is $7y$ (and we assume $y
eq0$ so that $7y
eq0$). So, $(7y)^0=1$
# Answer:
$1$

### Problem 2: $- 6(x^{3}y^{5})^{0}$
# Explanation:
## Step1: Apply zero - exponent rule to $(x^{3}y^{5})^{0}$
Using the zero - exponent rule $a^0 = 1$ (where $a=x^{3}y^{5}$ and $x^{3}y^{5}
eq0$ when $x
eq0$ or $y
eq0$), we have $(x^{3}y^{5})^{0}=1$
## Step2: Multiply by $-6$
$-6\times1=-6$
# Answer:
$-6$

### Problem 3: $-8^{0}x^{3}z^{9}$
# Explanation:
## Step1: Apply zero - exponent rule to $8^{0}$
By the zero - exponent rule, $8^{0}=1$ (since $8
eq0$)
## Step2: Multiply with the other terms
$-1\times x^{3}z^{9}=-x^{3}z^{9}$
# Answer:
$-x^{3}z^{9}$

### Problem 4: $-(-r^{8}y^{-1})^{0}$
# Explanation:
## Step1: Apply zero - exponent rule to $(-r^{8}y^{-1})^{0}$
Using the zero - exponent rule $a^0 = 1$ (where $a =-r^{8}y^{-1}$ and $-r^{8}y^{-1}
eq0$ when $r
eq0$ or $y
eq0$), we get $(-r^{8}y^{-1})^{0}=1$
## Step2: Multiply by $- 1$
$-1\times1=-1$
# Answer:
$-1$

### Problem 5: $5^{-2}x^{4}y^{9}$
# Explanation:
## Step1: Apply the negative - exponent rule
The negative - exponent rule states that $a^{-n}=\frac{1}{a^{n}}$ for $a
eq0$. So, $5^{-2}=\frac{1}{5^{2}}$
## Step2: Simplify $5^{2}$
Since $5^{2}=25$, we have $5^{-2}=\frac{1}{25}$
## Step3: Combine with the other terms
$5^{-2}x^{4}y^{9}=\frac{x^{4}y^{9}}{25}$
# Answer:
$\frac{x^{4}y^{9}}{25}$

### Problem 6: $\frac{3y^{4}}{12x^{-7}}$
# Explanation:
## Step1: Simplify the coefficient
Simplify $\frac{3}{12}=\frac{1}{4}$
## Step2: Apply the negative - exponent rule to $x^{-7}$
Using the rule $a^{-n}=\frac{1}{a^{n}}$ (or $\frac{1}{a^{-n}}=a^{n}$), we have $\frac{1}{x^{-7}}=x^{7}$
## Step3: Combine the terms
$\frac{3y^{4}}{12x^{-7}}=\frac{y^{4}x^{7}}{4}$
# Answer:
$\frac{x^{7}y^{4}}{4}$

### Problem 7: $\frac{2(a^{7}b^{-4})^{2}}{c^{1}}$
# Explanation:
## Step1: Apply the power - of - a - power rule $(a^{m})^{n}=a^{mn}$ to $(a^{7}b^{-4})^{2}$
For $(a^{7})^{2}$, using $(a^{m})^{n}=a^{mn}$, we get $a^{7\times2}=a^{14}$. For $(b^{-4})^{2}$, we get $b^{-4\times2}=b^{-8}$
## Step2: Apply the negative - exponent rule to $b^{-8}$
Using $a^{-n}=\frac{1}{a^{n}}$, $b^{-8}=\frac{1}{b^{8}}$, so $(a^{7}b^{-4})^{2}=a^{14}b^{-8}=\frac{a^{14}}{b^{8}}$
## Step3: Multiply by $2$ and divide by $c$
$\frac{2(a^{7}b^{-4})^{2}}{c}=\frac{2\times\frac{a^{14}}{b^{8}}}{c}=\frac{2a^{14}}{b^{8}c}$
# Answer:
$\frac{2a^{14}}{b^{8}c}$

### Problem 8: $3a^{-4}b^{3}$
# Explanation:
## Step1: Apply the negative - exponent rule to $a^{-4}$
Using $a^{-n}=\frac{1}{a^{n}}$, we have $a^{-4}=\frac{1}{a^{4}}$
## Step2: Combine with the other terms
$3a^{-4}b^{3}=\frac{3b^{3}}{a^{4}}$
# Answer:
$\frac{3b^{3}}{a^{4}}$

### Problem 9: $\frac{a^{16}b^{13}}{a^{6}b^{13}}$
# Explanation:
## Step1: Apply the quotient rule for exponents $\frac{a^{m}}{a^{n}}=a^{m - n}$ to the $a$ - terms
For the $a$ - terms, $\frac{a^{16}}{a^{6}}=a^{16 - 6}$
## Step2: Simplify the exponent of $a$
$16-6 = 10$, so $a^{16 - 6}=a^{10}$
## Step3: Apply the quotient rule for exponents to the $b$ - terms
For the $b$ - terms, $\frac{b^{13}}{b^{13}}=b^{13 - 13}$
## Step4: Simplify the exponent of $b$
$13-13 = 0$, and $b^{0}=1$ (for $b
eq0$)
## Step5: Combine the results
$\frac{a^{16}b^{13}}{a^{6}b^{13}}=a^{10}\times1=a^{10}$
# Answer:
$a^{10}$

### Problem 10: $\frac{3x^{-10}y^{12}}{8y^{10}}$
# Explanation:
## Step1: Apply the quotient rule for exponents to the $y$ - terms
Using $\frac{a^{m}}{a^{n}}=a^{m - n}$, for the $y$ - terms, $\frac{y^{12}}{y^{10}}=y^{12 - 10}$
## Step2: Simplify the exponent of $y$
$12 - 10=2$, so $y^{12 - 10}=y^{2}$
## Step3: Apply the negative - exponent rule to $x^{-10}$
Using $a^{-n}=\frac{1}{a^{n}}$, $x^{-10}=\frac{1}{x^{10}}$
## Step4: Combine the terms
$\frac{3x^{-10}y^{12}}{8y^{10}}=\frac{3y^{2}}{8x^{10}}$
# Answer:
$\frac{3y^{2}}{8x^{10}}$

### Problem 11: $\frac{-(d^{4}b^{3})^{2}}{d^{9}b^{6}}$
# Explanation:
## Step1: Apply the power - of - a - power rule to $(d^{4}b^{3})^{2}$
Using $(a^{m})^{n}=a^{mn}$, we have $(d^{4})^{2}=d^{8}$ and $(b^{3})^{2}=b^{6}$, so $(d^{4}b^{3})^{2}=d^{8}b^{6}$
## Step2: Substitute back into the fraction
The fraction becomes $\frac{-d^{8}b^{6}}{d^{9}b^{6}}$
## Step3: Apply the quotient rule for exponents to the $d$ - terms and $b$ - terms
For the $d$ - terms, $\frac{d^{8}}{d^{9}}=d^{8 - 9}=d^{-1}$, for the $b$ - terms, $\frac{b^{6}}{b^{6}}=b^{6 - 6}=b^{0}=1$ (for $b
eq0$)
## Step4: Apply the negative - exponent rule to $d^{-1}$
$d^{-1}=\frac{1}{d}$, so $\frac{-d^{8}b^{6}}{d^{9}b^{6}}=-\frac{1}{d}$
# Answer:
$-\frac{1}{d}$

### Problem 12: $25(g^{-11}h^{2})^{-3}\cdot g^{4}h^{-6}$
# Explanation:
## Step1: Apply the power - of - a - power rule to $(g^{-11}h^{2})^{-3}$
Using $(a^{m})^{n}=a^{mn}$, we have $(g^{-11})^{-3}=g^{(-11)\times(-3)}=g^{33}$ and $(h^{2})^{-3}=h^{2\times(-3)}=h^{-6}$
## Step2: Substitute back into the expression
The expression becomes $25g^{33}h^{-6}\cdot g^{4}h^{-6}$
## Step3: Apply the product rule for exponents to the $g$ - terms and $h$ - terms
For the $g$ - terms, $g^{33}\cdot g^{4}=g^{33 + 4}=g^{37}$, for the $h$ - terms, $h^{-6}\cdot h^{-6}=h^{-6+( - 6)}=h^{-12}$
## Step4: Apply the negative - exponent rule to $h^{-12}$
$h^{-12}=\frac{1}{h^{12}}$, so $25g^{33}h^{-6}\cdot g^{4}h^{-6}=25g^{37}\times\frac{1}{h^{12}}=\frac{25g^{37}}{h^{12}}$
# Answer:
$\frac{25g^{37}}{h^{12}}$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 1: \((7y)^0\)

Step1: Apply zero - exponent rule

Step1: Apply zero - exponent rule to \((x^{3}y^{5})^{0}\)

Step2: Multiply by \(-6\)

Step1: Apply zero - exponent rule to \(8^{0}\)

Step2: Multiply with the other terms

Answer:

Problem 2: \(- 6(x^{3}y^{5})^{0}\)