Question

Simplify. Assume no variable is 0.

#16-23 please

85 points!!!!! This is the jumbo pack!!!!

Answer 1

Step-by-step explanation:

Q16

$( {5x}^{3} {y}^{ - 5} )( {4xy}^{3} )$

First, we multiply the variable x together:

${x}^{3} \times x = {x}^{4}$

and remove brackets:

${5y}^{ - 5} \times {4x}^{4} \times {y}^{3}$

Now, we multiply the variable y:

${y}^{ - 5} \times {y}^{3} = {y}^{ - 2}$

Hence:

$5 \times 4 {x}^{4} \times {y}^{ - 2}$

Use the rule

${x}^{ - y} = \frac{1}{ {x}^{y} }$

Thus,

$5 \times 4 {x}^{4} \times \frac{1}{ {y}^{2} }$

Multiply:

$\frac{20 {x}^{4} }{ {y}^{2} }$

Q17

$( - 2 {b}^{3} c)(4 {b}^{2} {c}^{2} )$

$- 2 {b}^{3} c\times 4 {b}^{2} {c}^{2}$

Multiply the variable b together:

$- 2c \times 4 {b}^{5} {c}^{2}$

Multiply the variable c together:

$- 2 \times 4 {b}^{5} {c}^{3}$

Multiply.

$- 8 {b}^{5} {c}^{3}$

Q18:

$\frac{ {a}^{3} {n}^{7} }{a {n}^{4} }$

Divide the numerator and denominator by a:

$\frac{ {a}^{2} {n}^{7} }{ {n}^{4} }$

Apply the rule that

$\frac{ {x}^{y} }{ {x}^{z} } = {x}^{y - z}$

Hence, the equation will be

${a}^{2} {n}^{7 - 4}$

which is finally

${a}^{2} {n}^{3}$

Q19:

$\frac{ - {y}^{3} {z}^{5} }{ {y}^{2} {z}^{3} }$

Do the same rule said above:

$- \frac{ {z}^{5} {y}^{3 - 2} }{ {z}^{3} }$

$- \frac{y {z}^{5} }{ {z}^{3} }$

Do the rule again:

$- y {z}^{5 - 3}$

$- y {z}^{2}$

Q20:

$\frac{ - 7 {x}^{5} {y}^{5} {z}^{4} }{21 {x}^{7} {y}^{5} {z}^{2} }$

Cancel the common factor of 7:

$- \frac{ {x}^{5} {y}^{5} {z}^{4} }{3 {x}^{7} {y}^{5} {z}^{2} }$

$- \frac{ {y}^{5} {z}^{4} }{3 {y}^{5} {z}^{2} {x}^{7 - 5} }$

Thus,

$- \frac{ {y}^{5} {z}^{4} }{3 {y}^{5} {z}^{2} {x}^{2} }$

Cancel common factor of y^5

$- \frac{ {z}^{4} }{3 {z}^{2} {x}^{2} }$

$- \frac{ {z}^{4 - 2} }{3 {x}^{2} }$

$- \frac{ {z}^{2} }{3 {x}^{2} }$

Q21:

$\frac{9 {a}^{7} {b}^{5} {c}^{5} }{18 {a}^{5} {b}^{9} {c}^{3} }$

Cancel the common factor which is 9:

$\frac{{a}^{7} {b}^{5} {c}^{5} }{2 {a}^{5} {b}^{9} {c}^{3} }$

"Move" the variable a up using the rule:

$\frac{{a}^{7 - 5} {b}^{5} {c}^{5} }{2 {b}^{9} {c}^{3} }$

$\frac{{a}^{2} {b}^{5} {c}^{5} }{2 {b}^{9} {c}^{3} }$

"Move" the variable b down using the rule:

$\frac{{a}^{2} {c}^{5} }{2 {c}^{3} {b}^{9 - 5} }$

$\frac{{a}^{2} {c}^{5} }{2 {c}^{3} {b}^{4} }$

"Move" the variable c up using the rule

$\frac{{a}^{2} {c}^{5 - 3} }{2 {b}^{4} }$

$\frac{{a}^{2} {c}^{2} }{2 {b}^{4} }$

Q22:

$( {n}^{5} {)}^{4}$

Use the rule:

$({x}^{y} {)}^{z} = {x}^{yz}$

Hence,

${n}^{5 \times 4}$

${n}^{20}$

Q23:

$( {z}^{3} {)}^{6}$

Using the same rule above,

${z}^{3 \times 6}$

Therefore,

${z}^{18}$

Answer 2

Answer:

Q16

First, we multiply the variable x together:

and remove brackets:

Now, we multiply the variable y:

Hence:

Use the rule

Thus,

Multiply:

Q17

Multiply the variable b together:

Multiply the variable c together:

Multiply.

Q18:

Divide the numerator and denominator by a:

Apply the rule that

Hence, the equation will be

which is finally

Q19:

Do the same rule said above:

Do the rule again:

Q20:

Cancel the common factor of 7:

Thus,

Cancel common factor of y^5

Q21:

Cancel the common factor which is 9:

"Move" the variable a up using the rule:

"Move" the variable b down using the rule:

"Move" the variable c up using the rule

Q22:

Use the rule:

Hence,

Q23:

Using the same rule above,

Therefore,

 you get your answer.