Question

Directions:Simplify the following monomials.SHOW ALL STEPS!

Answer 1

Step-by-step explanation:

7. Ans;

\frac{6 {a}^{5} {b}^{7} }{ - 2 {a}^{3} {b}^{7} } = \frac{2 {a}^{3} {b}^{7}(3 {a}^{2} ) }{ - 2 {a}^{3} {b}^{7} } = - 3 {a}^{2}

−2a

3

b

7

6a

5

b

7

=

−2a

3

b

7

2a

3

b

7

(3a

2

)

=−3a

2

___o____o___

8. Ans;

\frac{ - 20 {x}^{3} {y}^{2} }{ - 5 {x}^{3}y } = \frac{ - 5 {x}^{3}y(4y) }{ - 5 {x}^{3}y } = 4y

−5x

3

y

−20x

3

y

2

=

−5x

3

y

−5x

3

y(4y)

=4y

___o___o___

9. Ans;

\frac{ - 16 a {b}^{4} }{4 {b}^{3} } = \frac{4 {b}^{3}( - 4ab) }{4 {b}^{3} } = - 4ab

4b

3

−16ab

4

=

4b

3

4b

3

(−4ab)

=−4ab

____o____o____

10. Ans;

\frac{21 {m}^{8} {n}^{5} }{27 {m}^{5} {n}^{4} } = \frac{3 {m}^{5} {n}^{4}(7 {m}^{3} n) }{3 {m}^{5} {n}^{4}(9) } = \frac{7 {m}^{3}n }{9}

27m

5

n

4

21m

8

n

5

=

3m

5

n

4

(9)

3m

5

n

4

(7m

3

n)

=

9

7m

3

n

___o___o___

11. Ans;

\frac{ - 15 {x}^{5} {y}^{4} }{45x {y}^{3} } = \frac{ 15x {y}^{3} ( - {x}^{4} y)}{15x {y}^{3}(3) } = - \frac{ {x}^{4}y }{3}

45xy

3

−15x

5

y

4

=

15xy

3

(3)

15xy

3

(−x

4

y)

=−

3

x

4

y

___o___o___

12.Ans;

\frac{7 {p}^{2} {q}^{2} }{14 {p}^{2} {q}^{2} } = \frac{7 {p}^{2} {q}^{2} }{7 {p}^{2} {q}^{2} (2) } = \frac{1}{2} = 0.5

14p

2

q

2

7p

2

q

2

=

7p

2

q

2

(2)

7p

2

q

2

=

2

1

=0.5

Answer 2

Answer:

7. Ans;

$\frac{6 {a}^{5} {b}^{7} }{ - 2 {a}^{3} {b}^{7} } = \frac{2 {a}^{3} {b}^{7}(3 {a}^{2} ) }{ - 2 {a}^{3} {b}^{7} } = - 3 {a}^{2}$

___o____o___

8. Ans;

$\frac{ - 20 {x}^{3} {y}^{2} }{ - 5 {x}^{3}y } = \frac{ - 5 {x}^{3}y(4y) }{ - 5 {x}^{3}y } = 4y$

___o___o___

9. Ans;

$\frac{ - 16 a {b}^{4} }{4 {b}^{3} } = \frac{4 {b}^{3}( - 4ab) }{4 {b}^{3} } = - 4ab$

____o____o____

10. Ans;

$\frac{21 {m}^{8} {n}^{5} }{27 {m}^{5} {n}^{4} } = \frac{3 {m}^{5} {n}^{4}(7 {m}^{3} n) }{3 {m}^{5} {n}^{4}(9) } = \frac{7 {m}^{3}n }{9}$

___o___o___

11. Ans;

$\frac{ - 15 {x}^{5} {y}^{4} }{45x {y}^{3} } = \frac{ 15x {y}^{3} ( - {x}^{4} y)}{15x {y}^{3}(3) } = - \frac{ {x}^{4}y }{3}$

___o___o___

12.Ans;

$\frac{7 {p}^{2} {q}^{2} }{14 {p}^{2} {q}^{2} } = \frac{7 {p}^{2} {q}^{2} }{7 {p}^{2} {q}^{2} (2) } = \frac{1}{2} = 0.5$

I hope I helped you^_^