Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key Exponential Notation

Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 1 Lesson 1.2 The Product and the Quotient of Powers to finish your assignments.

Math in Focus Grade 7 Course 3 A Chapter 1 Lesson 1.2 Answer Key Exponential Notation

Math in Focus Grade 8 Chapter 1 Lesson 1.2 Guided Practice Answer Key

Simplify each expression. Write your answer in exponential notation.

Question 1.
64 • 63 =
64 • 63 = Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 1 Use the Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 1 of powers property.
= Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 1 Simplify.
Answer:
67,

Explanation:
Given 64 • 63 using the Product of Powers Property states that when
multiplying two exponents with the same base, we add the exponents and
keep the base, So  64 • 63
= 64 + 3 = 67.

Question 2.
(-5) • (-5)5
(-5) • (-5)5 = Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 1 Use the Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 1 of powers property.
= Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 1 Simplify the exponent.
= Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 1 Simplify.
Answer:
(-5)6,

Explanation:
Given (-5) • (-5)5 using the Product of Powers Property states that when
multiplying two exponents with the same base, we add the exponents and
keep the base, So  (-5) • (-5)5
=(-5) 1 + 5 = (-5)6.

Question 3.
Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 2
Answer:
Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key Exponential Notation-1

Explanation:
Given \(\left(\frac{1}{5}\right)^{3}\) • \(\left(\frac{1}{5}\right)^{4}\)
using the Product of Powers Property states that when
multiplying two exponents with the same base, we add the exponents and
keep the base \(\left(\frac{1}{5}\right)^{3}\) • \(\left(\frac{1}{5}\right)^{4}\) =
\(\left(\frac{1}{5}\right)^{3 +4}\) = \(\left(\frac{1}{5}\right)^{7}\).

Simplify each expression. Write your answer in exponential notation.

Question 4.
p3 • p6
p3 • p6 = Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 1 Use the Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 1 of powers property.
= Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 1 Simplify.
Answer:
p3 • p6 = p3 + 6 = p9,

Explanation:
Given \({p}^{3}\) • \({p}^{6}\)
using the Product of Powers Property states that when
multiplying two exponents with the same base, we add the exponents and
keep the base \({p}^{3}\) • \({p}^{6}\) =
\({p}^{3 + 6}\) = \({p}^{9}\).

Question 5.
(-c)4 • (-c)
Answer:
(-c)4 • (-c)2 = (-c)4+2 = (-c)6,

Explanation:
Given \({-c}^{4}\) • \({-c}^{2}\)
using the Product of Powers Property states that when
multiplying two exponents with the same base, we add the exponents and
keep the base \({-c}^{4}\) • \({-c}^{2}\) =
\({-c}^{4 + 2}\) = \({-c}^{6}\).

Question 6.
(3s)5 • (3s)
Answer:
(3s)5 • (3s) = (3s)5+1 = (3s)6,

Explanation:
Given \({3s}^{5}\) • \({3s}^{1}\)
using the Product of Powers Property states that when
multiplying two exponents with the same base, we add the exponents and
keep the base \({3s}^{5}\) • \({3s}\) =
\({3s}^{5 + 1}\) = \({3s}^{6}\).

Simplify each expression. Write your answer in exponential notation.

Question 7.
pq3 • p5q2
Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 3

Answer:
Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key Exponential Notation-2
Explanation:
Given \({p}{q}^{1}{3}\) • \({p}{q}^{5}{2}\)
using the Product of Powers Property states that when
multiplying two exponents with the same base, we add the exponents and
keep the base \({p}^{1 + 5}\) • \({q}^{3 + 2}\) =
\({p}^{6}\) • \({q}^{5}\).

Question 8.
4s4t3 • 5s4t6
Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 4
Answer:
Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key Exponential Notation-3
Explanation:
Given 4. \({s}{t}^{4}{3}\) • 5 . \({s}{t}^{4}{5}\)
using the Product of Powers Property states that when
multiplying two exponents with the same base, we add the exponents and
keep the base 20 . \({s}^{4+ 4}\) •\({t}^{3 + 6}\) =
20 . \({s}^{8}\) • \({t}^{9}\).

Simplify each expression. Write your answer in exponential notation.

Question 9.
108 ÷ 105
108 ÷ 105 = Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 5 Use the Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 5 of powers property.
= Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 5 Simplify.
Answer:
108 ÷ 105 = 103,

Explanation:
Given to find 108 ÷ 105  dividing them with the exponent value in
the numerator (the top exponent) and subtract the exponent value of the
denominator (the bottom exponent). Here that means we take 108 ÷ 105 = 108-5= 103.

Question 10.
2.79 ÷ 2.76 = Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 6
= Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 6 simplify
Answer:
2.79 ÷ 2.76 = 2.79-6 = 2.73,

Explanation:
Given to find 2.79 ÷ 2.76 dividing them with the exponent value in
the numerator (the top exponent) and subtract the exponent value of the
denominator (the bottom exponent), means we take 2.79 ÷ 2.76 = 2.79-6= 2.73.

Question 11.
\(\left(\frac{5}{8}\right)^{6}\) ÷ \(\left(\frac{5}{8}\right)\)

= Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 6 Use the Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 6 of powers property.
= Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 6
Simplify.
Answer:
\(\left(\frac{5}{8}\right)^{5}\),

Explanation:
Given \(\left(\frac{5}{8}\right)^{6}\) ÷ \(\left(\frac{5}{8}\right)\)
dividing them with the exponent value in
the numerator (the top exponent) and subtract the exponent value of the
denominator (the bottom exponent) means we take as
\(\left(\frac{5}{8}\right)^{6-1}\) = \(\left(\frac{5}{8}\right)^{5}\).

Simplify each expression. Write your answer in exponential notation.

Question 12.
q7 ÷ q2
q7 ÷ q2 = Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 7
= Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 7 Simplify
Answer:
\({q}^{5}\),

Explanation:
Given to find q7 ÷  q2  dividing them with the exponent value in
the numerator (the top exponent) and subtract the exponent value of the
denominator (the bottom exponent). Here that means we take q7 ÷ q2 = q7-2= q5.

Question 13.
(-p)5 ÷ (-p)3
Answer:
\({-p}^{2}\),

Explanation:
Given to find (-p)5 ÷ (-p)2  dividing them with the exponent value in
the numerator (the top exponent) and subtract the exponent value of the
denominator (the bottom exponent). Here that means we take (-p)5 ÷ (-p)3 = (-p)5-3= (-p)2.

Question 14.
Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 8
Answer:
(r)3 . (s)2,  

Explanation:
Given to find (r)8 (s)6 ÷ (r)5 (s)4 dividing them with the exponent value in
the numerator (the top exponent) and subtract the exponent value of the
denominator (the bottom exponent). Here that means we take
(r)8 (s)6 ÷ (r)5 (s)4 = (r)8-5 . (s)6-4 = (r)3 . (s)2.  

Question 15.
Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 9
Answer:
Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key Exponential Notation-4
Explanation:
Given to find 63 (x)9 (y)7 ÷ 9.(x)3 (y)4 dividing them with the exponent value in
the numerator (the top exponent) and subtract the exponent value of the
denominator (the bottom exponent). Here that means we take
63 (x)9 (y)7 ÷ 9 (x)3 (y)4 = 7 . (x)9-3 . (y)7-4 = 7. (x)6 (y)3 .  

Simplify each expression. Write your answer in exponential notation.

Question 16.
Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 10
Answer:
\({6}^{2}\),

Explanation:
Given to find (6)7 (6)3 (6)2 ÷ (6)1 (6)4 (6)5 dividing them with the exponent value in
the numerator (the top exponent) and subtract the exponent value of the
denominator (the bottom exponent). Here that means we take
(6)7 (6)3 (6)2 ÷ (6)1 (6)4 (6)5 = (6)7+3+2÷ (6)1+4+5 =
(6)12 ÷ (6)10=(6)12-10
= (6)2.

Question 17.
Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 11
Answer:
\({7.5}^{2}\),

Explanation:
Given to find (7.5)5 (7.5)3 (7.5)1 ÷ (7.5)2 (7.5)1 (7.5)4 dividing them with the exponent value in
the numerator (the top exponent) and subtract the exponent value of the
denominator (the bottom exponent). Here that means we take
(7.5)5 (7.5)3 (7.5)1 ÷ (7.5)2 (7.5)1 (7.5)4 = (7.5)5+3+1÷ (7.5)2+1+4 =
(7.5)9 ÷ (7.5)7=(7.5)9-7
= (7.5)2.

Question 18.
Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 12
Answer:
Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key Exponential Notation-5
Explanation:
Given to find (b)5 (4a)4 (9a)3 ÷ (2a)2 (b)2 (6a)2 dividing them with the exponent value in
the numerator (the top exponent) and subtract the exponent value of the
denominator (the bottom exponent). Here that means we take
(9).(4).(a)4+3.(b)5 ÷ (2).(6).(a)2+2 (b)3 = ((36).(a)7 .(b)5 )÷ (12). (a)4.(b)2 =
(3)(a)7-4 (b)5-2=(3)(a)3
(b)3.

Solve. Show your work.

Question 19.
Jupiter is approximately 108 kilometers from the Sun.
The dwarf planet Eris is about 101° kilometers from the Sun.
How many times as far as Jupiter is Eris from the Sun?
Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 13
Answer:
Eris is 100 times as far as Jupiter from the Sun,

Explanation:
Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key Exponential Notation-6

Math in Focus Course 3A Practice 1.2 Answer Key

Question 1.
(-2)6 • (-2)2
Answer:
(-2)8 ,

Explanation:
Given (-2)6 • (-2)2 using the Product of Powers Property states that when
multiplying two exponents with the same base, we add the exponents and
keep the base, So  (-2)6 • (-2)2
=(-2) 6 + 2 = (-2)8.

Question 2.
7.23 • 7.24
Answer:
7.23 . 7.24 = 7.23+4 = 7.27,

Explanation:
Given to find 7.23 . 7.24 using the Product of Powers Property states that when
multiplying two exponents with the same base, we add the exponents and
keep the base, So
7.23 . 7.24 = 7.23+4= 7.27.

Question 3.
105 • 104
Answer:
109

Explanation:
Given 105 • 104 using the Product of Powers Property states that when
multiplying two exponents with the same base, we add the exponents and
keep the base, So 105 • 104 = 105+4 = 109.

Question 4.
\(\left(\frac{2}{3}\right) \cdot\left(\frac{2}{3}\right)^{5}\)
Answer:
\(\left(\frac{2}{3}\right)^{6}\),

Explanation:
Given \(\left(\frac{2}{3}\right)^{1}\) • \(\left(\frac{2}{3}\right)^{5}\)
using the Product of Powers Property states that when
multiplying two exponents with the same base, we add the exponents and
keep the base \(\left(\frac{2}{3}\right)^{1}\) • \(\left(\frac{2}{3}\right)^{5}\) =
\(\left(\frac{2}{3}\right)^{1 +5}\) = \(\left(\frac{2}{3}\right)^{6}\).

Question 5.
p • p8
Answer:
p1 • p8 = p1 + 8 = p9,

Explanation:
Given \({p}^{1}\) • \({p}^{8}\)
using the Product of Powers Property states that when
multiplying two exponents with the same base, we add the exponents and
keep the base \({p}^{1}\) • \({p}^{8}\) =
\({p}^{1 + 8}\) = \({p}^{9}\).

Question 6.
q8 ÷ q
Answer:
\({q}^{7}\),

Explanation:
Given to find q8 ÷  q1  dividing them with the exponent value in
the numerator (the top exponent) and subtract the exponent value of the
denominator (the bottom exponent). Here that means we take q8 ÷ q1 = q8-1= q7.

Question 7.
xy2 • x4y3
Answer:
\({x}^{5}\) • \({y}^{5}\),

Explanation:
Given xy2 • x4y3
using the Product of Powers Property states that when
multiplying two exponents with the same base, we add the exponents and
keep the base \({x}^{1 + 4}\) • \({y}^{2 + 3}\) =
\({x}^{5}\) • \({y}^{5}\).

Question 8.
2x2y4 • 5x5y
Answer:
10 X \({x}^{7}\) • \({y}^{5}\),

Explanation:
Given 2x2y4 • 5x5y using the Product of Powers Property states that when
multiplying two exponents with the same base, we add the exponents and
keep the base (2 X 5) X \({x}^{2 + 5}\) • \({y}^{4 + 1}\) =
10 X \({x}^{7}\) • \({y}^{5}\).

Question 9.
2.5x3y6 • 3x2y4
Answer:
7.5 X \({x}^{5}\) • \({y}^{10}\),

Explanation:
Given 2.5 X x3y6 • 5x2y4 using the Product of Powers Property states that when
multiplying two exponents with the same base, we add the exponents and
keep the base (2.5 X 3) X \({x}^{3 + 2}\) • \({y}^{6 + 4}\) =
7.5 X \({x}^{5}\) • \({y}^{10}\).

Question 10.
(-3)4 ÷ (-3)2
Answer:
(-3)2 ,

Explanation:
Given to find (-3)4 ÷  (-3)2  dividing them with the exponent value in
the numerator (the top exponent) and subtract the exponent value of the
denominator (the bottom exponent). Here that means we take (-3)4 ÷ (-3)2 = (-3)4-2= (-3)2.

Question 11.
210 ÷ 25
Answer:
25,

Explanation:
Given to find 210 ÷  25  dividing them with the exponent value in
the numerator (the top exponent) and subtract the exponent value of the
denominator (the bottom exponent). Here that means we take 210 ÷ 25 = 210-5= 25.

Question 12.
\(\left(-\frac{1}{6}\right)^{5} \div\left(-\frac{1}{6}\right)^{2}\)
Answer:
\(\left(-\frac{1}{6}\right)^{3}\),

Explanation:
Given \(\left(-\frac{1}{6}\right)^{5}\) ÷ \(\left(-\frac{1}{6}\right)^{2}\)
dividing them with the exponent value in
the numerator (the top exponent) and subtract the exponent value of the
denominator (the bottom exponent) means we take as
\(\left(-\frac{1}{6}\right)^{5-2}\) = \(\left(-\frac{1}{6}\right)^{3}\).

Question 13.
63y3z5 ÷ 9
Answer:
7y3z5,

Explanation:
Given to find 63 (y)3 (z)5 ÷ 9 dividing them with the exponent value in
the numerator (the top exponent) and subtract the exponent value of the
denominator (the bottom exponent). Here that means we take
63 (y)3 (z)5 ÷ 9 = 7. (y)3 (z)5 .  

Question 14.
h2k5 ÷ hk4
Answer:
hk,

Explanation:
Given to find  (h)2 (k)5 ÷ h.(k)4 dividing them with the exponent value in
the numerator (the top exponent) and subtract the exponent value of the
denominator (the bottom exponent). Here that means we take
(h)2 (k)5 ÷  h. (k)4 = (h)2-1 . (k)5-4 = hk.  

Question 15.
64a8b5 ÷ 4a3b2
Answer:
16a5b3,

Explanation:
Given to find 64 (a)8 (b)5 ÷ 4.(a)3 (b)2 dividing them with the exponent value in
the numerator (the top exponent) and subtract the exponent value of the
denominator (the bottom exponent). Here that means we take
64 (a)5 ( b)3 ÷ 4 (a)3 (b)2 = 16 . (a)8-3 . (b)5-2 = 16. a5b3

Question 16.
\(\frac{5^{9} \cdot 5^{7} \cdot 5^{8}}{5^{3} \cdot 5^{2} \cdot 5}\)
Answer:
518,

Explanation:
Given to find (5)9 (5)7 (5)8 ÷ (5)3 (5)2 (5)1 dividing them with the exponent value in
(5)9 (5)7 (5)8 ÷ (5)3 (5)2 (5)1 the numerator (the top exponent) and
subtract the exponent value of the denominator (the bottom exponent).
Here that means we take (5)9+7+8 ÷ (5)3+2+1 = (5)24 ÷ (5)6=(5)24-6
= (5)18.

Question 17.
Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 14
Answer:
\(\left(\frac{4}{9}\right)^{5}\),

Explanation:
Given to find \(\left(\frac{4}{9}\right)^{6}\). \(\left(\frac{4}{9}\right)^{5}\).
\(\left(\frac{4}{9}\right)^{4}\)  ÷ \(\left(\frac{4}{9}\right)^{3}\). \(\left(\frac{4}{9}\right)^{3}\). \(\left(\frac{4}{9}\right)^{4}\) dividing them with the exponent value in
the numerator (the top exponent) and subtract the exponent value of the
denominator (the bottom exponent). Here that means we take
\(\left(\frac{4}{9}\right)^{6+5+4}\) ÷ \(\left(\frac{4}{9}\right)^{3 +3+4}\),
\(\left(\frac{4}{9}\right)^{15}\) ÷ \(\left(\frac{4}{9}\right)^{10}\),
\(\left(\frac{4}{9}\right)^{15-10}\) = \(\left(\frac{4}{9}\right)^{5}\).

Question 18.
\(\frac{a^{9} \cdot a^{2} \cdot a^{3}}{a^{6} \cdot a^{3} \cdot a^{4}}\)
Answer:
a,

Explanation:
Given \(\frac{a^{9} \cdot a^{2} \cdot a^{3}}{a^{6} \cdot a^{3} \cdot a^{4}}\)
to find dividing them with the exponent value in
the numerator (the top exponent) and subtract the exponent value of the
denominator (the bottom exponent). Here that means we take
\(\frac{a^{9+2+3} \cdot}{a^{6+3+4} \cdot}\) =
\(\frac{a^{14} \cdot}{a^{13} \cdot}\) = (a)14-13 = = a.

Question 19.
\(\frac{b^{4} \cdot b^{6} \cdot b}{b^{3} \cdot b^{3} \cdot b^{3}}\)
Answer:
(b)2,

Explanation:
Given \(\frac{b^{4} \cdot b^{6} \cdot b}{b^{3} \cdot b^{3} \cdot b^{3}}\)
to find dividing them with the exponent value in
the numerator (the top exponent) and subtract the exponent value of the
denominator (the bottom exponent). Here that means we take
\(\frac{b^{4+6+1} \cdot}{a^{3+3+3} \cdot}\) =
\(\frac{b^{11} \cdot}{b^{9} \cdot}\) = (b)11-9 = (b)2.

Question 20.
\(\frac{3 x^{3} \cdot z^{4} \cdot 4 x^{3}}{2 x \cdot x \cdot 3 z}\)
Answer:
2x4z3,

Explanation:
Given to find 3.(x)3 (z)4 . 4(x)3 ÷ 2.(x)(x) (z) dividing them with the exponent value in
the numerator (the top exponent) and subtract the exponent value of the
denominator (the bottom exponent). Here that means we take
12 (x)6 (z)4 ÷ 6 (x)2 (z) = 2 . (x)6-2 . (z)4-1 = 2. (x)4 (z)3 .  

Question 21.
\(\frac{4 c^{6} \cdot 3 b^{4} \cdot 9 c^{5}}{b^{3} \cdot 6 c^{3} \cdot 2 c^{3}}\)
Answer:
9bc5,

Explanation:
Given to find \(\frac{4 c^{6} \cdot 3 b^{4} \cdot 9 c^{5}}{b^{3} \cdot 6 c^{3} \cdot 2 c^{3}}\)
dividing them with the exponent value in the numerator (the top exponent) and
subtract the exponent value of the denominator (the bottom exponent).
Here that means we take 4(c)6. 3(b)4 . 9(c)÷ (b)3 . 6 (c)3 . 2(c)=
9 . (b)4-3 . (c)11-6 = 9.(b).(c)5 .  

Solve. Show your work.

Question 22.
Pluto has a diameter of about 103 kilometers. The diameter of Saturn is
approximately 105 kilometers.
How many times as great as Pluto’s diameter is Saturn’s diameter?
Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 15
Answer:
100 times as great as Pluto’s diameter is Saturn’s diameter,

Explanation:
Given Pluto has a diameter of about (10)3 kilometers.
The diameter of Saturn is approximately (10)5 kilometers.
So number of times as great as Pluto’s diameter is Saturn’s diameter is
(10)5-3 = (10)2,
therefore 100 times as great as Pluto’s diameter is Saturn’s diameter.

Question 23.
Use the rectangular prism shown.
a) Express the volume of the rectangular prism using exponential notation.
Answer:
6(x)3,

Explanation:
Given width of the rectangular prism as 2x units,
height as x units and length as 3x units so the
volume of the rectangular prism using exponential notation is
w X h X l = 2x X x X 3x =6(x)1+1+1 = 6(x)3.

b) Another prism has dimensions that are twice the dimensions of the prism shown.
Express the volume of that prism using exponential notation.
Answer:

Math in Focus Grade 8 Chapter 1 Lesson 1.2 Answer Key The Product and the Quotient of Powers 16
Explanation:
Given another prism has dimensions that are twice the dimensions of the prism shown.
Expressing the volume of that prism using exponential notation is
width of the rectangular prism as 2x X 2x = 4x1+1  = 4x2 units,
height as 2 X x = 2x units and length as 3x X 3 x = 9x1+1 = 9x2 units,
So the volume of the rectangular prism using exponential notation is
w X h X l = 4x2 X 2x X 9x2 = 72 (x)2+1+2 = 72(x)5.

c) How many times greater is the volume of the larger prism than the volume of the smaller prism?
Answer:
12x2 times greater is the volume of the larger prism than the volume of the smaller prism,

Explanation:
To know number of times greater is the volume of the larger prism than the
volume of the smaller prism, we divide 72(x)÷ 6(x)3 = 12(x)5-3 = 12(x)2.

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