Respuesta :
Answer
1) c. 9 inches squared
2) c. By simplifying 8 to 23 to make both powers base two and subtracting the exponents
3) b. 2 to the 7 over 3 power
Explanation.
Question 1
The area of a rectangle is given by
area=l×w
=〖81〗^(1/3)×3^(2/3)
3^(4/3)×3^(2/3)=3^(4/3+2/3)
=3^(6/3)=3^2=9 inches squared
Question 2
Finding the 2^5/8=2^2
The steps for finding this is first to write 8 as 23. Then divide.
2^5/8=2^5/2^3
When dividing exponents with the same base you subtract the powers.
=2^(5-3)=2^2
Question 3
The question wants us to solve, ∛(3^7 ).the cube root of a number can be written as a power of a third.
So,
∛(3^7 )=3^(7×1/3 )
=3^(7/3)
1) c. 9 inches squared
2) c. By simplifying 8 to 23 to make both powers base two and subtracting the exponents
3) b. 2 to the 7 over 3 power
Explanation.
Question 1
The area of a rectangle is given by
area=l×w
=〖81〗^(1/3)×3^(2/3)
3^(4/3)×3^(2/3)=3^(4/3+2/3)
=3^(6/3)=3^2=9 inches squared
Question 2
Finding the 2^5/8=2^2
The steps for finding this is first to write 8 as 23. Then divide.
2^5/8=2^5/2^3
When dividing exponents with the same base you subtract the powers.
=2^(5-3)=2^2
Question 3
The question wants us to solve, ∛(3^7 ).the cube root of a number can be written as a power of a third.
So,
∛(3^7 )=3^(7×1/3 )
=3^(7/3)
Answer:
Ques 1)
Option: C is the correct answer.
c. 9 inches square.
Ques 2)
Option: c
c. By simplifying 8 to 2^3 to make both powers base two and subtracting the exponents.
Ques 3)
Option: b
b. 2 to the 7 over 3 power
Step-by-step explanation:
Ques 1)
A rectangle has a length of the cube root of 81 inches and a width of 3 to the 2 over 3 power inches.
i.e. let 'l' and 'b' denote the length and width of the rectangle.
i.e.[tex]l=\sqrt[3]{81}\\\\l=(3^4)^{\dfrac{1}{3}}\\\\l=3^{\dfrac{4}{3}[/tex]
since,
[tex](a^m)^n=a^{mn}[/tex]
and
[tex]w=3^{\dfrac{2}{3}}[/tex]
Hence, the area of rectangle is given by:
[tex]Area=l\times w\\\\\\Area=3^{\dfrac{1}{3}}\times3^{\dfrac{4}{3}}\\\\\\Area=3^({\dfrac{1}{3}+\dfrac{4}{3}})\\\\\\Area=3^{\dfrac{6}{3}}\\\\\\Area=3^2\\\\\\Area=9\ square\ inches[/tex]
As we know that:
[tex]a^m\times a^n=a^{m+n}[/tex]
Hence, Area=9 square inches.
Ques 2)
2 to the fifth power, over 8 = 2 to the 2nd power
i.e. we need to prove that:
[tex]\dfrac{2^5}{8}=2^2[/tex]
As we know that:
[tex]\dfrac{2^5}{8}=\dfrac{2^5}{2^3}=2^{5-3}=2^2[/tex]
( since
[tex]\dfrac{a^m}{a^n}=a^{m-n}[/tex] )
Hence, the correct answer is: Option: c
Ques 3)
The cube root of 2 to the seventh power.
i.e.
[tex]\sqrt[3]{2^7}\\\\=(2^7)^{\dfrac{1}{3}}\\\\=2^{\dfrac{7}{3}}[/tex]
Since,
[tex]\sqrt[n]{x}=x^{\dfrac{1}{n}}[/tex]
and
[tex](a^m)^n=a^{mn}[/tex]
Hence, the correct answer is: option: b
