Let's solve step by step each problem:
A)
Here is important to use the multiplication property. When the base is the same you can sum the exponents:
[tex]8^7\times8^{-12}=8^{7+(-12)}=8^{7-12}=8^{-5}[/tex]
Now you can see the exponent is negative and you can use another property to turn him positive. This can be done by inverting the fraction:
[tex]8^{-5}=\frac{1}{8^5}[/tex]
8^5 is such a huge number and definitely, this answer will be between 0 and 1. Just showing you the value:
[tex]\frac{1}{8^5}=0.00003[/tex]
B)
Here you have to use the division property. When the base is the same you can subtract the exponents:
[tex]\frac{7^4}{7^{-3}}=7^{4-(-3)}=7^7=823543[/tex]
(not between 0 and 1)
C) Use the multiplication property that we used on letter A)
[tex](\frac{1}{3})^2\times(\frac{1}{3})^9=(\frac{1}{3})^{2+9}=(\frac{1}{3})^{11}[/tex][tex](\frac{1}{3})^{11}=\frac{1^{11}}{3^{11}}=0.000005^{}[/tex]
Between 0 and 1!
D) Use the division property that we used on letter B)
[tex]\frac{(-5)^6}{(-5)^{10}}=(-5)^{(6-10)}=(-5)^{-4}[/tex][tex](-5)^{-4}=\frac{1}{(-5)^4}=0.0016[/tex]
Between 0 and 1! Therefore:
A, C and D are between 0 and 1.
B is not between 0 and 1.
