Which value is equivalent to 8 multiplied by 4 multiplied by 2 whole over 8 multiplied by 7, the whole raised to the power of 2 multiplied by 8 to the power of 0 over 7 to the power of negative 3, whole to the power of 3 multiplied by 7 to the power of negative 9
Answers
Answer:
Answer: First Option is correct.
Step-by-step explanation:
Since we have given that
8 multiplied by 4 multiplied by 2 whole over 8 multiplied by 7, the whole raised to the power of 2 multiplied by 8 to the power of 0 over 7 to the power of negative 3, whole to the power of 3 multiplied by 7 to the power of negative 9.
It means
(\frac{8\times 4\times 2}{8\times 7})^{2} \times (\frac{8^0}{7^{-3}})^3 \times 7^{-9}(8×78×4×2)2×(7−380)3×7−9
So, when we cancel out 8 from numerator and denominator form the first expression, we get
(\frac{ 4\times 2}{7})^{2} \times (\frac{8^0}{7^{-3}})^3 \times 7^{-9}(74×2)2×(7−380)3×7−9
=\frac{8^2}{7^2}\times \frac{8^0}{7^{-3}}\times 7^{-9}=7282×7−380×7−9
As we know that the exponential rule :
(\frac{a}{b})^m=\frac{a^m}{b^m}(ba)m=bmam
and
(a^m)^n=a^{mn}(am)n=amn
And
a^0=1a0=1
By applying such rule, we get,
\begin{gathered}\frac{8^2}{7^2}\times \frac{8^0}{7^{-3}}\times 7^{-9}\\\\=\frac{64}{49}\times \frac{1}{7^{-9}}\times 7^{-9}\\\\=\frac{64}{49}\times \frac{7^{-9}}{7^{-9}}\\\\=\frac{64}{49}\end{gathered}7282×7−380×7−9=4964×7−91×7−9=4964×7−97−9=4964
Hence, First Option is correct
Step-by-step explanation:
Answer: First Option is correct.
Step-by-step explanation:
Since we have given that
8 multiplied by 4 multiplied by 2 whole over 8 multiplied by 7, the whole raised to the power of 2 multiplied by 8 to the power of 0 over 7 to the power of negative 3, whole to the power of 3 multiplied by 7 to the power of negative 9.
It means
(\frac{8\times 4\times 2}{8\times 7})^{2} \times (\frac{8^0}{7^{-3}})^3 \times 7^{-9}(8×78×4×2)2×(7−380)3×7−9
So, when we cancel out 8 from numerator and denominator form the first expression, we get
(\frac{ 4\times 2}{7})^{2} \times (\frac{8^0}{7^{-3}})^3 \times 7^{-9}(74×2)2×(7−380)3×7−9
=\frac{8^2}{7^2}\times \frac{8^0}{7^{-3}}\times 7^{-9}=7282×7−380×7−9
As we know that the exponential rule :
(\frac{a}{b})^m=\frac{a^m}{b^m}(ba)m=bmam
and
(a^m)^n=a^{mn}(am)n=amn
And
a^0=1a0=1
By applying such rule, we get,
\begin{gathered}\frac{8^2}{7^2}\times \frac{8^0}{7^{-3}}\times 7^{-9}\\\\=\frac{64}{49}\times \frac{1}{7^{-9}}\times 7^{-9}\\\\=\frac{64}{49}\times \frac{7^{-9}}{7^{-9}}\\\\=\frac{64}{49}\end{gathered}7282×7−380×7−9=4964×7−91×7−9=4964×7−97−9=4964
Hence, First Option is correct