Find the numerical equivalent of 99⋅9−6

Question

. Your answer should be written without exponents

Answer 1

To find the numerical equivalent of \( 99 \cdot 9^{-6} \), we first need to evaluate \( 9^{-6} \).

The expression \( 9^{-6} \) is the same as \( \frac{1}{9^6} \).

Next, we need to calculate \( 9^6 \): \[ 9^6 = (3^2)^6 = 3^{12} \] Calculating \( 3^{12} \): \[ 3^2 = 9, \quad 3^3 = 27, \quad 3^4 = 81, \quad 3^5 = 243, \quad 3^6 = 729 \] Continuing, we compute \( 3^{12} \) as follows: \[ 3^6 = 729 \quad \Rightarrow \quad 3^{12} = 729 \times 729 = 531441 \] So, \( 9^6 = 531441 \) and thus: \[ 9^{-6} = \frac{1}{531441} \] Now, we multiply: \[ 99 \cdot 9^{-6} = 99 \cdot \frac{1}{531441} = \frac{99}{531441} \]

Next, we can simplify \( \frac{99}{531441} \):

• Factor \( 99 \): \[ 99 = 3^2 \cdot 11 \]
• Factor \( 531441 \) (as calculated previously, \( 531441 = 9^6 = (3^2)^6 = 3^{12} \)): \[ 531441 = 3^{12} \] Now we can rewrite the fraction: \[ \frac{99}{531441} = \frac{3^2 \cdot 11}{3^{12}} = \frac{11}{3^{10}} \] Calculating \( 3^{10} \): \[ 3^{10} = 59049 \quad (\text{since } 3^5 = 243 \text{ and } 3^{10} = 243^2 = 59049) \]

Thus, the final answer is: \[ \frac{11}{59049} \]

So, the numerical equivalent of \( 99 \cdot 9^{-6} \) is: \[ \boxed{\frac{11}{59049}} \]

Answer 2

Find the numerical equivalent of 9 to the power of 9 ⋅ 9 to the power of −6 Your answer should be written without exponents

Answer 3

To find the numerical equivalent of \( 9^9 \cdot 9^{-6} \), we can use the property of exponents that states \( a^m \cdot a^n = a^{m+n} \).

Applying this rule, we have: \[ 9^9 \cdot 9^{-6} = 9^{9 + (-6)} = 9^{9 - 6} = 9^3 \]

Now, we calculate \( 9^3 \): \[ 9^3 = 9 \cdot 9 \cdot 9 \]

Calculating step-by-step:

1. First, \( 9 \cdot 9 = 81 \).
2. Then, \( 81 \cdot 9 = 729 \).

Thus, \( 9^3 = 729 \).

Therefore, the numerical equivalent of \( 9^9 \cdot 9^{-6} \) is: \[ \boxed{729} \]

Answer 4

What is the missing number in this simplified exponential expression: 7 to the power of ?÷7 to the power of 5=7 to the power of 12

Answer 5

To solve the equation

\[ \frac{7^?}{7^5} = 7^{12} \]

we can use the property of exponents that says \(\frac{a^m}{a^n} = a^{m-n}\). Thus, we rewrite the left-hand side:

\[ 7^{? - 5} = 7^{12} \]

Since the bases are the same (both are base 7), we can set the exponents equal to each other:

\[ ? - 5 = 12 \]

Now, we solve for \( ? \):

\[ ? = 12 + 5 = 17 \]

Thus, the missing number is

\[ \boxed{17} \]

Answer 6

Apply the Property of Negative Integer Exponents to solve 7 to the power of −3 The answer should not contain any exponent expression or decimals

Answer 7

To express \( 7^{-3} \) using the property of negative integer exponents, we use the rule that states:

\[ a^{-n} = \frac{1}{a^n} \]

Applying this to \( 7^{-3} \), we have:

\[ 7^{-3} = \frac{1}{7^3} \]

Now, we need to calculate \( 7^3 \):

\[ 7^3 = 7 \cdot 7 \cdot 7 \]

Calculating step-by-step:

1. \( 7 \cdot 7 = 49 \)
2. \( 49 \cdot 7 = 343 \)

Thus,

\[ 7^3 = 343 \]

Now substituting back, we find:

\[ 7^{-3} = \frac{1}{7^3} = \frac{1}{343} \]

Since the answer should not contain any exponent expression or decimals, the final result is:

\[ \boxed{\frac{1}{343}} \]