Apply the Property of Negative Integer Exponents to solve 7−3 . The answer should not contain any exponent expression or decimals.(1 point)7−3=

Answer 1

To solve $7^{-3}$ using the property of negative integer exponents, we can use the rule that states $a^{-n}=\frac{1}{a^n}$.

Applying this property:

$$7^{-3}=\frac{1}{7^3}$$

Now, we need to calculate $7^3$:

$$7^3=7\times 7\times 7=49\times 7=343$$

Now substituting back, we get:

$$7^{-3}=\frac{1}{343}$$

Thus, the final answer is:

$$\boxed{{\displaystyle \frac{1}{343}}}$$

Answer 2

Apply the Property of Negative Integer Exponents to generate an expression equivalent to 3−178−14 with positive exponents only.(1 point)3−178−14=

Answer 3

To simplify the expression $3^{-1}{7}^{-14}$ using the property of negative integer exponents, we can apply the rule that states $a^{-n}=\frac{1}{a^n}$.

1. Rewrite $3^{-1}$: $$3^{-1}=\frac{1}{3^1}=\frac{1}{3}$$

2. Rewrite $7^{-14}$: $$7^{-14}=\frac{1}{7^{14}}$$

Putting it all together, we get: $$3^{-1}{7}^{-14}=\frac{1}{3}\cdot \frac{1}{7^{14}}=\frac{1}{3\cdot 7^{14}}$$

Thus, the expression $3^{-1}{7}^{-14}$ with positive exponents only is: $$\boxed{{\displaystyle \frac{1}{3\cdot 7^{14}}}}$$

Answer 4

Apply the Properties of Integer Exponents to generate equivalent expressions to (7−3)7 ⋅ 7206−2 with only positive exponents, then solve the expression. Provide the answer as an improper fraction.(1 point)(7−3)7 ⋅ 7206−2=

Answer 5

To simplify the expression $(7^{-3}{)}^7\cdot 7^{{206}^{-2}}$ with positive exponents, we'll start by applying the properties of integer exponents.

1. Simplifying $(7^{-3}{)}^7$: According to the power of a power property, $(a^m{)}^n=a^{m\cdot n}$: $$(7^{-3}{)}^7=7^{-3\cdot 7}=7^{-21}$$

2. Simplifying $7^{{206}^{-2}}$: We first need to simplify ${206}^{-2}$: $${206}^{-2}=\frac{1}{{206}^2}$$ Therefore, $$7^{{206}^{-2}}=7^{\frac{1}{{206}^2}}$$

To combine these, we need to express $7^{-21}$ with a positive exponent. Using the property $a^{-n}=\frac{1}{a^n}$: $$7^{-21}=\frac{1}{7^{21}}$$

So our expression becomes: $$(7^{-3}{)}^7\cdot 7^{{206}^{-2}}=\frac{1}{7^{21}}\cdot 7^{\frac{1}{{206}^2}}$$

3. Combining the exponents: Using the property $a^m\cdot a^n=a^{m+n}$: $$\frac{1}{7^{21}}\cdot 7^{\frac{1}{{206}^2}}=\frac{7^{\frac{1}{{206}^2}}}{7^{21}}=7^{\frac{1}{{206}^2}-21}$$

4. Expressing the result simply: We can leave it as: $$7^{\frac{1}{{206}^2}-21}$$

5. Finding a common denominator: Now, we need a common denominator to combine $\frac{1}{{206}^2}-21$: $$21=\frac{21\cdot {206}^2}{{206}^2}=\frac{21\cdot {206}^2}{{206}^2}$$

Thus, $$\frac{1}{{206}^2}-21=\frac{1-21\cdot {206}^2}{{206}^2}$$

The entire expression can now be simplified to: $$7^{\frac{1-21\cdot {206}^2}{{206}^2}}$$

6. Expressing as an improper fraction: To return to the form of the initial expression, we can write: $$=\frac{1}{7^{21-\frac{1}{{206}^2}}}$$

Thus, the final answer depending on the simplification yields: \[ \boxed{\frac{1}{7^{21 - \frac{1}{206^2}}} \]

If we wish to keep all terms strictly without roots or fractions that are non-integer based, we generally keep them in simple forms as they arise within the operations.

Let me know if you require numeric output or full numeric simplification for such computed values!

Answer 6

Apply the product rule of integer exponents to find the numerical equivalent of 12−5⋅127 . Show your work and explain how you used the product rule of integer exponents.(2 points)

Answer 7

To find the numerical equivalent of the expression ${12}^{-5}\cdot {12}^7$, we will use the product rule of integer exponents. The product rule states that when multiplying two expressions with the same base, you can add their exponents. In mathematical terms:

$$a^m\cdot a^n=a^{m+n}$$

Step-by-Step Calculation:

Step 1: Identify the base and exponents

• Here, the base is $12$.
• The exponents are $-5$ and $7$.

Step 2: Apply the product rule Using the product rule: $${12}^{-5}\cdot {12}^7={12}^{-5+7}$$

Step 3: Simplify the exponent Now we calculate the exponent: $$-5+7=2$$ So: $${12}^{-5}\cdot {12}^7={12}^2$$

Step 4: Calculate ${12}^2$ Now we compute ${12}^2$: $${12}^2=12\times 12=144$$

Conclusion

The numerical equivalent of ${12}^{-5}\cdot {12}^7$ is: $$\boxed{{\displaystyle 144}}$$

Explanation of the Product Rule

In this calculation, we applied the product rule of integer exponents effectively by recognizing that both terms share the same base $12$. By adding their exponents, we simplified the multiplication of the two terms into a single term with an easily computed exponent.

Using this method allowed us to convert a more complex expression into a simpler form, facilitating the calculation of the final result efficiently.

Answer 8

How many times is 4 × 104 larger than 2 × 102 ?(1 point) Responses 200 200 2 2 20 20 2,000 2,000

Answer 9

To compare how many times $4\times {10}^4$ is larger than $2\times {10}^2$, we can first calculate the values of each expression.

1. Calculate $4\times {10}^4$: $$4\times {10}^4=4\times 10,000=40,000$$

2. Calculate $2\times {10}^2$: $$2\times {10}^2=2\times 100=200$$

3. Now, to find out how many times $4\times {10}^4$ is larger than $2\times {10}^2$, we divide the two results: $$\frac{40,000}{200}$$

4. Perform the division: $$\frac{40,000}{200}=200$$

Thus, $4\times {10}^4$ is 200 times larger than $2\times {10}^2$.

The answer is: $$\boxed{{\displaystyle 200}}$$

Answer 10

How many times less is 102 than 106 ? Express your answer as an integer power of 10.(1 point) Responses 0.0001 0.0001 104 10 superscript 4 baseline 10−4 10 superscript negative 4 baseline 10,000

Answer 11

To determine how many times less ${10}^2$ is than ${10}^6$, we can set up the calculation as a division:

$$\frac{{10}^2}{{10}^6}$$

Using the property of exponents that states $\frac{a^m}{a^n}=a^{m-n}$, we can simplify this:

$$\frac{{10}^2}{{10}^6}={10}^{2-6}={10}^{-4}$$

Thus, ${10}^2$ is ${10}^{-4}$ times less than ${10}^6$.

The answer expressed as an integer power of 10 is: $$\boxed{{\displaystyle {10}^{-4}}}$$