Fill in the blanks with two consecutive integers to complete the following inequality.

Mathematics

2 answers:

Andreyy893 years ago

8 0

Answer:

9 < √97 < 10

Step-by-step explanation:

√97 is a little less than √100

Which is 10

9 < √97 < 10

baherus [9]3 years ago

4 0

<h3>Answer: 9 < √97 < 10 </h3>

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Explanation:

Consider that 81 < 97 < 100 is a true statement. Note the 81 and 100 are perfect squares such that 81 = 9^2 and 100 = 10^2

We can apply the square root to all three sides to get the following

$81 < 97 < 100\\\\\sqrt{81} < \sqrt{97} < \sqrt{100}\\\\\sqrt{9^2} < \sqrt{97} < \sqrt{10^2}\\\\9 < \sqrt{97} < 10\\\\\text{The number }\sqrt{97} \text{ is between 9 and 10}\\\\$

Using a calculator, we can help verify this

$\sqrt{97} \approx 9.85$

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Alex Ar [27]

Start by distributing the exponent to each of the terms in $(2u)^{3}$ . This will become $2^{3} u^{3}$ , and $2^{3} =8$ . Now, the expression is:

$\frac{2u^{3} v^{2} }{8u^{3} * u^{2} }$ .

We can now simplify the bottom to read: $8u^{5}$ because when multiplying variables raised to an exponent, we add the exponents. The expression now looks like:

$\frac{2u^{3} v^{2} }{8u^{5} }$

The 2/8 simplifies to 1/4:

$\frac{u^{3} v^{2} }{4u^{5} }$

Now, we have two u terms on the top and bottom. When dividing variables raised to an exponent, we subtract the exponents. However, since 3-5=-2, the term will be on the bottom to avoid the negative exponent. The final answer is:

$\frac{v^{2} }{4u^{2} }$