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yanalaym [24]
3 years ago
12

There is a Chick-fil-a exactly 6 miles due east of Berkmar Middle School. There is also a Wal-Mart 6 miles due north of Berkmar

Middle School. How far is the Chick-fil-a from the Wal-Mart? Leave your answer in its simplest radical form.
​
Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
5 0

Answer:

6√2 miles

Step-by-step explanation:

Using the solution diagram drawn :

We have a right angled triangle, the distance between Walmart mad chick-FIL is a measure of the hypotenus of the right angled triangle ;

The hypotenus, = √(opposite² + adjacent²)

Hypotenus = √6² + 6²

Hypotenus = √(36 + 36)

Hypotenus = √72

Hypotenus = √36 * 2 = √36 * √2 = 6√2 miles

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45 feet

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4 years ago
Represent the arithmetic series using the recursive formula 94, 87, 80, 73,
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Answer:

a_1 = 94 \\ a_n = a_ {n - 1}  - 7

Step-by-step explanation:

The given arithmetic sequence is 94, 87, 80, 73,...

The first term is

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The common difference is

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The recursive formula is given by:

a_n = a_ {n - 1} + d

We substitute the common difference to obtain:

a_n = a_ {n - 1}  - 7

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3 years ago
What is the intercepts for these ?
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How many envelopes are there in 7 packages
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5 0
4 years ago
Evaluate cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.
Andre45 [30]

Answer:

Option d)  5 to the power of negative 5 over 6 is correct.

\dfrac{\sqrt[3]{\bf 5} \times \sqrt{\bf 5}}{\sqrt[3]{\bf 5^{\bf 5}}}= 5^{\frac{\bf -5}{\bf 6}}

Above equation can be written as 5 to the power of negative 5 over 6.

ie, 5^\frac{\bf -5}{\bf 6}

Step-by-step explanation:

Given that cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.

It can be written as below

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}} \times 5^{\frac{1}{2}}}{5^{\frac{5}{3}}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}+\frac{1}{2}}}{5^{\frac{5}{3}}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{2+3}{6}}}{5^{\frac{5}{3}}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5}{6}} \times 5^{\frac{-5}{3}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5-10}{6}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{5^5}= 5^{\frac{-5}{6}}

Above equation can be written as 5 to the power of negative 5 over 6.

7 0
4 years ago
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