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2 years ago

14

A bookstore sells used books for $8 each. Let n represent the number of books. The cost for n books is 8 xn or 8n.

1 answer:

zalisa [80]2 years ago

6 0

The unknown quantity is the books

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A cone has a height of 12 millimeters and a radius of 10 millimeters. What is it’s volume. Use 3.14 and round to the nearest hun

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2 years ago

Miguel earns 2,456.75 every month he also earns an extra 4.75 every time he sells a new gym membership last month Miguel sold 32

zalisa [80]

Answer: Total amount of money earned last month = $2608.75

Step-by-step explanation:

Miguel earns 2,456.75 every month. This is his constant pay for the month. He also earns an extra 4.75 every time he sells a new gym membership.

Last month, Miguel sold 32 new gym membership. This means that the extra money that he earned for last month will be the number of new gym membership sold times the amount her earns per new gym membership sold. It becomes

32 × 4.75 = 152

Total amount of money earned last month will be sum of his monthly salary + the extra earned. It becomes

2456.75 + 152 = $2608.75

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3 years ago

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Step-by-step explanation:

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1 year ago

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Darius has a 24-ounce sports drink he drinks 18 ounces

katovenus [111]

Answer: 6 ounces

Explaintion: 24 minus 18 equals 6

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3 years ago

Match each series with the equivalent series written in sigma notation

PIT_PIT [208]

Answer:

$3 + 12 + 48 + 192 + 768 = \sum\limits^4_{n=0} 3 * 4^n$

$4 + 32 + 256 + 2048 + 16384 = \sum\limits^4_{n=0} 4 * 8^n$

$2 + 6 + 18 + 54 + 162 = \sum\limits^4_{n=0} 2* 3^n$

$3 + 15 + 75 + 375 + 1875 = \sum\limits^4_{n=0} 3* 5^n$

Step-by-step explanation:

Given

See attachment for complete question

Required

Match equivalent expressions

Solving (a):

$3 + 12 + 48 + 192 + 768$

The expression can be written as:

$3 \to 3*4^{0$ --- 0

$12 \to 3 * 4^{1$ ---- 1

$48 \to 3 * 4^{2$ --- 2

$192 \to 3 * 4^{3$ ---- 3

$768 \to 3 * 4^{4$ ---- 4

For the nth term, the expression is:

$Term = 3 * 4^{n$ ---- n

So, the summation is:

$3 + 12 + 48 + 192 + 768 = \sum\limits^4_{n=0} 3 * 4^n$

Solving (b):

$4 + 32 + 256 + 2048 + 16384$

The expression can be written as:

$4 \to 4 * 8^0$ --- 0

$32 \to 4 * 8^1$ ---- 1

$256 \to 4 * 8^2$ --- 2

$2048 \to 4 * 8^3$ ---- 3

$16384 \to 4 * 8^4$ ---- 4

For the nth term, the expression is:

$Term \to 4 * 8^n$ ---- n

So, the summation is:

$4 + 32 + 256 + 2048 + 16384 = \sum\limits^4_{n=0} 4 * 8^n$

Solving (c):

$2 + 6 + 18 + 54 + 162$

The expression can be written as:

$2 \to 2 * 3^0$ --- 0

$6 \to 2 * 3^1$ ---- 1

$18 \to 2 * 3^2$ --- 2

$54 \to 2 * 3^3$ ---- 3

$162 \to 2 * 3^4$ ---- 4

For the nth term, the expression is:

$Term \to 2 * 3^n$ ---- n

So, the summation is:

$2 + 6 + 18 + 54 + 162 = \sum\limits^4_{n=0} 2* 3^n$

Solving (d):

$3 + 15 + 75 + 375 + 1875$

The expression can be written as:

$3 \to 3 * 5^0$ --- 0

$15 \to 3 * 5^1$ ---- 1

$75 \to 3 * 5^2$ --- 2

$375 \to 3 * 5^3$ ---- 3

$1875 \to 3 * 5^4$ ---- 4

For the nth term, the expression is:

$Term \to 3 * 5^n$ ---- n

So, the summation is:

$3 + 15 + 75 + 375 + 1875 = \sum\limits^4_{n=0} 3* 5^n$

5 0

2 years ago