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Illusion [34]
3 years ago
7

What 6,589 to the nearest thousand

Mathematics
2 answers:
raketka [301]3 years ago
6 0

6,589 to the nearest thousand would be <u>7,000</u> since you have high numbers as 5,8, and 9 that allows you to round up your answer to 7,000.

BlackZzzverrR [31]3 years ago
6 0

6,589 to the nearest thousand is 7,000. Look to the number right of the hundreds place and if its below 4 you round down and if its above 5 you round up.

<h2><em>-Rhear</em></h2>
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gladu [14]

Answer:

There can be 9 arrays.

Step-by-step explanation:

We are asked that the number of arrays that Courtney can make with 36 photos.

The dimensions of the possible arrays are  

(1 \times 36), (2 \times 18), (3 \times 12), (4 \times 9), (6 \times 6), (9 \times 4), (12 \times 3), (18 \times 2) \textrm { and } (36 \times 1)

Here, the array with 1 row by 36 columns and 36 rows by 1 column are different. (Answer)

5 0
3 years ago
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Answer:

35

Step-by-step explanation:

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bagirrra123 [75]
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If a=2b^3 and b= -1/2c ^-2,express a in terms of c.
ElenaW [278]

Answer:

Part 1) a=-\frac{1}{4c^6}

Part 2) a=-\frac{1}{4c^{-6}}

Step-by-step explanation:

Part 1) we have

a=2b^{3} ----> equation A

b=-\frac{1}{2}c^{-2} ----> equation B

substitute equation B in equation A

a=2(-\frac{1}{2}c^{-2})^{3}

Applying property of exponents

(x^{m})^{n}=x^{m*n}

x^{-m} =\frac{1}{x^{m}}

a=2(-\frac{1}{2}c^{-2})^{3}=2(-\frac{1}{2})^3(c^{-2})^{3}=2(-\frac{1}{8})(c^{-6})=-\frac{1}{4c^6}

therefore

a=-\frac{1}{4c^6}

Part 2) we have

a=2b^{3} ----> equation A

b=-\frac{1}{2c^{-2}}=-\frac{c^{2}}{2} ----> equation B

substitute equation B in equation A

a=2(-\frac{c^{2}}{2})^{3}

Applying property of exponents

(x^{m})^{n}=x^{m*n}

x^{-m} =\frac{1}{x^{m}}

a=2(-\frac{c^{2}}{2})^{3}=2(-\frac{c^{6}}{8})

simplify

a=-\frac{c^{6}}{4}

therefore

a=-\frac{1}{4c^{-6}}

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Answer:

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Step-by-step explanation:

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