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gladu [14]
3 years ago
15

Can you help with this? Length x width x height

Mathematics
2 answers:
lyudmila [28]3 years ago
6 0

Answer:

So first you want to multiply the first two fractions (on the first shape) 5/8 x 6/8= 15/32 and then you multiply that by your second number which

is 5/8. 5/8 x 15/32 is  75/256in^3

(And on the second shape) 4/5 x 4/5 is 16/25. 16/25 x 4/5 which is 64/125in^3

Gemiola [76]3 years ago
5 0

Answer:

75/256 in³    <    64/125 in³

Step-by-step explanation:

First, find the volumes of the two 3-D figures:

For Figure 1:

Using the volume formula:

V = l × w × h

   = 6/8 × 5/8 × 5/8

   = (6/8 × 5/8) × 5/8

   = 30/64 × 5/8

V = 150/512

Simplify:

V = 75/256

So Figure 1's volume is 75/256 in³.

For Figure 2:

V = l × w × h

   = 4/5 × 4/5 × 4/5

   = (4/5 × 4/5) × 4/5

   = 16/25 × 4/5

   = 64/125

So Figure 2's volume is 64/125 in³.

Now, comparing the two volumes:

You can cross multiply the fractions to figure out which fraction is greater:

Figure 1                       Figure 2

<u> 75 </u>                =                <u> 64 </u>

256                                  125

75 (125)          =             256 (64)

9375               =             16384

So Figure 2 has the greater volume.

Your final result should be:

75/256 in³    <    64/125 in³

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Write 0.216 as a rational number in lowest terms
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Step-by-step explanation:

0.216

\mathrm{Multiply\:and\:divide\:by\:10\:for\:every\:number\:after\:the\:decimal\:point.}

\mathrm{There\:are\:}3\mathrm{\:digits\:to\:the\:right\:of\:the\:decimal\:point,\:therefore\:multiply\:and\:divide\:by\:}1000

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\mathrm{The\:GCD\:of\:}a,\:b\mathrm{\:is\:the\:largest\:positive\:number\:that\:divides\:both\:}a\mathrm{\:and\:}b\mathrm{\:without\:a\:remainder}

2,\:3\mathrm{\:are\:all\:prime\:numbers,\:therefore\:no\:further\:factorization\:is\:possible}

=2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:3

2,\:5\mathrm{\:are\:all\:prime\:numbers,\:therefore\:no\:further\:factorization\:is\:possible}

=2\cdot \:2\cdot \:2\cdot \:5\cdot \:5\cdot \:5

\mathrm{The\:prime\:factors\:common\:to\:}216,\:1000\mathrm{\:are\:}

=2\cdot \:2\cdot \:2

\mathrm{Multiply\:the\:numbers:}\:2\cdot \:2\cdot \:2=8

=8

=\frac{8\cdot \:27}{8\cdot \:125}

\mathrm{Cancel\:the\:common\:factor:}\:8

=\frac{27}{125}

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