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2 years ago

Use a ladder diagram to find the prime factorization of 64 i need answer quick

Mathematics

2 answers:

choli [55]2 years ago

8 0

Answer:

2^5

Step-by-step explanation:

64 = 2*16 = 2*2*8 = 2*2*2*4 = 2*2*2*2*2 = 2^5

natima [27]2 years ago

3 0

Answer:

128

Step-by-step explanation:

2 | 64

2 | 32

2 | 16

2 | 8

2 | 4

2 | 2

2 | 1

Prime factorization = 2 x 2 x 2 x 2 x 2 x 2 x 2 = $2^{7\\}$ = 128

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The "Let's Roll" game uses a number cube with the numbers 2,4,6,8,10, and 12. There are prizes for rolling any number less than

deff fn [24]

There is a 33.3% chance of rolling a number less than 6. It’s 2/6, which is reduced to 1/3. = that to x/100, multiply 1 x 100, and divide that by 3. So it’s 100 divided by 3, and that’s 33.3, which is the percentage.

8 0

2 years ago

Read 2 more answers

Which term is correct when simplifying √128? A.) 16 B.) 12 C.) 10 D.) 8√2

alina1380 [7]

When dealing with roots, we want to factor to see if it's possible

$128 = 2 \times 64$
$= 2 \times 2 \times 32$
$= 2 \times 2 \times 2 \times 16$
$= 2 \times 2 \times 2 \times 2 \times 8$
$= 2 \times 2 \times 2 \times 2 \times 2 \times 4$
$= 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

now since we're dealing with square roots, we circle every pair of numbers we see in the final factoring
so..
$= (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times 2$

now for every circled pair, we take only 1 of the 2 circled and bring it outside
anything left uncircled stays inside the root
so..
$= 2 \times 2 \times 2 \times \sqrt{2}$

now simply multiply
$8 \sqrt{2}$

3 0

3 years ago

A box contains 5 red balls, 6 white balls and 9 black balls. Two balls are drawn at

valina [46]

Answer:

$P(Same)=\frac{61}{190}$

Step-by-step explanation:

Given

$Red = 5$

$White = 6$

$Black = 9$

Required

The probability of selecting 2 same colors when the first is not replaced

The total number of ball is:

$Total = 5 + 6 + 9$

$Total = 20$

This is calculated as:

$P(Same)=P(Red\ and\ Red) + P(White\ and\ White) + P(Black\ and\ Black)$

So, we have:

$P(Same)=\frac{n(Red)}{Total} * \frac{n(Red) - 1}{Total - 1} + \frac{n(White)}{Total} * \frac{n(White) - 1}{Total - 1} + \frac{n(Black)}{Total} * \frac{n(Black) - 1}{Total - 1}$