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3 years ago

0.000000000051304 in scientific notation? Ixl

Mathematics

1 answer:

Grace [21]3 years ago

5 0

Answer:

5.1304 x 10^14

Step-by-step explanation:

there is 10 zeros behind the decimal point plus 4 other decimal places

in a scientific notation the number has to be less then 10 but more then 1

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I need the answers for these two questions

Sindrei [870]

Answer:

$29.14

$86.68

Step-by-step explanation:

$30 before tax

Discount: 20%

100% - 20% = 80% = 0.8

0.8 * $30 = $24 after discount

Tax: 6.4%

100% + 6.4% = 106.4% = 1.064

1.604 * $24 = 25.54 after tax and discount

Tip: 15% = 0.15

0.15 * $24 = $3.60

Total = $25.54 + $3.60 = $29.14

$75 before tax

10% discount

100% - 10% = 90%

90% of $75 = 0.9 * $75 = $67.50 after discount

Tax: 7%

100% + 7% = 107% = 1.07

1.07 * $67.5 = $72.23 after discount and tax

Tip: 20% = 0.2

0.2 * $67.50 = $13.50

Total = $72.23 + $13.50 = $85.73

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3 years ago

Chance knows the length of a football field (including end zones) is 120 yards. The width of the field is 53.33 yards. What is t

givi [52]

Answer = about 6,400 yards squared

Step-by-step explanation:

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4 years ago

Please Help the image is below with the question!

jok3333 [9.3K]

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3 years ago

Aaron wants to buy a bicycle that costs $128. So far he has saved $56 the equation of 8 + 56 equal 128 can be used to find the a

lutik1710 [3]

Answer:

Aaron still needs to save $9

Step-by-step explanation:

Assuming the equation is 8a + 56 = 128

8a + 56 = 128

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8a = 72

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4 years ago

List all the ways to select two members from S with repetition. The order in which the members are selected is not important. Fo

nexus9112 [7]

Answer:

15 ways

Step-by-step explanation:

We are given that a set S={E,F,G,H,J}

We have to find the number of ways to select two members from S with repetition.

Combination formula

$\binom{n}{r}=\frac{n!}{r!(n-r)!}$

We have n=5 , r= 2

Number of ways in which two members from S can be selected when repetition is not allowed= $5C_2=\frac{5!}{2!(5-2)!}$

Number of ways in which two members from S can be selected when repetition is not allowed= $\frac{5\times4\times3!}{2\times1 3!]$

Number of ways in which two members from S can be selected when repetition is not allowed= $5\times 2=10$

When a member repeat then combination

{E,E},{F,F},{G,G},{H,H},{J,J}

There are 5 combination when a member is repeat and select two members from S.

Total number of ways in which to select two members from S with repetition =10+5=15 ways

3 0

4 years ago