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Kamila [148]
3 years ago
6

What's 63,889 rounded to the nearest hundred

Mathematics
2 answers:
levacccp [35]3 years ago
4 0
63,889 rounded to the nearest hundred is 63,900
Jlenok [28]3 years ago
4 0
The hundred's place here is 8, if you look at the number before 8 it's also 8 which is larger than 5, so you round the hundred's place 8 up to a 9. Now you have your answer. 63,900
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Help!!!!!!!!!!!!!!!!!!!!!
stealth61 [152]
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7 0
3 years ago
If y= x+3/6-x, what is the value of y when x =7i?
djverab [1.8K]

Answer:

Option (2)

Step-by-step explanation:

Given :  y = \frac{x+3}{6-x}

If x = 7i

y = \frac{7i+3}{6-7i}

By simplifying denominator of the given rational expression,

y = \frac{7i+3}{6-7i}\times \frac{6+7i}{6+7i}

y = \frac{(7i+3)(6+7i)}{6^2-(7i)^2}

y = \frac{7i(6+7i)+3(6+7i)}{36-49i^2}

y = \frac{42i+49i^2+18+21i}{36+49} [Since, i² = (-1)]

y = \frac{63i-49+18}{85}

y = \frac{63i-31}{85}

y = -\frac{31}{85}+\frac{63}{85}i

Therefore, Option (2) is the correct option.

3 0
3 years ago
let m represent the number of children playing soccer. those children are separated into four equal groups. write an expression
Mkey [24]

Answer:

m/4

Step-by-step explanation:

m is how many children are playing.

They are separated into 4 groups. Making it m(how many people are playing), out of 4.

For example, if m=12, then it would be 12/4, so 3 children will be in each group.

Hope this helps, you can change this to you own desire.

4 0
3 years ago
Find the 52nd number in the sequence<br> 9, 16, 23....
BartSMP [9]

Answer:

366

starts at 9, jumps by 7 each time.

so it's 9 plus 7 times the number if remaining jumps.

51x7= 357

add the origiinal 9, and you get 366

3 0
3 years ago
Use the principle of inclusion and exclusion to find the number of positive integers less than 1,000,000 that are not divisable
wel
This is a simple problem based on combinatorics which can be easily tackled by using inclusion-exclusion principle.
We are asked to find number of positive integers less than 1,000,000 that are not divisible by 6 or 4.
let n be the number of positive integers.
∴ 1≤n≤999,999
Let c₁ be the set of numbers divisible by 6 and c₂ be the set of numbers divisible by 4.
Let N(c₁) be the number of elements in set c₁ and N(c₂) be the number of elements in set c₂.

∴N(c₁) = \frac{999,999}{6} = 166666
N(c₂) = \frac{999,999}{4} = 250000
∴N(c₁c₂) = \frac{999,999}{24} = 41667
∴ Number of positive integers that are not divisible by 4 or 6,

N(c₁`c₂`) = 999,999 - (166666+250000) + 41667 = 625000
Therefore, 625000 integers are not divisible by 6 or 4
8 0
4 years ago
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