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Lady_Fox [76]
3 years ago
6

Which numbers are a distance of 4 units from 7 on the number line?

Mathematics
1 answer:
Dominik [7]3 years ago
3 0

Answer:

11

Step-by-step explanation:

The distance is 4 unit so we should count 4 counts starting from number 7 on the number line

Depending on that the answer should be :

7+4=11

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Answer:

2 is the opposite of-2 and -2 is the opposite of 2

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A jet ski rental charges A flat rate of $75 ,plus an additional $60 per hour you want to spend less than $200. for how many hour
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The equation to solve would be setting

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8 0
3 years ago
4(5x + 9)^2 – 33 = -21
ozzi

Answer:

x = sqrt(3)/5 - 9/5 or x = -9/5 - sqrt(3)/5

Step-by-step explanation by completing the square:

Solve for x over the real numbers:

4 (5 x + 9)^2 - 33 = -21

Add 33 to both sides:

4 (5 x + 9)^2 = 12

Divide both sides by 4:

(5 x + 9)^2 = 3

Take the square root of both sides:

5 x + 9 = sqrt(3) or 5 x + 9 = -sqrt(3)

Subtract 9 from both sides:

5 x = sqrt(3) - 9 or 5 x + 9 = -sqrt(3)

Divide both sides by 5:

x = sqrt(3)/5 - 9/5 or 5 x + 9 = -sqrt(3)

Subtract 9 from both sides:

x = sqrt(3)/5 - 9/5 or 5 x = -9 - sqrt(3)

Divide both sides by 5:

Answer:  x = sqrt(3)/5 - 9/5 or x = -9/5 - sqrt(3)/5

7 0
3 years ago
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1+2-3+4+5-6+7+8-9...+97+98-99
zaharov [31]

Answer:

  1584

Step-by-step explanation:

The sum of this sequence can be found a number of ways. One way is to recast it as the series whose terms are groups of three terms of the given series.

__

<h3>series of partial sums</h3>

The partial sums, taken 3 terms at a time, are

  1+2-3 = 0

  4+5-6 = 3

  7+8-9 = 6

...

  97+98-99 = 96

So the original series is equivalent to ...

  0 +3 +6 +... +96 = 3×1 +3×2 +... +3×32 = 3×(1 +2 +... +32)

That is, the sum is 3 times the sum of the consecutive integers 1..32.

__

<h3>consecutive integers</h3>

The sum of integers 1..n is given by the equation ...

  s(n) = n(n+1)/2

__

<h3>series sum</h3>

Using this to find the sum of our series, we find it to be ...

  series sum = 3 × (32)(33)/2 = 1584

_____

<em>Alternate solution</em>

The given series is the sum of integers 1-99, with 6 times the sum of integers 1-33 subtracted. That is, ...

  1 + 2 - 3 + 4 + 5 - 6 = 1+2+3+4+5+6 -2(3 +6) = 1+2+3+4+5+6 -6(1+2)

Continuing on to ...97 +98 -99 gives the result s(99) -6s(33).

Computed that way, we find the sum to be ...

  (99)(100)/2 -6(33)(34)/2 = 4950 -3366 = 1584

3 0
2 years ago
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