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snow_lady [41]
3 years ago
6

What to do when u have improper fraction

Mathematics
2 answers:
siniylev [52]3 years ago
5 0
Divide the top number with the bottom number look at my example for help

Elina [12.6K]3 years ago
4 0
Turn it into a mixed fraction
Hope this helped :)
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For the arthritic sequence 5,9,13,17,21,... how many terms would you need to get to 129
Vlad1618 [11]

9514 1404 393

Answer:

  32

Step-by-step explanation:

The first term is 5 and the common difference is 9-5=4, so the n-th term is ...

  an = a1 +d(n -1)

  an = 5 +4(n -1) = 4n +1

Then the term number for the term 129 is ...

  129 = 4n +1 . . . . . put the given value in the formula

  128 = 4n  . . . . . . . subtract 1

  32 = n . . . . . . . . . .divide by 4

The 32nd term is 129.

8 0
3 years ago
How much money left after spending $8, $7 and $13, then received $10 and spent $15. now $12 is left. how much money did one have
Inessa [10]
$8 + $7 + $13 + 15 = $43
$x - $43 = $12
x = $55

i am a mathematics teacher. if anything to ask please pm me
4 0
3 years ago
Your school district cancels schools when more than 2.5 inches of snow accumulate on the ground each hour. Snow is currently acc
kari74 [83]

Answer:

Extra snow needed > 0. 3 inch

Step-by-step explanation:

Given that:

Cancelation occurs when > 2.5 inch of snow accumulates each hour

Accumulation rate = 11/5 inch per hour

Extra snow Accumulation needed to cancel. School ;

Extra snow needed > cancelation volume - current rate

Extra snow needed > (2.5 - 11/5)

Extra snow needed > 0. 3 inch

5 0
3 years ago
It took Jaheim 2 hours to read the first 100 pages of a book what was his speed?
jeka94
If in 2 hours he read 200 pgs then in 1 hour will be 50 pgs
4 0
3 years ago
Read 2 more answers
Find the area between the two functions
Inga [223]

The area between the two functions is 0

<h3>How to determine the area?</h3>

The functions are given as:

f₁(x)= 1

f₂(x) = |x - 2|

x ∈ [0, 4]

The area between the functions is

A = ∫[f₂(x) - f₁(x) ] dx

The above integral becomes

A = ∫|x - 2| - 1 dx (0 to 4)

When the above is integrated, we have:

A = [(|x - 2|(x - 2))/2 - x] (0 to 4)

Expand the above integral

A = [(|4 - 2|(4 - 2))/2 - 4] - [(|0 - 2|(0 - 2))/2 - 0]

This gives

A = [2 - 4] - [-2- 0]

Evaluate the expression

A = 0

Hence, the area between the two functions is 0

Read more about areas at:

brainly.com/question/14115342

#SPJ1

5 0
1 year ago
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