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NeTakaya
3 years ago
5

What are the last 2 digits of 7 to the power of 2018. This question was on a past exam paper.

Mathematics
1 answer:
Xelga [282]3 years ago
6 0

Answer:

49

Step-by-step explanation:

  • Try finding pattern of last two digits of 7 to the power of some numbers.

7^1=07

7^2=49

7^3=343

7^4= 2301

7^5= 16807

  • so , the pattern is, 07,49,43,01
  • so, if we start writing all numbers from 1 to 2018 , four number in each line, 2018 will fall in the second column.
  • so it will have 49 as the  last 2 digits [by seeing this pattern : 07,49,43,01 ]
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Step-by-step explanation:

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Jim wishes to buy 3 gifts that cost 15 dollars, 9 dollars, and 12 dollars. he has 1/4 of the money he needs, how much more money
Archy [21]
In order to solve this, you have to plug in all of the costs of the gifts and add them, which in all equals $36. since jim only has 1/4 of the money he needs to buy the gifts, you need to divide 36 by 4. this gives you 9, so now you know he has 9 dollars.
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3 years ago
Solve the system of equations by the addition method <br> { 5x-y=-3<br> { 4x+3y=-48
Svetach [21]

Answer:

(-3,-12)

Step-by-step explanation:

Multiply the first equation by three to match the second equations y-term

  • 3(5x-y=-3)
  • 4x+3y=-48

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  • 4x+3y=-48

The y-terms cancel out

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  • 4x=-48

Add the x-terms with each other and the terms after the equal sign as well

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Divide -57 by 19

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Input your x-term (-3) into one of the two first equations and multiply

Either:

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  • -15-y=-3 or
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Simplify your terms

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Divide

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3 years ago
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Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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