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mixas84 [53]
3 years ago
14

Find three consecutive even integers such that the sum of the smallest number and twice the middle number is 20 more than the la

rgest number.
Mathematics
1 answer:
motikmotik3 years ago
6 0

Answer:

<h3>              10, 12, 14</h3>

Step-by-step explanation:

Consecutive even integers increase by 2.

x   ← an integer

2x  ← an even integer (the smallest)

2x+2   ← next even integer consecutive to 2x (the middle one)

2x+2+2=2x+4  ← the last even integer consecutive to 2x+2 (the largest)

2(2x+2)    ← twice the middle number

2x+2(2x+2)   ←  the sum of the smallest number and twice the middle

2x+4+20   ←  20 more than the largest number

2x + 2(2x + 2) = 2x + 4 + 20

2x + 4x + 4 = 2x + 24

            -4           -4

            6x = 2x + 20

         -2x       -2x

            4x = 20

            ÷2     ÷2

            2x = 10

2x+2 = 10+2 = 12

2x+4 = 10+4 = 14

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2 years ago
twice the number of donuts was only 6 less than 24 times the number of cream puffs. 10 times the number of cream puffs was only
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donuts and cream puffs were  45 & 4 respectively !

<u>Step-by-step explanation:</u>

Here we have , twice the number of donuts was only 6 less than 24 times the number of cream puffs. 10 times the number of cream puffs was only 5 less than the number of donuts   We need to find how many donuts and cream puffs were there . Let's find out:

Let number of donuts and cream puffs are x & y respectively ! So ,

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According to this statement equation is :

⇒ 2x=24y-6

⇒ x=12y-3            ................(1)

  • 10 times the number of cream puffs was only 5 less than the number of donuts

According to this statement equation is :

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3 years ago
10bf + 25bg – 21p – 14pq<br> Quadratic factorisation
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The problem 10bf + 25bg – 21p – 14pq cannot be factored because it cannot be simplifed further. This is because there are no common factors between in any of the terms.

6 0
3 years ago
It’s easy please help please please
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Step-by-step explanation:

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