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Katen [24]
4 years ago
13

2(m + 2h - 2) = Calling all helpers for this question lol?

Mathematics
2 answers:
Alekssandra [29.7K]4 years ago
6 0
The answer is 2m+4h-4. there are no liked terms so you can’t simplify anymore

AURORKA [14]4 years ago
3 0

Answer:

4h+2m−4

Step-by-step explanation:

=(2)(m+2h+−2)

=(2)(m)+(2)(2h)+(2)(−2)

=2m+4h−4

=4h+2m−4

 

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If eggs in the basket are removed two at a time, one egg will remain. If the eggs are removed three at a time, 2 eggs will remai
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<h2>Answer:</h2>

<em><u>Smallest number of eggs in the basket = 119</u></em>

<h2>Step-by-step explanation:</h2>

In the question,

We know that the number of eggs in the basket should be a a multiple of 7.

But it can not be a multiple of 2, 3, 4, 5 and 6 because every time we pick up the eggs 2, 3, 4, 5 or 6 at a time we are left with some eggs with us.

Therefore, the number of eggs can not be a multiple of these numbers.

Now,

Let us say the number of eggs in the basket be 7x.

So,

Let us take the LCM of 2, 3, 4, 5 and 6.

So,

LCM = 60

Now, the number would be greater than 60 and the multiple of 7.

So, checking on all the multiples of 7 above 60 and checking the condition that, the remainder left on dividing by,

2 is 1. (2x + 1)

by 3 is 2. (3x + 2)

by 4 is 3. (4x + 3)

by 5 is 4. (5x + 4)

by 6 is 5. (6x + 5)

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On Checking the multiples of 60 which are divisible by 7 are ,

60 + 1 , 120 + 1, 60 - 1, 120 - 1.

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For 61 it is not satisfying all the conditions.

121 is also not satisfying all the conditions.

59 is also not satisfying all the conditions.

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The number 119 on checking its divisibility by 2. Leaves remainder 1 as, (2(59) + 1).

Divisibility by 3 leaves remainder 2 as, {3(39) + 2}.

Divisibility by 4 leaves remainder 3 as, {4(29) + 3}.

Divisibility by 5 leaves remainder 4 as, {5(23) + 4}.

Divisibility by 6 leaves remainder 5 as, {6(19) + 5}.

Divisibility by 7 leaves remainder 0 as, {7(17) + 0}.

<em><u>Therefore, the minimum number of eggs in the basket are 119.</u></em>

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4 years ago
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