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F1Krazy
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The answer is

! Letter e).Look at each figure individually. Pattern: If figure is repeated in same figure(appears twice) you use subtraction .If figure isnt repeated you use adition.

E. Look at each figure individually. The pattern is: If a symbol is repeated in the same figure (appears twice), you use subtraction. If a symbol isn't repeated, you use addition.

The answer is

! Letter e).Look at each figure individually. Pattern: If figure is repeated in same figure(appears twice) you use subtraction .If figure isnt repeated you use adition.

The answer is

E. Look at each figure individually. The pattern is: If a symbol is repeated in the same figure (appears twice), you use subtraction. If a symbol isn't repeated, you use addition.

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The answer is

Letter e).

Because in each row or column, the following rule is followed:

When a symbol appears more than once in any box, the numbers of that symbol in the first and second boxes are subtracted. In the first row, 2 clubs minus 1 club = 1 club (because in this row, club IS repeated). Next row, same thing: clubs repeated and their numbers subtracted.

When a symbol appears only once, the numbers of that symbol in the first and second boxes are added. In the first and second rows, spades appear only once in each box, so it's addition: 1 spade +1 spade = 2 spades.

This same rule works in each row and each column, so:

look at the last row. Spade appears repeated, so it's subtracted: 2 spades - 2 spades = 0 spades. Club appears only once in figure, so it's added: 1 club + 1 club = 2 clubs.

! Letter e).Look at each figure individually. Pattern: If figure is repeated in same figure(appears twice) you use subtraction .If figure isnt repeated you use adition.

The answer is

Letter e).

Because in each row or column, the following rule is followed:

When a symbol appears more than once in any box, the numbers of that symbol in the first and second boxes are subtracted. In the first row, 2 clubs minus 1 club = 1 club (because in this row, club IS repeated). Next row, same thing: clubs repeated and their numbers subtracted.

When a symbol appears only once, the numbers of that symbol in the first and second boxes are added. In the first and second rows, spades appear only once in each box, so it's addition: 1 spade +1 spade = 2 spades.

This same rule works in each row and each column, so:

look at the last row. Spade appears repeated, so it's subtracted: 2 spades - 2 spades = 0 spades. Club appears only once in figure, so it's added: 1 club + 1 club = 2 clubs.

The answer is

! Letter e).Look at each figure individually. Pattern: If figure is repeated in same figure(appears twice) you use subtraction .If figure isnt repeated you use adition.

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Kate Gregory
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The answer is

Letter e).

Because in each row or column, the following rule is followed:

! When a symbol appears more than once in any box, the numbers of that symbol in the first and second boxes are subtracted. In the first row, 2 clubs minus 1 club = 1 club (because in this row, club IS repeated). Next row, same thing: clubs repeated and their numbers subtracted.
Thank you Vassilis ! When a symbol appears only once, the numbers of that symbol in the first and second boxes are added. In the first and second rows, spades appear only once in each box, so it's addition: 1 spade +1 spade = 2 spades.

When a symbol appears more than once in any box, the numbers of that symbol in the first and second boxes are subtracted. In the first row, 2 clubs minus 1 club = 1 club (because in this row, club IS repeated). Next row, same thing: clubs repeated and their numbers subtracted.

When a symbol appears only once, the numbers of that symbol in the first and second boxes are added. In the first and second rows, spades appear only once in each box, so it's addition: 1 spade +1 spade = 2 spades.

This same rule works in each row and each column, so:

! look at the last row. Spade appears repeated, so it's subtracted: 2 spades - 2 spades = 0 spades. Club appears only once in figure, so it's added: 1 club + 1 club = 2 clubs. We are heare for puzzling,not tô talk...its a complexo problem and people wanna simple answers.Lol hilarious

look at the last row. Spade appears repeated, so it's subtracted: 2 spades - 2 spades = 0 spades. Club appears only once in figure, so it's added: 1 club + 1 club = 2 clubs.

The answer is

Letter e).

Because in each row or column, the following rule is followed:

! When a symbol appears more than once in any box, the numbers of that symbol in the first and second boxes are subtracted. In the first row, 2 clubs minus 1 club = 1 club (because in this row, club IS repeated). Next row, same thing: clubs repeated and their numbers subtracted.
Thank you Vassilis ! When a symbol appears only once, the numbers of that symbol in the first and second boxes are added. In the first and second rows, spades appear only once in each box, so it's addition: 1 spade +1 spade = 2 spades.

This same rule works in each row and each column, so:

! look at the last row. Spade appears repeated, so it's subtracted: 2 spades - 2 spades = 0 spades. Club appears only once in figure, so it's added: 1 club + 1 club = 2 clubs. We are heare for puzzling,not tô talk...its a complexo problem and people wanna simple answers.Lol hilarious

The answer is

Letter e).

Because in each row or column, the following rule is followed:

When a symbol appears more than once in any box, the numbers of that symbol in the first and second boxes are subtracted. In the first row, 2 clubs minus 1 club = 1 club (because in this row, club IS repeated). Next row, same thing: clubs repeated and their numbers subtracted.

When a symbol appears only once, the numbers of that symbol in the first and second boxes are added. In the first and second rows, spades appear only once in each box, so it's addition: 1 spade +1 spade = 2 spades.

This same rule works in each row and each column, so:

look at the last row. Spade appears repeated, so it's subtracted: 2 spades - 2 spades = 0 spades. Club appears only once in figure, so it's added: 1 club + 1 club = 2 clubs.

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Rand al'Thor
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