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kodGreya [7K]
3 years ago
5

Hey yall like dat sweet algeba? then 5ab - 2c where a = 2 b = 3 and c is 0. gud luck daddio

Mathematics
1 answer:
nordsb [41]3 years ago
7 0

Answer:

30

Step-by-step explanation:

5ab-2c

5(2)(3)-2(0)

5(6)-2(0)

30-0

30

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The square with side length 2 cm is dilated by a scale factor of 7/3 is the dilated image larger or smaller than the original im
qwelly [4]
Length of square side= 2 cm
Dilation factor= 7/3
Simplify 7/3= 2.33
Apply the dilation factor:
Length of side= 2.33 x 2 = 4.67 cm
As the length of the side of the square is increased in length, which means the dilated image is larger than the original image.

Answer: Larger than then original. 
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In a class there are 15 students. 8 of them like playing soccer, 6 of them like swimming, and 2 like both and swimming and playi
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Answer:

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When using rational expectations, forecast errors will, on average, be ________ and ________ be predicted ahead of time. A) zero
mafiozo [28]

Answer:

A) zero; cannot

Step-by-step explanation:

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3 years ago
According to DeMorgan 's theorem, the complement of W · X + Y · Z is W' + X' · Y' + Z'. Yet both functions are 1 for WXYZ = 1110
sergeinik [125]

Answer:

The parenthesis need to be kept intact while applying the DeMorgan's theorem on the original equation to find the compliment because otherwise it will introduce an error in the answer.

Step-by-step explanation:

According to DeMorgan's Theorem:

(W.X + Y.Z)'

(W.X)' . (Y.Z)'

(W'+X') . (Y' + Z')

Note that it is important to keep the parenthesis intact while applying the DeMorgan's theorem.

For the original function:

(W . X + Y . Z)'

= (1 . 1 + 1 . 0)

= (1 + 0) = 1

For the compliment:

(W' + X') . (Y' + Z')

=(1' + 1') . (1' + 0')

=(0 + 0) . (0 + 1)

=0 . 1 = 0

Both functions are not 1 for the same input if we solve while keeping the parenthesis intact because that allows us to solve the operation inside the parenthesis first and then move on to the operator outside it.

Without the parenthesis the compliment equation looks like this:

W' + X' . Y' + Z'

1' + 1' . 1' + 0'

0 + 0 . 0 + 1

Here, the 'AND' operation will be considered first before the 'OR', resulting in 1 as the final answer.

Therefore, it is important to keep the parenthesis intact while applying DeMorgan's Theorem on the original equation or else it would produce an erroneous result.

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