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kvasek [131]
3 years ago
9

Ata yayincilik 4. deneme 8.sinif (2021) cevap anhtari vasa atabilirmisiniz​

Mathematics
1 answer:
aev [14]3 years ago
7 0

Answer:

what

Step-by-step explanation:

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What needs to be done to 21/10 to complete the problem? If you can, try to fix it yourself.<br><br>​
Olin [163]

Answer:

This is an improper fraction so we need to make it a mix fraction.

Step-by-step explanation:

21/10 = 10/10+10/10+1/10

10/10=1

So your final answer will be 2\frac{1}{10}

8 0
2 years ago
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What is the value of x a. 8.125 b .7.25c.6.125 d.3.25
ahrayia [7]
What's the equation with x? I can't solve it without knowing where to find x....
3 0
3 years ago
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Last one i hope (no pictures).
-BARSIC- [3]

Answer:

9

Step-by-step explanation:

RST = 90

90 - (22 + 41) = 27.

3x = 27

27/3 = 9

x = 9

8 0
3 years ago
(1 point) A tank contains 2640 L of pure water. A solution that contains 0.09 kg of sugar per liter enters the tank at the rate
OverLord2011 [107]

Answer:

t=0      Sugar = 0 Kg

t=1min Sugar=0.27 Kg

Step-by-step explanation:

Data

Tank = 2640 L (pure water)

Sol=0.09kg Sugar per liter

Vin = Vout = 3L/min

Sugar in the beginning = ?

if beginning is t = 0min Amount of sugar = 0, this is due to the fact that at the moment of entering the tank the content is only water , but if beginning is t= 1min then;

\frac{3L}{min}*\frac{0.09Kg}{L}=0.27kg/min

7 0
3 years ago
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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.

3 0
3 years ago
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