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stellarik [79]
3 years ago
14

What's the answer to the math problem 6y=8-9+6y

Mathematics
2 answers:
garik1379 [7]3 years ago
8 0
<span>6y=8-9+6y
Subtract 9 from 8
6y=-1+6y
Subtract 6y from both sides
Final Answer: -1</span>
Ad libitum [116K]3 years ago
6 0
The answer is -1 (6y=8-9+6y
6y-6y=8-9)since when we remove a num the second side we change its sign then we eliminate 6y since it became 0 and at last 8-9=-1which is the solution
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A restaurant sells curried cashew soup. On Monday the cook uses 3/5 of a bag of cashew. On Tuesday the cook uses ½ as any cashew
Vadim26 [7]

Answer:

The restaurant cooked 3/10 as many cashews on Tuesday

Step-by-step explanation:

If the word problem says that they used 3/5 of a bag of cashews on Monday, and they used 1/2 AS MANY.

So "as many" gives us a clue to multiply.

3/5 * 1/2 = 3/10

So the restaurant cooks 3/10 bag of cashews on Tuesday

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4 0
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zhuklara [117]

Answer:

m<EBD = 130 degrees

Step-by-step explanation:

We can say that Line EC is bisected by Line AD at point B.  The angle CBD = 50 degrees is a supplement of the bisected line EC.  With this in mind we can say:

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5 0
4 years ago
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A deck of ordinary cards is shuffled and 13 cards are dealt. What is the probability that the last card dealt is an ace?
alexandr1967 [171]

Answer:

The probability that the last card dealt is an ace is \frac{1}{13}.

Step-by-step explanation:

Given : A deck of ordinary cards is shuffled and 13 cards are dealt.

To find : What is the probability that the last card dealt is an ace?

Solution :

There are total 52 cards.

The total arrangement of cards is 52!.

There is 4 ace cards in total.

Arrangement for containing ace as the 13th card is 4\times 51!.

The probability that the last card dealt is an ace is

P=\frac{4\times 51!}{52!}

P=\frac{4\times 51!}{52\times 51!}

P=\frac{4}{52}

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Therefore, the probability that the last card dealt is an ace is \frac{1}{13}.

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3 years ago
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