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DiKsa [7]
3 years ago
5

Solve 2x - 8 < 7.

Mathematics
1 answer:
professor190 [17]3 years ago
7 0

Answer:

<h2>{x | x < 15/2}</h2>

Step-by-step explanation:

2x - 8 < 7            <em>add 8 to both sides</em>

2x - 8 + 8 < 7 + 8

2x < 15       <em>divide both sides by 2</em>

2x : 2 < 15 : 2

x < 15/2

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I need help fast I will give brainliest ​
almond37 [142]

Answer:

7

Step-by-step explanation:

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8 0
3 years ago
Read 2 more answers
Please answer the question from the attachment.
grin007 [14]
Since all of the variables are y^3, all we have to do is the addition and subtraction of the whole numbers...

2y^3 - 3y^3 + 3y^3 - 4y^3 =

-2y^3

***Notice that the -3y^3 and 3y^3 just cancel out so all you have to do is the
2-4 = -2

Piece of cake :)
8 0
3 years ago
HELP PLEASE!!! I need help with 94 if you could show the steps that would be very helpful!
aksik [14]
A combination is an unordered arrangement of r distinct objects in a set of n objects. To find the number of permutations, we use the following equation:

n!/((n-r)!r!)

In this case, there could be 0, 1, 2, 3, 4, or all 5 cards discarded. There is only one possible combination each for 0 or 5 cards being discarded (either none of them or all of them). We will be the above equation to find the number of combination s for 1, 2, 3, and 4 discarded cards.

5!/((5-1)!1!) = 5!/(4!*1!) = (5*4*3*2*1)/(4*3*2*1*1) = 5

5!/((5-2)!2!) = 5!/(3!2!) = (5*4*3*2*1)/(3*2*1*2*1) = 10

5!/((5-3)!3!) = 5!/(2!3!) = (5*4*3*2*1)/(2*1*3*2*1) = 10

5!/((5-4)!4!) = 5!/(1!4!) = (5*4*3*2*1)/(1*4*3*2*1) = 5

Notice that discarding 1 or discarding 4 have the same number of combinations, as do discarding 2 or 3. This is being they are inverses of each other. That is, if we discard 2 cards there will be 3 left, or if we discard 3 there will be 2 left.

Now we add together the combinations

1 + 5 + 10 + 10 + 5 + 1 = 32 choices combinations to discard.

The answer is 32.

-------------------------------

Note: There is also an equation for permutations which is:

n!/(n-r)!

Notice it is very similar to combinations. The only difference is that a permutation is an ORDERED arrangement while a combination is UNORDERED.

We used combinations rather than permutations because the order of the cards does not matter in this case. For example, we could discard the ace of spades followed by the jack of diamonds, or we could discard the jack or diamonds followed by the ace of spades. These two instances are the same combination of cards but a different permutation. We do not care about the order.

I hope this helps! If you have any questions, let me know :)








7 0
3 years ago
In the problem 19.28 - 1.546, first round each number to the nearest tenth. Then subtract.​
goblinko [34]

Answer:

17.8

Step-by-step explanation:

First, round the two. umbers to the nearest the th, or the first number past the decimal.  19.28 rounds to 19.3 (round up because 8 is bigger than 5) and 1.546 rounds to 1.5 (round down because 4 is smaller than 5).  Then subtract.  19.3 - 1.5 = 17.8

8 0
3 years ago
Someone help please!! :)
LUCKY_DIMON [66]

Answer:

First one

Step-by-step explanation:

BECAUSE I JUST KNOW OKAY!!! STOP ASKING HARD QUESTIONS BRAINLY!

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