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Leokris [45]
3 years ago
9

Solve this simultaneous equation using subsitution 5x-4y=-1 2x+y=10

Mathematics
1 answer:
Morgarella [4.7K]3 years ago
3 0

Answer:

HIVNRDHGFHEHGDUOWHEDWUGUPEJKGODRFHYEHFUJDJBVCN SUFCEKGGUFVEUVFGVDBCUEGYYHDYGWEGYDCBEOGE

Step-by-step explanation:

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Solve x^2 - 3x - 10 = 0
Pachacha [2.7K]

Answer:

5 and -2

Step-by-step explanation:

{x}^{2}  - 3x - 10 = 0

{x}^{2}  - 5x + 2x - 10 = 0

x(x - 5) + 2(x - 5) = 0

(x - 5)(x + 2) = 0

x - 5 = 0

x = 5

or

x + 2 = 0

x =  - 2

6 0
2 years ago
Geoffrey's babysitter stayed from 5:15 P.M. until 11:35 P.M. For how much time
RUDIKE [14]

Answer:

6 Hours and 20 minutes

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
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
You collect 140 total pieces of candy trick or treating and 35 of them are m&m's. What fraction of the candy is not m&m'
tekilochka [14]

\frac{3}{4} of the candy is not m&m's.

Step-by-step explanation:

Given,

Total pieces of candy = 140

Number of m&m's = 35

Fraction = \frac{Number\ of\ m&m}{Total\ pieces\ of\ candy}

Fraction = \frac{35}{140}

Both 35 and 140 are multiples of 7, therefore,

Fraction of m&m's = \frac{5}{20} = \frac{1}{4}

As the number of m&m's fraction and not m&m's fraction will make a total of 1, therefore

fraction of m&m's + fraction of not m&m's = 1

fraction of not m&m's = 1 - fraction of m&m's

fraction of not m&m's = 1-\frac{1}{4}=\frac{4-1}{4}

fraction of not m&m's = \frac{3}{4}

\frac{3}{4} of the candy is not m&m's.

Keywords: fraction, subtraction

Learn more about fractions at:

  • brainly.com/question/8929610
  • brainly.com/question/8908016

#LearnwithBrainly

7 0
3 years ago
9.3x10^7 - 3.4x10^6=
Salsk061 [2.6K]
Loook on google and find the answers
7 0
3 years ago
Read 2 more answers
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