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kenny6666 [7]
3 years ago
7

Solve by elimination 6x-5y=27 3x+10y=-24

Mathematics
1 answer:
Anna [14]3 years ago
4 0

Answer:

point form: (-34/3, -19)

equation form: X= - 34/3, Y = -19

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20 POINTS !!!!<br> The roots of the equation x^2+bx+c=0 are −4 and 2. Find b.
Irina18 [472]

first, love the killua profile picture ! i love hunterxhunter i am currently on season 5, episode 44 or 45 aka 119 or 120 but anyways,

answer:

b = 2

step-by-step explanation:

because its roots are -4 and 2,

and a = 1 [The coefficient of x^2] we know that x^2 + bx + c = (x - (-4)) (x-2)=(x+4)(x-2) = 0.

using the distributive property we get,

x^2 + 2x - 8 = 0.

by equating coefficients (common strategy for partial fractions and undetermined coefficient method of solving ODE), we get b = 2 and c = -8 ( solved c just for fun guess :) ) .

hope this helps :)  

<em>-audrey <3</em>

5 0
3 years ago
-27=2x^2-4 PLEASE HELP ASAP
Nesterboy [21]
X = [ (i sqrt46) / (2) ] ; - [ (i sqrt46) / (2) ]
6 0
3 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
1. Convert the quinary number 42 into binary number.​
choli [55]

Answer:

42 in binary number is 101010.

7 0
3 years ago
Find the factors of 9,24,30,48
Alenkasestr [34]
3*3
6*4
6*5
6*8? i guessss
3 0
3 years ago
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