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dlinn [17]
3 years ago
13

4x^2 +28x -32 =0 solve this problem?

Mathematics
2 answers:
Softa [21]3 years ago
5 0
4x^2+28x-32=0 \ \ \ |:4 \\\\ x^2+7x-8=0 \\\\ x^2+8x-x-8=0 \\\\ x(x+8)-(x+8)=0 \\\\ (x+8)(x-1)=0 \\\\ \\x+8=0 \\\\ \boxed{x=-8} \\\\\\ x-1=0 \\\\ \boxed{x=1} \\\\\\ \boxed{\boxed{S=\{-8;1 \}}}
erastovalidia [21]3 years ago
4 0
We know that if xy=0 then x or/and y=0


4x^2+28x-32=0
factor out the 4
4(x^2+7-8)=0

factor out the x^2+7-8
we must find out what two numbers multiply to get -8 and add to get 7
the numbers are 8 and -1

so 4[(x-1)(x+8)]=0

we know that if exg 4y=0 then y=0
so
therefor
(x-1)(x+8)=0
set each to zero
x-1=0
x=1
x+8=0
x=-8

x could be 1 or -8
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4 0
3 years ago
Eights rooks are placed randomly on a chess board. What is the probability that none of the rooks can capture any of the other r
erastova [34]

Answer:

The probability is \frac{56!}{64!}

Step-by-step explanation:

We can divide the amount of favourable cases by the total amount of cases.

The total amount of cases is the total amount of ways to put 8 rooks on a chessboard. Since a chessboard has 64 squares, this number is the combinatorial number of 64 with 8, 64 \choose 8 .

For a favourable case, you need one rook on each column, and for each column the correspondent rook should be in a diferent row than the rest of the rooks. A favourable case can be represented by a bijective function  f : A \rightarrow A , with A = {1,2,3,4,5,6,7,8}. f(i) = j represents that the rook located in the column i is located in the row j.

Thus, the total of favourable cases is equal to the total amount of bijective functions between a set of 8 elements. This amount is 8!, because we have 8 possibilities for the first column, 7 for the second one, 6 on the third one, and so on.

We can conclude that the probability for 8 rooks not being able to capture themselves is

\frac{8!}{64 \choose 8} = \frac{8!}{\frac{64!}{8!56!}} = \frac{56!}{64!}

7 0
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Answer:

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