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Mandarinka [93]
3 years ago
11

Explain how to solve 5x2-3x=25 by completing the square. What are the solutions

Mathematics
1 answer:
Vilka [71]3 years ago
4 0

5x2-3x=25

Divide each term by 5:

x^2 - 3/5x = 5

Make a trinomial square of the left side by taking the square of half of the coefficient of x.

The coefficient of x is -3, so the square would be (-3/10)^2

Add that to both sides of the equation.

Now you have:

x^2 - 3/5x + (-3/10)^2 = 5 + (-3/10)^2

Simplify both sides:

x^2 -3/5x +9/100 = 509/100

Factor the trinomial:

(x-3/10)^2 = 509/100

Solve for x, , take the square root of both sides:

x-3/10 = +/-√509 / 100

Simplify the right side:

x-3/10 = +/-√509/10

Add 3/10 to both sides:

X = +/-√509/10 + 3/10 ( this is the exact form)

You can then find the decimal answer if needed.

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Answer:

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2. He will be able to make 8 pompoms

Step-by-step explanation:

1.

4 x 1.6 = 5.6 meters were used to make decorations.

Then we do 12 - 5.6 = 6.4

2.

per one pompom it takes 0.8 meters of streamers.

6.4 is the left over so we do

6.4 divided by 0.8 = 8 pompoms you can make

6 0
2 years ago
There are 24 pairs of shoes in a closet. If 25% of the pairs of shoes are black, how many pairs are NOT black?
Elden [556K]
25%=0.25
24*0.25=6
24-6=18
18 pairs are not black
7 0
3 years ago
Read 2 more answers
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
3 years ago
How much do you make annually at 15 dollars an hour 23 hours a week?
ahrayia [7]
You multiply $15 by 23 which is 345. Then you multiply 345 by 52 for there is 52 in a year. The answer $17940.
4 0
3 years ago
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Among 500 freshmen pursuing a business degree at a university, 315 are enrolled in an economics course, 213 are enrolled in a ma
scoundrel [369]

Answer:

Step-by-step explanation:

Given that among 500 freshmen pursuing a business degree at a university, 315 are enrolled in an economics course, 213 are enrolled in a mathematics course, and 123 are enrolled in both an economics and a mathematics course.

From the above we find that

a) either economics of Math course is

315+213-123 =405

Out of 500 students 405 have taken either Math or Economics

Hence

c) student who have taken neither = 500-405 =95

Exactly one course is either math or economics - both

= 405-123 = 282

3 0
3 years ago
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