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3 years ago

jason wants to choose 9 players for his track team ,there are 12 players to choose from.how many different teams can jason make?

2 answers:

6 0

He could only make one team, considering that if you do 12-9, you are left with 3, and Jason wants to have his team consisted of 9 players. However, if you make 3 teams with 4 players on each team, or 4 teams with 3 players on each team, you could do that as well.

marin [14]3 years ago

5 0

Answer: 220 team can Jason make.

Step-by-step explanation:

Since, the total numbers of players = 12

The number of players in one team = 9,

Hence, the total combination of choosing 9 players out of 12 players

$=^12C_9$

$=\frac{12!}{9!\times (12-9)!$

$=\frac{12\times 11\times 10\times 9!}{9!\times 3!}$

$=\frac{12\times 11\times 10}{3\times 2}$

$=\frac{1320}{6}=220$

Thus, 220 team can Jason make.

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Solving Rational Functions Hello I'm posting again because I really need help on this any help is appreciated!!

Greeley [361]

Answer:

x = √17 and x = -√17

Step-by-step explanation:

We have the equation:

$\frac{3}{x + 4} - \frac{1}{x + 3} = \frac{x + 9}{(x^2 + 7x + 12)}$

To solve this we need to remove the denominators.

Then we can first multiply both sides by (x + 4) to get:

$\frac{3*(x + 4)}{x + 4} - \frac{(x + 4)}{x + 3} = \frac{(x + 9)*(x + 4)}{(x^2 + 7x + 12)}$

$3 - \frac{(x + 4)}{x + 3} = \frac{(x + 9)*(x + 4)}{(x^2 + 7x + 12)}$

Now we can multiply both sides by (x + 3)

$3*(x + 3) - \frac{(x + 4)*(x+3)}{x + 3} = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}$

$3*(x + 3) - (x + 4) = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}$

$(2*x + 5) = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}$

Now we can multiply both sides by (x^2 + 7*x + 12)

$(2*x + 5)*(x^2 + 7x + 12) = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}*(x^2 + 7x + 12)$

$(2*x + 5)*(x^2 + 7x + 12) = (x + 9)*(x + 4)*(x+3)$

Now we need to solve this:

we will get

$2*x^3 + 19*x^2 + 59*x + 60 = (x^2 + 13*x + 3)*(x + 3)$

$2*x^3 + 19*x^2 + 59*x + 60 = x^3 + 16*x^2 + 42*x + 9$

Then we get:

$2*x^3 + 19*x^2 + 59*x + 60 - ( x^3 + 16*x^2 + 42*x + 9) = 0$

$x^3 + 3x^2 + 17*x + 51 = 0$

So now we only need to solve this.

We can see that the constant is 51.

Then one root will be a factor of 51.

The factors of -51 are:

-3 and -17

Let's try -3

p( -3) = (-3)^3 + 3*(-3)^2 + +17*(-3) + 51 = 0

Then x = -3 is one solution of the equation.

But if we look at the original equation, x = -3 will lead to a zero in one denominator, then this solution can be ignored.

This means that we can take a factor (x + 3) out, so we can rewrite our equation as:

$x^3 + 3x^2 + 17*x + 51 = (x + 3)*(x^2 + 17) = 0$

The other two solutions are when the other term is equal to zero.

Then the other two solutions are given by:

x = ±√17

And neither of these have problems in the denominators, so we can conclude that the solutions are:

x = √17 and x = -√17

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3 years ago

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DIA [1.3K]

Stepstep explanation:

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3 years ago

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Igoryamba

Answer:

C 2 and 7 are interior angles

Step-by-step explanation:

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3 years ago

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blsea [12.9K]

Y = 18( 0.58) ^t

It would be written as :
Y = 18(0.11) ^ (t/4)
Y = 18 (0.11 ^ (1/4)) ^ t
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Y = 18 (0.5759) ^ t

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3 years ago

A piece of wire 11 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria

forsale [732]

Answer:

11 meters

Step-by-step explanation:

First, we can say that the square has a side length of x. The perimeter of the square is 4x, and that is how much wire goes into the square. To maximize the area, we should use all the wire possible, so the remaining wire goes into the triangle, or (11-4x).

The area of the square is x², and the area of an equilateral triangle with side length a is (√3/4)a². Next, 11-4x is equal to the perimeter of the triangle, and since it is equilateral, each side has (11-4x)/3 length. Plugging that in for a, we get the area of the equilateral triangle is

(√3/4)((11-4x)/3)²

= (√3/4)(11/3 - 4x/3)²

= (√3/4)(121/9 - 88x/9 + 16x²/9)

= (16√3/36)x² - (88√3/36)x + (121√3/36)

The total area is then

(16√3/36)x² - (88√3/36)x + (121√3/36) + x²

= (16√3/36 + 1)x² - (88√3/36)x + (121√3/36)

Because the coefficient for x² is positive, the parabola would open up and the derivative of the parabola would be the local minimum. Therefore, to find the maximum area, we need to go to the absolute minimum/maximum points of x (x=0 or x=2.75)

When x=0, each side of the triangle is 11/3 meters long and its area is

(√3/4)a² ≈ 5.82

When x=2.75, each side of the square is 2.75 meters long and its area is

2.75² = 7.5625

Therefore, a maximum is reached when x=2.75, or the wire used for the square is 2.75 * 4 = 11 meters

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3 years ago