Pls help will give brainliest and thanks and 5star pls help
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For x-intercepts "b" and "c", a quadratic can be written as
y = a(x -b)(x -c)
You know the x-intercepts, so you can write the equation as
y = a(x +2)(x -3) . . . . . . eliminates the first two answer choices
Substituting (x, y) = (-1, 2), you have
2 = a(-1+2)(-1-3) = -4a
Then dividing by -4 gives
-2/4 = a = -1/2 . . . . . corresponds to the third answer choice
The appropriate selection is
y = -1/2 (x + 2)(x - 3)
y = a(x -b)(x -c)
You know the x-intercepts, so you can write the equation as
y = a(x +2)(x -3) . . . . . . eliminates the first two answer choices
Substituting (x, y) = (-1, 2), you have
2 = a(-1+2)(-1-3) = -4a
Then dividing by -4 gives
-2/4 = a = -1/2 . . . . . corresponds to the third answer choice
The appropriate selection is
y = -1/2 (x + 2)(x - 3)
First, let's make a two-way table using a matrix:
![\left[\begin{array}{cccc}&Tennis&No&Total\\Male&7&5&12\\Female&9&6&15\\Total&16&11&27\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D%26Tennis%26No%26Total%5C%5CMale%267%265%2612%5C%5CFemale%269%266%2615%5C%5CTotal%2616%2611%2627%5Cend%7Barray%7D%5Cright%5D%20)
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