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never [62]
3 years ago
7

In basketball you score 2 points for a field goal and 1 point for a free throw. Suppose that you have scored at least 3 points i

n every game this season, and have a high season score of 15 points in one game. How many field goals and free throws could you have made in any one game?
2x + y ≤ 15
x + y ≥ 3

write one possible solution to the problem in sentence form

Mathematics
2 answers:
Kaylis [27]3 years ago
5 0

Answer:

There are 2 points for a field goal, and 1 point for a free throw.

 x=Points scored for a field goal

 y=Points scored for a free throw

It is also given that, you have scored at least 3 points in every game this season, and have a high season score of 15 points in one game.

 2x+y≤15

x+y≥3

Plotting the inequality on two dimensional xy plane

The solution set through graph is attached below.

One possible solution is

x=5 field goals

y=5 free throw

⇒In a game I have made 4 field goals and 3 free throws. So total number of points scored in game will be 4×2+3=11 points.

olga55 [171]3 years ago
4 0
2x+y>3 describes the low end, presuming that you have defined to mean the number of field goals and to mean the number of free throws. yet your best game was 15, so the values must also satisfy 2x+y
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The coordinates of the vertices of a rectangle are (-4, 3), (-1, 5),(3, – 1), and (0, – 3).
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Answer:

Perimeter of rectangle is 23.34

Step-by-step explanation:

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The formula to find perimeter of rectangle is: Perimeter=2(l+w)

We need to find length and width first using the coordinates (-4, 3), (-1, 5),(3, – 1), and (0, – 3).

Let A=(-4, 3) B=(-1, 5) C=(3, – 1) and D=(0, – 3) according to the figure attached.

So, If we find distance of point AB it can be considered length of rectangle and distance of point AC can be considered width of rectangle

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Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Distance=\sqrt{(-1-(-4))^2+(5-3)^2}\\Distance=\sqrt{(-1+4)^2+(5-3)^2}\\Distance=\sqrt{(3)^2+(2)^2}\\Distance=\sqrt{9+4}\\Distance=\sqrt{13}\\Distance=3.61

So distance between point A and B is 3.61

Finding distance AC, we have x_1=-4, y_1=3, x_2=3, y_2=-1

Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Distance=\sqrt{(3-(-4))^2+(-1-3)^2}\\Distance=\sqrt{(3+4)^2+(-1-3)^2} \\Distance=\sqrt{(7)^2+(-4)^2} \\Distance=\sqrt{49+16} \\Distance=\sqrt{65}\\Distance=8.06

So, distance between point A and C is 8.06

We have length = 3.61 and width = 8.06

Using formula of perimeter

Perimeter=2(l+w)\\Perimeter=2(3.61+8.06)\\Perimeter=2(11.67)\\Perimeter=23.34

So, Perimeter of rectangle is 23.34

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