Coach Pettaway is buying basketball equipment for his team. He gets a

1 answer:

soldier1979 [14.2K]3 years ago

3 0

First, let’s all acknowledge that whoever comes up with problems like this WANTS kids to hate math...smh

I’m sure there is a prettier way to solve this, but here’s what I did:

8(2.25) + 3(22.50) =
18 + 67.50 = 85.50 per “set” of balls/jerseys
400/85.50 = 4.678 = number of “sets” he can buy. Round down to 4 so we have room for tax.

85.5 x 4 “sets”= $342
Tax on 342 is 0.06 x 342 = 20.52

$342 + 20.52 = $362.52 spent

Basketballs = 4 sets x 8 balls per set= 32
Jerseys = 4 sets x 3 jerseys per set= 12

32 basketballs, 12 jerseys, $362.52 spent

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write a linear equation that intersects y=x^2 at two points. Then write a second linear equation that intersects y=x^2 at just o

Liula [17]

We know that $y = x^2$ is a parabola, concave up, with vertex in the origin $(0,0)$

So, we may use three horizontal lines for our purpose: any horizontal line above the x axis will intersect the parabola twice. The x axis itself intersects the parabola once on the vertex, while any horizontal line below the x axis won't intercept the parabola.

Here's the examples:

The horizontal line $y = 4$ intercepts the parabola twice: the system $y = x^2,\ y = 4$ is solved by $x^2=4 \implies x = \pm 2$
The horizontal line $y=0$ intercepts the parabola only once: the system is $y=x^2,\ y=0$ which yields $x^2=0\implies x=0$ which is a repeated solution
The horizontal line $y=-5$ intercepts the parabola only once: the system is $y=x^2,\ y=-5$ which yields $x^2=-5$ which is impossible, because a squared number can't be negative.