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Alexxx [7]
2 years ago
12

The number of customers in a grocery store is modeled by the function y -X? + 10x + 50, where y is

Mathematics
1 answer:
vova2212 [387]2 years ago
7 0

Using the vertex of the quadratic function, it is found that:

a) The maximum number of customers in the store is at 12 P.M.

b) 75 customers are in the store at this time.

The number of customers in x hours after 7 AM is given by:

y = -x^2 + 10x + 50

Which is a quadratic equation with coefficients a = -1, b = 10, c = 50

Item a:

The maximum value, considering that a < 0, happens at:

x_v = -\frac{b}{2a}

Hence:

x_v = -\frac{10}{2(-1)} = 5

5 hours after 7 A.M, hence, the maximum number of customers in the store is at 12 P.M.

Item b:

The value is y(5), hence:

y(5) = -(5)^2 + 10(5) + 50 = 75

75 customers are in the store at this time.

A similar problem is given at brainly.com/question/24713268

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See attached graphic:
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Read 2 more answers
a volley ball player servers the ball. The ball follows a path given by the equation y=-0.01x^2+0.5x+3 where x and y are measure
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Answer:

(a)

Distance from player should be 13.82 feet or 36.2 feet

(b)

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Step-by-step explanation:

we are given

The ball follows a path given by the equation

y=-0.01x^2+0.5x+3

where

x and y are measured in feet and the origin is on the court directly below where the player hits the ball

(a)

net height is 8 ft

so, we can set y=8

and then we can solve for x

8=-0.01x^2+0.5x+3

8\cdot \:100=-0.01x^2\cdot \:100+0.5x\cdot \:100+3\cdot \:100

800=-x^2+50x+300

-x^2+50x-500=0

x^2-50x+500=0

we can use quadratic formula

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

x=\frac{-50\pm \sqrt{50^2-4\left(-1\right)\left(-500\right)}}{2\left(-1\right)}

x=5\left(5-\sqrt{5}\right),\:x=5\left(5+\sqrt{5}\right)

x=13.82,x=36.2

So, distance from player should be 13.82 feet or 36.2 feet

(b)

we can plug x=30 and check whether y=8 ft

y=-0.01(30)^2+0.5(30)+3

y=9ft

we know that

height of net is 8 ft

so, the ball will go over the net


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