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Hatshy [7]
4 years ago
6

Determine if the following table represents a quadratic function.

Mathematics
1 answer:
snow_tiger [21]4 years ago
7 0

Answer:

<em>The options are not visible enough (See Explanation)</em>

Step-by-step explanation:

Given

x:-  1   || 2   || 3   || 4   || 5

y:-  13 || 22 || 37 || 58 || 85

Required

Determine if the function is quadratic

Calculate the difference between the values of y

Difference = 22 - 13 = 9

Difference = 37 - 22 = 15

Difference = 58 - 37 = 21

Difference = 85 - 58 = 27

<em>The resulting difference are: 9 || 15 || 21 ||  27</em>

Next; Calculate the difference between the difference of values of y

Difference = 15 - 9 = 6

Difference = 21 - 15 = 6

Difference = 27 - 21 = 6

<em>The resulting difference are: 6 || 6 || 6</em>

<em>For the function to be quadratic, the above difference must be the same and since they are the same (6), then the function represents a quadratic function.</em>

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The coordinates of the vertices of ∆PQR are P(-2,5), Q(-1,1), and R(7,3). Determine whether ∆PQR is a right triangle. Show your
Mashutka [201]

Given

∆PQR points are P(-2,5), Q(-1,1), and R(7,3)

Determine whether ∆PQR is a right triangle

To proof

As given ∆PQR points are P(-2,5), Q(-1,1), and R(7,3)

First find out the sides of triangle

FORMULA

Distance formula between two points

D^{2}= (x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}

 Distance   between two points P(-2,5) and Q(-1,1)

PR = \sqrt{(-1+2)^{2}+(1-5)^{2}  }

PR = \sqrt{17}

Distance between two points Q(-1,1)and  R(7,3)

QR = \sqrt{(7+1)^{2} +(3-1)^{2}  }

QR =\sqrt{68}

Distance between two points  R(7,3) and P(-2,5)

RP =\sqrt{(-2-7)^{2} + (5-3)^{2}  }

RP=\sqrt{85}

now show that ∆PQR is a right triangle

RP^{2} = PQ^{2} +QR^{2}

Putting the value given above

(\sqrt{85}) ^{2} = \sqrt{17} ^{2} +\sqrt{68} ^{2}

85 = 17 +68

85 =85

In the right triangle

HYPOTENUSE² = BASE² + PERPENDICULAR²

This is prove above

Hence ∆PQR is a right triangle

Hence proved










7 0
4 years ago
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