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Aleks04 [339]
3 years ago
12

What is the answer to this question?

Mathematics
1 answer:
Leviafan [203]3 years ago
6 0

Answer:

Quadratic: y = -x^2 + 7

Step-by-step explanation:

Diff                       Diff         Diff in

in x     x       y       in y         Differences

         -2      3

1        -1       6       3

1         0      7        1               -2            

1         1       6       -1               -2

1         2      3       -3               -2

The differences in x are all 1. The differences in y are not equal, so it is not linear. The differences in the differences in y are all 2. It is quadratic.

Try: y = -x^2 + 7

x = -2:

y = -(-2)^2 + 7 = -4 + 7 = 3

You get (-2, 3) which is a given point.

Now try the same x value with the other quadratic equation.

y = x^2 + 7

x = -2:

y = (-2)^2 + 7 = 4 + 7 = 11

You get (-2, 11) which is not a given point.

Answer:

Quadratic: y = -x^2 + 7

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Step-by-step explanation:

The equation is linear and expressed in slope- intercept form

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m = (y₂ - y₁ ) / (x₂ - x₁ )

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3 years ago
2. Given a quadrilateral with vertices (−1, 3), (1, 5), (5, 1), and (3,−1):
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<h2>Explanation:</h2>

In every rectangle, the two diagonals have the same length. If a quadrilateral's diagonals have the same length, that doesn't mean it has to be a rectangle, but if a parallelogram's diagonals have the same length, then it's definitely a rectangle.

So first of all, let's prove this is a parallelogram. The basic definition of a parallelogram is that it is a quadrilateral where both pairs of opposite sides are parallel.

So let's name the vertices as:

A(-1,3) \\ \\ B(1,5) \\ \\ C(5,1) \\ \\ D(3,-1)

First pair of opposite sides:

<u>Slope:</u>

\text{For AB}: \\ \\ m=\frac{5-3}{1-(-1)}=1 \\ \\ \\ \text{For CD}: \\ \\ m=\frac{1-(-1)}{5-3}=1 \\ \\ \\ \text{So AB and CD are parallel}

Second pair of opposite sides:

<u>Slope:</u>

\text{For BC}: \\ \\ m=\frac{1-5}{5-1}=-1 \\ \\ \\ \text{For AD}: \\ \\ m=\frac{-1-3}{3-(-1)}=-1 \\ \\ \\ \text{So BC and AD are parallel}

So in fact this is a parallelogram. The other thing we need to prove is that the diagonals measure the same. Using distance formula:

d=\sqrt{(y_{2}-y_{1})^2+(x_{2}-x_{1})^2} \\ \\ \\ Diagonal \ BD: \\ \\ d=\sqrt{(5-(-1))^2+(1-3)^2}=2\sqrt{10} \\ \\ \\ Diagonal \ AC: \\ \\ d=\sqrt{(3-1)^2+(-5-1)^2}=2\sqrt{10} \\ \\ \\

So the diagonals measure the same, therefore this is a rectangle.

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3 years ago
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