Quadrilateral ABCD has vertices A(1, -3) , B(2, 1), C(6, 3), and D(7, -1). Prove or disprove that the Quad ABCD is a rectangle.

1 answer:

7 0

If AC=BD, than Quad ABCD is a rectangle
AC
√(3+3)²+(6-1)²
√6²+5²
√36+25
√61
BD
√(1+1)²+(2-7)²
√2²+(-5)²
√4+25
√29
Quad ABCD is not a rectangle because √61 and √29 are not equal.

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4. Name the first five terms of the arithmetic sequence. a1 = -35, d=4

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Answer:

Step-by-step explanation:

We first have to write the equation for the sequence, then finding the first five terms will be easy. Follow the formatting:

$a_n=a_1+d(n-1)$ and we are given enough info to fill in:

$a_n=-35+4(n-1)$ and

$a_n=-35+4n-4$ and

$a_n=-39+4n$ or in linear format:

$a_n=4n-39$ where n is the position of the number in the sequence. We already know the first term is -35.

The second term:

$a_2=4(2)-39$ so

$a_2=8-39$ and

$a_2=-31$

The third term:

$a_3=4(3)-39$ and

$a_3=12-39$ so

$a_3=-27$ and we could go on like this forever, but the nice thing about this is when we know the difference all we have to do is add it to each number to get to the next number.