7,14,28,56 11th term

A Turing machine with doubly inﬁnite tape is similar to an ordinary Turing machine, but its tape is inﬁnite to the left as wella

atroni [7]

Answer:

Use multitape Turing machine to simulate doubly infinite one

Explanation:

It is obvious that Turing machine with doubly infinite tape can simulate ordinary TM. For the other direction, note that 2-tape Turing machine is essentially itself a Turing machine with doubly (double) infinite tape. When it reaches the left-hand side end of first tape, it switches to the second one, and vice versa.

8 0

3 years ago

Complete the chart below to show the properties that hold true for the quadrilateral given in each row.

puteri [66]

Hey there!

Opposite sides are congruent:

all of them

opposite angles are congruent

rectangle and square

all sides are congruent

rhombus and square

diagonals congruent:

rectangle, rhombus, and square

diagonals are perpendicular

square and rhombus

Have a terrificly amazing day!

6 0

2 years ago

Determine the most precise name for ABCD (parallelogram, rhombus, rectangle, or square). Explain how you determined your answer.

Sonja [21]

<h3>Answer: Rhombus</h3>

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Reason:

Let's find the distance from A to B. This is equivalent to finding the length of segment AB. I'll use the distance formula.

$A = (x_1,y_1) = (3,5) \text{ and } B = (x_2, y_2) = (7,6)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(3-7)^2 + (5-6)^2}\\\\d = \sqrt{(-4)^2 + (-1)^2}\\\\d = \sqrt{16 + 1}\\\\d = \sqrt{17}\\\\d \approx 4.1231\\\\$

Segment AB is exactly $\sqrt{17}$ units long, which is approximately 4.1231 units.

If you were to repeat similar steps for the other sides (BC, CD and AD) you should find that all four sides are the same length. Because of this fact, we have a rhombus.

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Let's see if this rhombus is a square or not. We'll need to see if the adjacent sides are perpendicular. For that we'll need the slope.

Let's find the slope of AB.

$A = (x_1,y_1) = (3,5) \text{ and } B = (x_2,y_2) = (7,6)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{6 - 5}{7 - 3}\\\\m = \frac{1}{4}\\\\$

Segment AB has a slope of 1/4.

Do the same for BC

$B = (x_1,y_1) = (7,6) \text{ and } C = (x_2,y_2) = (6,2)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{2 - 6}{6 - 7}\\\\m = \frac{-4}{-1}\\\\m = 4\\\\$

Unfortunately the two slopes of 1/4 and 4 are not negative reciprocals of one another. One slope has to be negative while the other is positive, if we wanted perpendicular lines. Also recall that perpendicular slopes must multiply to -1.

We don't have perpendicular lines, so the interior angles are not 90 degrees each.

Therefore, this figure is not a rectangle and by extension it's not a square either.

The best description for this figure is a rhombus.