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3 years ago

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A square on a coordinate plane has the points A(0, 0), B(0, 5), C(5, 5) and D(5, 0). If Square ABCD is rotated about the origin

180 degrees clockwise to form Square A'B'C'D', what would be the length of one of the sides? A. 0 units B. 5 units C. 10 units D. 20 units

1 answer:

Georgia [21]3 years ago

3 0

Answer:

B. 5 units

Step-by-step explanation:

Transformation is the movement of a point from its initial location to a new location. Types of transformation are reflection, dilation, rotation or translation.

If a point X(x,y) is rotated about the origin 180 degrees clockwise the new point is at X'(-x, -y).

If the square with vertices at A(0, 0), B(0, 5), C(5, 5) and D(5, 0) is rotated about the origin 180 degrees clockwise the new points are at A'(0, 0), B'(0, -5), C'(-5, -5), D'(-5, 0)

A square has four equal sides. The distance between two points $X(x_1,y_1)\ and\ Y(x_2,y_2)$ is:

$|XY|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Hence:

$|A'B'|=\sqrt{(0-0)^2+(-5-0)^2} =5$

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Step-by-step explanation:

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The point P(7, −2) lies on the curve y = 2/(6 − x). (a) If Q is the point (x, 2/(6 − x)), use your calculator to find the slope

NARA [144]

Answer:

a) (i) $m = 2.22$ , (ii) $m = 2$ , (iii) $m = 2$ , (iv) $m = 2$ , (v) $m = 1.82$ , (vi) $m = 2$ , (vii) $m = 2$ , (viii) $m = 2$ ; b) $m \approx 2$ ; c) The equation of the tangent line to curve at P (7, -2) is $y = 2\cdot x + 12$ .

Step-by-step explanation:

a) The slope of the secant line PQ is represented by the following definition of slope:

$m = \frac{\Delta y}{\Delta x} = \frac{y_{Q}-y_{P}}{x_{Q}-x_{P}}$

(i) $x_{Q} = 6.9$ :

$y_{Q} =\frac{2}{6-6.9}$

$y_{Q} = -2.222$

$m = \frac{-2.222 + 2}{6.9-7}$

$m = 2.22$

(ii) $x_{Q} = 6.99$

$y_{Q} =\frac{2}{6-6.99}$

$y_{Q} = -2.020$

$m = \frac{-2.020 + 2}{6.99-7}$

$m = 2$

(iii) $x_{Q} = 6.999$

$y_{Q} =\frac{2}{6-6.999}$

$y_{Q} = -2.002$

$m = \frac{-2.002 + 2}{6.999-7}$

$m = 2$

(iv) $x_{Q} = 6.9999$

$y_{Q} =\frac{2}{6-6.9999}$

$y_{Q} = -2.0002$

$m = \frac{-2.0002 + 2}{6.9999-7}$

$m = 2$

(v) $x_{Q} = 7.1$

$y_{Q} =\frac{2}{6-7.1}$

$y_{Q} = -1.818$

$m = \frac{-1.818 + 2}{7.1-7}$

$m = 1.82$

(vi) $x_{Q} = 7.01$

$y_{Q} =\frac{2}{6-7.01}$

$y_{Q} = -1.980$

$m = \frac{-1.980 + 2}{7.01-7}$

$m = 2$

(vii) $x_{Q} = 7.001$

$y_{Q} =\frac{2}{6-7.001}$

$y_{Q} = -1.998$

$m = \frac{-1.998 + 2}{7.001-7}$

$m = 2$

(viii) $x_{Q} = 7.0001$

$y_{Q} =\frac{2}{6-7.0001}$

$y_{Q} = -1.9998$

$m = \frac{-1.9998 + 2}{7.0001-7}$

$m = 2$

b) The slope at P (7,-2) can be estimated by using the following average:

$m \approx \frac{f(6.9999)+f(7.0001)}{2}$

$m \approx \frac{2+2}{2}$

$m \approx 2$

The slope of the tangent line to the curve at P(7, -2) is 2.

c) The equation of the tangent line is a first-order polynomial with the following characteristics:

$y = m\cdot x + b$

Where:

$x$ - Independent variable.

$y$ - Depedent variable.

$m$ - Slope.

$b$ - x-Intercept.

The slope was found in point (b) (m = 2). Besides, the point of tangency (7,-2) is known and value of x-Intercept can be obtained after clearing the respective variable:

$-2 = 2 \cdot 7 + b$

$b = -2 + 14$

$b = 12$

The equation of the tangent line to curve at P (7, -2) is $y = 2\cdot x + 12$ .

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3 years ago

Subtract using the number line. <br><img src="https://tex.z-dn.net/?f=%20-%201%20%5Cfrac%7B1%7D%7B3%7D%20%20-%20%20%5Cfrac%7B1%7

Allisa [31]

Answer:

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- 1 3/6

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Step-by-step explanation:

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