The <em>correct answer</em> is:
D) reflect over the y axis and then reflect again over the y axis.
Explanation:
Logically, if we reflect a figure across the y-axis and then reflect across the y-axis again, we have undone what we originally did, and the figure is back in its original position.
Algebraically, reflecting across the y-axis maps every point (x, y) to (-x, y). Reflecting this point across the y-axis maps (-x, y) to (x, y); this is our original point.
The diagonal is the hypotenuse of a right triangle with the sides of the square as legs.
By Pythagoras (s = side)
s^2 + s^2 = 15^2
2s^2 = 225
s^2 = 112.5
s = sqrt(112.5)
s = 10.6 cm (to the nearest tenth)
Answer:
Part one: The function rule for the area of the rectangle is A(x) = 3x² - 2x
Part two: The area of the rectangle is 8 feet² when its width is 2 feet
Step-by-step explanation:
Assume that the width of the rectangle is x
∵ The width of the rectangle = x feet
∵ The length of the rectangle is 2 ft less than three times its width
→ That means multiply the width by 3, then subtract 2 from the product
∴ The length of the rectangle = 3(x) - 2
∴ The length of the rectangle = (3x - 2) feet
∵ The area of the rectangle = length × width
∴ A(x) = (3x - 2) × x
→ Multiply each term in the bracket by x
∵ A(x) = x(3x) - x(2)
∴ A(x) = 3x² - 2x
∴ The function rule for the area of the rectangle is A(x) = 3x² - 2x
∵ The rectangle has a width of 2 ft
∵ The width = x
∴ x = 2
→ Substitute x by 2 in A(x)
∵ A(2) = 3(2)² - 2(2)
∴ A(2) = 3(4) - 4
∴ A(2) = 12 - 4
∴ A(2) = 8
∴ The area of the rectangle is 8 feet² when its width is 2 feet
