1. N is a midpoint of the segment KL, then N has coordinates
2. To find the area of △KNM, the length of the base MK is 2b, and the length of the height is a. So an expression for the area of △KNM is
3. To find the area of △MNL, the length of the base ML is 2a and the length of the height is b. So an expression for the area of △MNL is
4. Comparing the expressions for the areas you have that the area is equal to the area . This means that the segment from the midpoint of the hypotenuse of a right triangle to the opposite vertex forms two triangles with equal areas.
B. (x + 2)(x + 12)
C. (x + 3)(x + 8)
D. (x + 1)(x + 11)
Factor 2m2 + 27m + 70
A. (2m + 7)(m + 10)
B. (m + 10)(2m + 7)
C. (2m + 3)(m + 20)
D. (2m + 7)(m – 10)
Factor g2 – 64
A. (g – 1)(g + 64)
B. (g + 8)(g – 8)
C. (g – 8)(g – 8)
D. (g + 8)(g + 8)
Factor 49t2 – 9
A. (7t + 3)(7t + 3)
B. (7t + 3)(–7t – 3)
C. (7t + 3)(7t – 3)
D. 7t – 3
The area of a rectangular field is given by the trinomial t2 – 4t – 45. The length of the rectangle is t + 5. What is the expression for the width of the field?
A. t – 9
B. t – 5
Answer:
y = -0.5x + 6, Domain [0,12] and Range [0, 6]
Step-by-step explanation:
The slope in a linear function represents the rate of change. Since the amount of snow is decreasing (indicating a negative slope) at 0.5 inches an hour, our slope would be -0.5.
Our y intercept, 6, would be the starting amount of snow. At 0 hours, no snow has been melted.
Our domain is [0,12] these are the x values from where at zero there is 6 inches of snow and at 12 there is 0 inches of snow.
Finally, our range is [0,6]. Since there cannot be no more than 6 or less than 0 inches of snow, our range is [0,6]
Hope this helped!
______revolutions/min