Answer:
l(l - 2) = 168
Step-by-step explanation:
Let w = width
Let l = length
Let A = area
w = l - 2
A = l * w
also, A = 168
equating the two values of A
l * w = 168
l * (l - 2) = 168
l^2 - 2l = 168
l(l - 2) = 168
Answer for Question 1 is option 2
Answer for Question 2 is option 3
Answer for Question 3 is option 3
Step-by-step explanation:
For question 1 by plotting the two points we get the coordinates as (-2, 3) and (2, 1) where the x values are -2, 2 and the y values are 3, 1 respectively. By substituting values of x in the formula given we get that equation 2 satisfies the values of y as in the graph.
For question 2 the plotted line is split into two parts where one half is for values of x <2 and the other half is for values of x>2. By finding out the coordinates we get them as (2,-2), (1,-2.5) and (0,-3) for x<2 and by substituting the values of x in the formula we get option 3 as our answer.
For question 3 we need to find the value of y when x=3 and the first condition is only for values greater than 3 so we take the second condition and substitute the value of x=3 and we get the value of y as 5 so the answer is option 3
Answer:
C. The circumference of circle 1 is less than the circumference of circle 2.
Step-by-step explanation:
6<7
We know that the formula for the area of a circle is πr². This means the greater the radius, the greater the area.
We also know that the formula for circumference is 2πr. It is also dependent on radius. The greater the radius, the greater the circumference.
Given that Circle 2 has a greater area, this means that it has a greater radius which in turn means that it has a greater circumference.
C. The circumference of circle 1 is less than the circumference of circle 2.
Answer: The diagonals of a parallelogram bisect each other. Steps (a), (b), and (c) outline proof of this theorem. (See Exercise 25 for a particular instance of this theorem.) (a) In the parallelogram OA BC shown in the figure, check that the coordinates of must be (b) Use the midpoint formula to calculate the midpoints of diagonals and (c) The two answers in part (b) are identical. This shows that the two diagonals do indeed bisect each other, as we wished to prove. (FIGURE CAN'T COPY)
