Let t, h, b represent the weighs of tail, head, and body, respectively.
t = 4 . . . . given
h = t + b/2 . . . . the head weighs as much as the tail and half the body
b/2 = h + t . . . . half the body weighs as much as the head and tail
_____
Substituting for b/2 in the second equation using the expression in the third equation, we have
... h = t + (h + t)
Subtracting h from both sides gives
... 0 = 2t . . . . . . in contradiction to the initial statement about tail weight.
Conclusion: there's no solution to the problem given here.
Answer:
A = 1/2(2πr)(r)
Explanation:
The circumference of the circle, 2πr, is the measure completely around the circle. When the pieces of the circle are rearranged, half of this circumference will be on the top of the parallelogram and half will be on the bottom. This means the base will be 1/2(2πr).
The approximate height of the parallelogram is the radius of the circle; this makes the area
A = 1/2(2πr)(r)
Answer:
-52
Have a great day! Brainliest?
Answer:
Second option: On a coordinate plane, rectangle A'B'C'D' prime has points
(See the graph attached)
Step-by-step explanation:
For this exercise it is importnat to know that a Dilation is defined as a transformation in which the Image (The figure obtained after the transformation) has the same shape as the Pre-Image (which is the original figure before the transformation), but they have different sizes.
In this case, you know that the vertices of the rectangle ABCD ( The Pre-Image) are the following:
Therefore, to find the vertices of the rectangle A'B'C'D' (The Image) that results of dilating the rectangle ABCD by a factor of 4 about the origin, you need to multiply the coordinates of each original vertex by 4. Then, you get:
Finally, knowing those points, you can identify that the graph that shows the result of that Dilation, is the one attached.
Step 1. Solve both inequalities for
:
Step 2. To check a point in the solution of the given system of inequalities, look for the intercepts of the lines
and
:
(1)
(2)
Replace (1) in (2):
Solve for
:
(3)
Replace (3) in (1):
We can conclude that the point (-2,3) is in the solution of the system if <span>
inequalities</span>
; also any point inside the dark shaded area of the graph of the system of inequalities is also a solution of the system.