Answer:
Mark point E where the circle intersects segment BC
Step-by-step explanation:
Apparently, Bill is using "technology" to perform the same steps that he would use with compass and straightedge. Those steps involve finding a point equidistant from the rays BD and BC. That is generally done by finding the intersection point(s) of circles centered at D and "E", where "E" is the intersection point of the circle B with segment BC.
Bill's next step is to mark point E, so he can use it as the center of one of the circles just described.
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<em>Comment on Bill's "technology"</em>
In the technology I would use for this purpose, the next step would be "select the angle bisector tool."
Answer:
15 - 5x = 5 · 3 - 5 · x = 5(3 - x)
Like terms are going to have the EXACT same variables....or they can just be constants with no variables.
x^2 and 3x^2 are like terms
x^2 and x^3 are not like terms
8 and 9 are like terms
8x and 9y are not like terms
so ur like terms in ur problem are : 2y^3 and y^3
Answer:
yes, you got it correct
Step-by-step explanation:
Let's call the width: w
the lenght is then 3w+4 ("4 more than 3 times the width")
and the parameter would be 2(w+3w+4), that is 2*(4w+4), that is 8w+8.
this is also equal to 18.4:
8w+8=18.4
8w=10.4
w=1.3
this is the width, and the lenght is:
4+3*1,3=4+3.9=7.9
and the area is their product:
1.3*7.9=10.27
