Answer: The segment EF of the isoceles triangle ΔEFP is 10 units
Solution
If point P is equidistant from the vertices of ΔDEF, the segment PE must measure the same that the segment PF, then these two segments are congruent:
PE=PF
Then ΔEFP is an isisceles triangle, because it has two congruent sides (PE=PF), and the height of the side EF (PJ perpendicular to EF) divides ΔEFP into two congruent right triangles (ΔEJP and ΔFJP). The legs EJ and FJ must be congruent:
FJ=EJ
3x-1=x+3
Solving for x: Subtracting x and adding 1 both sides of the equation:
3x-1-x+1=x+3-x+1
2x=4
Dividing both sides of the equation by 2:
2x/2=4/2
x=2
Then:
FJ=3x-1→FJ=3(2)-1→FJ=6-1→FJ=5
EJ=x+3→EJ=2+3→EJ=5
EF=EJ+FJ→EF=5+5→EF=10
Let A = {1, 2, 3, 4,}, B = {3, 4, 5, 6, 7}, and C = {2, 3, 5, 7} Find: B∩C
natka813 [3]
The notation
means "B intersect C" telling you to look at the intersecting, or overlapping, region of the Venn Diagram. This is where the common values that are found in BOTH set B and set C.
The values {3, 5, 7} are in both set B and set C.
So that is why
Answer:
quadratic
Step-by-step explanation:
Answer:
the surface area of the rectangle is 40
the surface area of the triangles is 6
the base rectangle is 24
Step-by-step explanation: