P=2(l+w)
190=2[(2l+5)+l)
190=4l+10+2l
190=6l+10
180=6l
l=30
190=2(30+w)
190=60+2w
130=2w
w=65
The width of the rectangle is 65.
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
See the answer here is overspending. In order to discuss this you must repeat
Answer:
y=-3x-5
Step-by-step explanation:
y = mx+b
m= -11 - 1
2 - -2
y=-3x+b
(-2,1) and (2,-11)
is
y=-3x-5
Hope this Helps:)