Hello,
1)Let's call A=inter (4x-y-7=0, x+3y-31=0)
4x-y=7
x+3y=31
==>13x=52==>x=4
and y=4*4-7=9
A=(4;9)
2)Let's call B=inter (4x-y-7=0, x+5y-7=0)
4x-y=7
x+5y=7
==>21x=42
==>x=2 and y=4*2-7=1
B=(2,1)
3) Slope of BD: y=-x/3+31/3 ==> slope perpendicular=3
Line BD:y-1=3(x-2)==>y=3x+5
4) Slope AE: y=-x/5+7/5 ==>slope perpendicular=5
Line AE: y-9=5(x-4)==> y=5x-11
5) intersection (BD;AE):
y=5x-11
y=3x-5
==>2x=6==>x=3
and y=3*3-5=4
H=(3,4)
Answer:
(see below)
Step-by-step explanation:
First, to make it easier for yourself, "flip" the triangles so that they "match." To see what I'm talking about, refer to IMAGE.A.
Now that you can tell that the congruent side and angles are corresponding, you have to prove them congruent.
There is one side and two angles, so it's AAS or SAA.
Answer:
Step-by-step explanation:
There is nothing I can solve
You use a closed dot with the greater than or less than with the line under the symbol or the greater than or equal to or the less than or equal to symbols. The line indicates a closed dot.