Which of the following segments is an angle bisector? 1. AL 2. JB 3. KC
1 answer:
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A.
(0,00
(2.6,7.9)
(4.8,12.4)
(9.7,15.1)
b.
well, the points don't look like they are on the line but they actually are (plot twist)
so
since (0,0) is on th egarph
0=a(0)²+b(0)+c
0=c
so
f(x)=ax²+bx+0 or
f(x)=ax²+bx
find a and b
sub points
(2.6,7.9)
7.9=a(2.6)²+b(2.6)
7.9=6.76a+2.6b
(4.8,12.4)
12.4=a(4.8)²+b(4.8)
12.4=23.04a+4.8b
use those 2 equations
7.9=6.76a+2.6b
12.4=23.04a+4.8b
eliminate b
multiply first equation by -4.8 and 2nd by 2.6 and add them
-37.92=-32.448a-12.48b
32.24=59.904a-12.48b +
-5.68=27.456a+0b
-5.68=27.456a
divide both sides by 27.456
(-5.68/27.456)=a
find b
12.4=23.04a+4.8b
12.4=(-130.8672/27.456)+4.8b
12.4+(130.8672/27.456)=4.8b
(12.4+(130.8672/27.456))/4.8=b
da equation is
c. the roots are found with your calculator
the roots are at x=0 and x=17.287323943662
so very close to the target but not exactly on it
if the target has a radius of 0.287323943662 or more then it will hit the target
(0,00
(2.6,7.9)
(4.8,12.4)
(9.7,15.1)
b.
well, the points don't look like they are on the line but they actually are (plot twist)
so
since (0,0) is on th egarph
0=a(0)²+b(0)+c
0=c
so
f(x)=ax²+bx+0 or
f(x)=ax²+bx
find a and b
sub points
(2.6,7.9)
7.9=a(2.6)²+b(2.6)
7.9=6.76a+2.6b
(4.8,12.4)
12.4=a(4.8)²+b(4.8)
12.4=23.04a+4.8b
use those 2 equations
7.9=6.76a+2.6b
12.4=23.04a+4.8b
eliminate b
multiply first equation by -4.8 and 2nd by 2.6 and add them
-37.92=-32.448a-12.48b
32.24=59.904a-12.48b +
-5.68=27.456a+0b
-5.68=27.456a
divide both sides by 27.456
(-5.68/27.456)=a
find b
12.4=23.04a+4.8b
12.4=(-130.8672/27.456)+4.8b
12.4+(130.8672/27.456)=4.8b
(12.4+(130.8672/27.456))/4.8=b
da equation is
c. the roots are found with your calculator
the roots are at x=0 and x=17.287323943662
so very close to the target but not exactly on it
if the target has a radius of 0.287323943662 or more then it will hit the target