Points equidistant from DE EF are in the bisector of angle DEF
points equidistant from EF DF are in the bisector of angle EFD
the sought after point is the intersection of bisectricess of triangle
I have a feeling I've seen this question before
anyway
A=number of hours that plan A is
B=number of hours that plan B is
so
on wednesday, 7hr
5A and 6B so
5A+6B=7
on thursday, 3 hours
3 of A and 2 of B
3A+2B=3
so we gots
5A+6B=7
3A+2B=3
elimination
eliminate B's
multiply 2nd equation by -3 and add to 1st equation
5A+6B=7
<u>-9A-6B=-9 +</u>
-4A+0B=-2
-4A=-2
divide both sides by -4
A=1/2
A=0.5
sub back
3A+2B=3
3(0.5)+2B=3
1.5+2B=3
minus 1.5 both sides
2B=1.5
divide by 2 both sides
B=0.75
plan A lasts 1/2 hour or 0.5 hour or 30 mins
plan B lasts 3/4 hour or 0.75 hour or 45 mins
Answer:
900cm^2
Step-by-step explanation:
24, 10, 30 in half= 12, 5, 15 so, 12*5*15=900cm^2
(-3y^2 − 8) − (-5y^2<span> + 1)
= </span>-3y^2 − 8 + 5y^2<span> - 1
= 2</span>y^2 <span>- 9
answer is
</span>2y^2 - 9<span>
</span>
