A. terminating
b. repeating
c. repeating
d. terminating
e. repeating
i think
Answer:
Idk if its multiple choice but you can do 1 of 2 ways theres using distance formula =√(5-3)sq+(1-4)sq
=√4+9
=√13 <--
or
3.6 units
Given: The two points that are P(5,1) and Q(3,4).
To find: The distance between these two points.
Solution: It is given that there are two points that are P(5,1) and Q(3,4).
The distance between these two points can be found out as using the distance formula that is: 3.6
Thus, the distance between the given two points is 3.6 units.
So you choose 13 or 3.6 Hope this helps :)
Answer:
Here is the full proof:
AC bisects ∠BCD Given
∠CAB ≅ ∠CAD Definition of angle bisector
DC ⊥ AD Given
∠ADC = 90° Definition of perpendicular lines
BC ⊥ AB Given
∠ABC = 90° Definition of perpendicular lines
∠ADC ≅ ∠ABC Right angles are congruent
AC = AC Reflexive property
ΔCAB ≅ ΔCAD SAA
BC = DC CPCTC
Answer:
Fn= 174.9 N : Magnitude of the net force the people exert on the donkey.
Step-by-step explanation:
We find the components of the forces in x-y-z
Force of Jack in z =F₁z=90.5 N in direction (+z)
Force of Jill in x = F₂x= -82.3*cos45°= - 58.19 N (-x)
Force of Jill in y =F₂y=-82.3*sin45°= + 58.19 N (+y)
Force of Jane in x =F₃x=125*cos45°= + 88.4 N (+x)
Force of Jane in y =F₃y=125*sin45°= + 88.4 N (+y)
Calculating of the components of the net force the people exert on the donkey.
Fnx= F₂x+F₃x=( - 58.19+ 88.4 )N=30.2N (+x)
Fny= F₂y+F₃y=( 58.19+88.4 ) = 146.59 N (+y)
Fnz =F₁z=90.5 N (+z)
Calculating of the magnitude of the net force the people exert on the donkey.
Answer: 150-170 free throws
Step-by-step explanation:
200 times 80% or .8 will give you 160
5% of 200 is 10 so add and subtract that to/from 160 and you'll get your range
