Answer:
This is proved by ASA congruent rule.
Step-by-step explanation:
Given KLMN is a parallelogram, and that the bisectors of ∠K and ∠L meet at A. we have to prove that A is equidistant from LM and KN i.e we have to prove that AP=AQ
we know that the diagonals of parallelogram bisect each other therefore the the bisectors of ∠K and ∠L must be the diagonals.
In ΔAPN and ΔAQL
∠PNA=∠ALQ (∵alternate angles)
AN=AL (∵diagonals of parallelogram bisect each other)
∠PAN=∠LAQ (∵vertically opposite angles)
∴ By ASA rule ΔAPN ≅ ΔAQL
Hence, by CPCT i.e Corresponding parts of congruent triangles PA=AQ
Hence, A is equidistant from LM and KN.
Answer:
Charlie is incoerre
Step-by-step explanation:
Given that :
Line of sight to top of flagpole = 16 feets
Angle of elevation = 60°
The height of the flagpole :
Applying trigonometry to the drawing in the attached picture :
We use:
Sinα = opposite / hypotenus
This enables us to use the value of the hypotenus (line of sight) to obtain the missing height value (h)
Sin(60°) = h / 16
h = 16 * sin 60
h = 16 * 0.8660254
h = 13.856406
Hence flagpole is about 13.86 feets high
Charlie's reasoning is incorrect, the line of sight constitutes the hypotenus and not Adjacent and hence, shoul
Answer:
the domain is -5 to infinity
Explanation:
f
(
x
)
=
√
x
+
5
A square root is
≥
0
so
x
+
5
≥
0
Here is a graph of the function
x
≥
−
5
g
r
a
p
h
{
√
x
+
5
[
−
10
,
10
,
−
5
,
5
]
}
