<span>Consider a angle â BAC and the point D on its defector
Assume that DB is perpendicular to AB and DC is perpendicular to AC.
Lets prove DB and DC are congruent (that is point D is equidistant from sides of an angle â BAC
Proof
Consider triangles ΔADB and ΔADC
Both are right angle, â ABD= â ACD=90 degree
They have congruent acute angle â BAD and â CAD( since AD is angle bisector)
They share hypotenuse AD
therefore these right angle are congruent by two angle and sides and, therefore, their sides DB and DC are congruent too, as luing across congruent angles</span>
Answer:
k = - 6
Step-by-step explanation:
Since (x - 2) is a factor then x = 2 is a root
Substitute x = 2 into the polynomial and equate to zero, that is
2³ + k(2)² + 12(2) - 8 = 0
8 + 4k + 24 - 8 = 0
4k + 24 = 0 ( subtract 24 from both sides )
4k = - 24 ( divide both sides by 4 )
k = - 6
Answer:
228.42 ft
Step-by-step explanation:
To solve for this question, we would be applying the Trigonometric function of Sine.
Sin theta = Opposite/Hypotenuse
Theta = 52°
Opposite = 180ft
Hypotenuse = Length of the wire = x
Hence:
sin 52 = 180/x
x = 180/sin 52
x = 228.42327871 ft
Approximately = 228.42 ft
55/9 = 6.11
57/7 = 8.14
41/5 = 8.2
65/8 = 8.125
The answer is 41/5 is larger than 8.15
