<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>
The equation of a circle is: (x - h)² + (y - k)² = radius² where (h, k) is the center. Now just plug in your given information and simplify.
(x - 2)² + (y - 0)² = 9²
Your equation is:
(x - 2)² + y² = 81
Answer: y = 5d + 7 / c - 3
Step-by-step explanation:
Step 1: Add -3y to both sides
cy - 7 + -3y = 5d + 3y + -3y
= cy - 3y - 7 = 5d
Step 2: Now, add 7 on both sides
cy - 3y - 7 +7 = 5d +7
= cy - 3y = 5d +7
Step 3: Factor out y
y(c - 3) = 5d + 7
Step 4: Divide both sides by c-3
y(c - 3) / c - 3 = 5d+7 / c-3
Your answer for this should be
y = 5d + 7 / c - 3
Hope this helped!
Answer:
So basically anything higher than .5 would be x>.5 so it has to be greater than or equal to .51
Step-by-step explanation:
.51 .75 .89 .56 .78 .86 .61 it goes on and on as long as it is greater than .50
