Answer:
brainliest please
Step-by-step explanation:
x/3-2<4
x/3<6
x<18
<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:
32r + 39
Step-by-step explanation:
<u>Step 1: Distribute</u>
4(5 + 8r) + 19
<em>20 + 32r + 19</em>
<em />
<u>Step 2: Combine like terms</u>
<em>32r + 39</em>
<em />
Answer: 32r + 39
44-2x=x-10
subtract 44 from both sides
-2x= x-54
subtract x from both sides
-3x= -54
divide by -3
x= 18
square root of 18= 4.24
in radical form 3 square root of 2
hope this helps
you will need to find the area
A=height multiplied by the base divided by 2
