Answer:
Her adjusted body weight is 78.26kg = 78.3kg, rounded to the nearest tenth.
Step-by-step explanation:
To find the Adjusted Body Weight(AjBW), first we have to consider the Ideal Body Weight(IBW).
We have the following formulas:
, in which i is the number of inches that the woman has above 5 feet.
, in which 
is her Actual Body Weight
So, in this problem:
Each pound has 0.45kg. So her weight is 
kg.
The woman is 8 inches above 5 feet, so 
.
kg
Her ideal body weight is 63.9kg. Her adjusted body weight is:
kg
Her adjusted body weight is 78.26kg = 78.3kg.
Answer: (1,-1)
Step-by-step explanation:
Midpoint of BC=(6+4)/2, (3–1)/2. =(5,1)
Slope of BC is (3+1)/4–6)= 4/-2 = -2
Slope of perpendicular bisector of BC =+1/2
Eqn of perpendicular bisector is : Y-1 =1/2 (x-5)
Y=1/2 •(x-5) +1
Midpoint of AB. (6–2)/2, (3–1)/2 ={2,1)
Slope of AB is(3+1)/(-2–6) = 4/-8 =-1/2
Slope of perpendicular bisector = +2
Eqn of perpendicular bisector is Y-1. =2( X-2)
Y=2X-4+1 = 2X -3
Solving Y=(X-5)/2 +1
& Y=2X-3
2X-3 =(x-5/2)+1
2X-4 =(x-5/2)
4X-8 = x-5
3X =3
X=1
Y= 2×1–3= -1
Circumcentre is(1,-1)
The answer is Megan
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The domain is x and the range is f(x)
