Our aim is to calculate the Radius so that to use the formula related to the area of a segment of a circle, that is: Aire of segment = Ф.R²/2
Let o be the center of the circle, AB the chord of 8 in subtending the arc f120°
Let OH be the altitude of triangle AOB. We know that a chord perpendicular to a radius bisects the chord in the middle. Hence AH = HB = 4 in
The triangle HOB is a semi equilateral triangle, so OH (facing 30°)=1/2 R. Now Pythagoras: OB² = OH² + 4²==> R² = (R/2)² + 16
R² = R²/4 +16. Solve for R ==> R =8/√3
OB² = OH² +
Answer:No, because it has a constant rate of change.
Step-by-step explanation:
I don't know specifically what the sum and product of an equation is but
factor using quadratic formula
if you have an equation in ax^2+bx+c=0 then you can solve for x using

or

so input
a=1
b=7
c=2



=aprox -0.2984378812836



=aprox -6.7015621187165
Answer:
D. ∠B = ∠B' = 74° and ∠C = ∠C' = 66°
Step-by-step explanation:
Given,

We know,sum of all 3 angles of a triangle is
°
Doing so ,

Simplifying all like terms,

Dividing both side by 12


Now we plug this in each angles and find their exact values.
°
°
°
Now substituting the
in angles of dilated triangle.
°
°
°
We see that all the corresponding angles are equal and as far as the options are concerned only option (D) matches.