Answer:
Step-by-step explanation:
given
m∠A=(9y-15)°,
m∠B=(y^2+20)°, and
m∠C=(y+31)°.
In a triangle
m∠A+m∠B+∠C=(9y-15)°+(y^2+20)°+(y+31)°
m∠A+m∠B+∠C =10y+y^2-15+20+31
m∠A+m∠B+∠C =10y+y^2+36
m∠A+m∠B+∠C=y^2+10y+36
m∠A+m∠B+∠C=180°
y^2+10y+36-180=0
Y^2+10y-144=0
Y^2+(18-8)y-144=0
Y^2+18y-8Y-144=0
Y(Y+18)-8(Y+18)=0
(Y+18)(Y-8)=0
We should take positive value of Y so we take Y=8
So value of angle c is
m∠C=(y+31)°.
=(8+31)°
=39°
So the value of angle c is 39°.
(2x100)x9, is one possible way.
20/45 can be reduced to 4/9. That is the simplest form.
Answer: b = p - a -c
Step-by-step explanation:
p= a + b + c Solving for b means getting b on one side by itself.
p = a + b + c To solve for b first subtract a from both sides to get,
-b -b
p - a = b + c Now subtract c from both sides to get
-c -c
p - a - c = b
b = p - a -c
