Replace x with -1.
G(-1) = -5(-1) + 2
= 5 + 2
= 7
I hope this Helps!
It has not been indicated whether the figure in the questions is a triangle or a quadrilateral. Irrespective of the shape, this can be solved. The two possible shapes and angles have been indicated in the attached image.
Now, from the information given we can infer that there is a line BD that cuts angle ABC in two parts: angle ABD and angle DBC
⇒ Angle ABC = Angle ABD + Angle DBC
Also, we know that angle ABC is 1 degree less than 3 times the angle ABD, and that angle DBC is 47 degree
Let angle ABD be x
⇒ Angle ABC = 3x-1
Also, Angle ABC = Angle ABD + Angle DBC
Substituting the values in the above equations
⇒ 3x-1 = x+47
⇒ 2x = 48
⇒ x = 24
So angle ABD = 24 degree, and angle ABC = 3(24)-1 = 71-1 = 71 degree
Answer:
Perron–Frobenius theorem for irreducible matrices. Let A be an irreducible non-negative n × n matrix with period h and spectral radius ρ(A) = r. Then the following statements hold. The number r is a positive real number and it is an eigenvalue of the matrix A, called the Perron–Frobenius eigenvalue.
Answer:
perimeter of circle=2πr
16=2πr
r=16/2π=8/πcm
now.
area of circle=πr²=π(8/π)²=π×64/π²=64/(22/7)=64×7/22=20.36cm²