For each of the following inner product spaces v and linear operators t on v, evaluate t∗ at the given vector in v .
(a) v=r2,t(a, b)=(2a+b, a−3b),x=(3,5)
(b) v=c2,t(z1,z2)=(2z1+iz2,(1−i)z1),x=( 3−i,1+2i)
(c) v=p1(r) with ⟨f, g⟩=∫1−1f(t)g(t)dt, t(f)=f′+3f f(t)=4−2t

the answer is:

the answer is:

to solve the problem, we need to remember the quotient of power property, it's defined by the following relation:

if we have a quotienf of powers that have the same base, we need to keep the same base and subtract the exponent of the denominator power to the exponent of the numerator power.

so, we are given the expression:

then, calculating we have:

hence, the answer is:

have a nice day!

answer: 100

step-by-step explanation:

the answer would be d.

step-by-step explanation:

move all variables to the left side and all constants to the right side

(x−3)^2+(y−1)^2=16

this is the form of a circle. use this form to determine the center and radius of the circle.

( x − h ) 2 + ( y − k ) 2 = r 2

match the values in this circle to those of the standard form. the variable   r   represents the radius of the circle,   h   represents the x-offset from the origin, and   k   represents the y-offset from origin.

r = 4

h = 3

k = 1

the center of the circle is found at   ( h , k ) .

center:   ( 3 , 1 )

these values represent the important values for graphing and analyzing a circle.

center:   ( 3 , 1 ).

radius:   4