HELP In triangle JKL, sin(b°) = three fifths and cos(b°) = four fifths. If triangle JKL is dilated by a scale factor of 2, what is tan(b°)? triangle KL in which angle K is a right angle and angle L measures b degrees tan(b°) = five thirds tan(b°) = four thirds tan(b°) = three fourths tan(b°) = five fourths

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!

the answer is 1/4

step-by-step explanation:

the domain of a function is the set of input values, x , for which a function is defined. the domain is shown in the left oval in the picture below. the function provides an output value, f(x) , for each member of the domain.

the answer is z7+8/5