g(x)=2x+9 › g(-1)=-1.2+9=7

f(x)=x-5 ›f(7)=7-5=2 (we used the output of the function g(x), as the input of the function f(x))

y=2x-3, (3, 4) is not on the line y=2x-3 because if x=3 then y=2*3-3=3. The answer is D.

Given these intervals, (A ? B) is [0, 7). Then, (2, 8) \ [0, 7) is the required result. It is [7, 8), and D is the correct answer. 

Only the last argument is false. log_a(x)^y=ylog_ax. So the remaining 3 arguments are true.

f(6)=2-6=-4 hence 6>5

f(3)=3.3=9 hence 1<3<5

f(-1)=(-1)²+4=5 hence -1<1

so, 2f(6)-3f(3)+6f(-1)=2.(-4)-3.9+6.5=-8-27+30=-5

2000 +2000 × 10/100 + 2000 × 10/100 = 2000 + 2×200 = 2400 TL

The correct answer is B.

Since A represents the infinite whole real line interval, so does A ∪ B. Then, C  \ (A ∪ B) must be an empty set.  

The correct answer is D.

The vertex of y=x² +2x +1 is (-1, – 0) because - (b / 2a)= -(2 / 2) = -1 and y=(-1)² + 2.-1 + 1 = 1 - 2 + 1 = 0

The answer is A.

 k(x)=3/(x-2)›k(1)=3/(1-2)=-3

m(x)=(3/x)+7›m(-3)=(3/-3)+7=6

m(k(1))=(mok)(1)=6

log3(1/(x+1))=2› 3²=1/(X+1)›9=1/(X+1) then 9x+9=1›x=-8/9

(f °g)(1)=f(g(1)) hence g(1)=0 f(0)=5.(0)³=0

Constant functions map every element from their domain to the same element of the range. We may therefore define a constant function for every element in the range. Since there are 11 different elements in the range we may write 11 different constant functions.

using the change-of-base formula, log₅2=ln2/ln5

log₄3=ln3/ln4=ln3/ln2²=ln3/2ln2

log₉5=ln5/ln9=ln5/3ln3

then,log₅2 . log₄3 . log₉5=(ln2/ln5) . (ln3/2ln2) . (ln5/3ln3)=1/6

We write 81x⁴ = (3x)⁴ and use the law
ln ab = b ln a,

then, 4ln(3x)=20›ln(3x)=5›e⁵=3x›e⁵/3=x

An exponential function moves multiplicative. However  increases arithmetically. Hence the solution is B.

The intersection interval is (2, 4] because 2 is open in the second interval and 4 is closed in both intervals.

4(4 + 9) - 3(12-6) = 52 - 18 = 34