Answer:
11
Explanation:
Given h(t) = t²+t+ 12 and k(t) = √t-1, we are to find k(k.h)(10)
k{h(t)} = k{ t²+t+ 12}
Since k(t)= √t-1, we will replace the variable t in the function with t²+t+ 12
k(h(t)) = √{(t²+t+ 12)-1}
k(h(t)) = √t²+t+12-1
k(h(t)) = √t²+t+11
Substituting t = 10 into the resulting function;
k(h(10)) = √(10)²+(10)+11
k(h(10)) = √100+10+11
k(h(10)) = √121
k(h(10))= 11
hence the value of (k compose h) (10) is 11