Iven h (t) = t squared + t + 12 andk (t) = StartRoot t minus 1 EndRoot, evaluate (k compose h) (10)

Step-by-step 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

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