Iven h (t) = t squared + t + 12 andk (t) = StartRoot t minus 1 EndRoot, evaluate (k compose h) (10)
1 answer:
Answer:
<h2>11</h2>
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
<em>k(h(10))= 11</em>
<em></em>
<em>hence the value of (k compose h) (10) is 11</em>
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