There are two <em>true</em> statements:
- When the function is composed with r, the <em>composite</em> function is V(t) = (1/48) · π · t⁶.
- V(r(6)) shows that the volume is 972π cubic inches after 6 seconds.
<h3>How to use composition between two function</h3>
Let be <em>f</em> and <em>g</em> two functions, there is a composition of <em>f</em> with respect to <em>g</em> when the domain of <em>f</em> is equal to the range of <em>g</em>. In this question, the <em>domain</em> variable of the function V(r) is replaced by substitution.
If we know that V(r) = (4/3) · π · r³ and r(t) = (1/4) · t², then the composite function is:
V(t) = (4/3) · π · [(1/4) · t²]³
V(t) = (4/3) · π · (1/64) · t⁶
V(t) = (1/48) · π · t⁶
There are two <em>true</em> statements:
- When the function is composed with r, the <em>composite</em> function is V(t) = (1/48) · π · t⁶.
- V(r(6)) shows that the volume is 972π cubic inches after 6 seconds.
To learn on composition between functions: brainly.com/question/12007574
#SPJ1
Answer:
the answer is c, (-10,4) and the radius is 10
Step-by-step explanation:
Hope this helps!❆
All real numbers above -4
