Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Connect with professionals ready to provide precise answers to your questions on our comprehensive Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

Given the functions:

[tex]\[
\begin{array}{l}
g(a) = 3a + 4 \\
h(a) = a^2 + a
\end{array}
\][/tex]

Find [tex]\(\left(\frac{g}{h}\right)\left(\frac{a}{2}\right)\)[/tex].

Sagot :

To find \( \left( \frac{g}{h} \right) \left( \frac{a}{2} \right)\), we'll proceed by evaluating the given functions \(g(a)\) and \(h(a)\) at a specific value of \(a\). Here's the step-by-step process:

1. Define the functions \( g(a) \) and \( h(a) \):
- \( g(a) = 3a + 4 \)
- \( h(a) = a^2 + a \)

2. Substitute \(\frac{a}{2}\) into the functions \( g \) and \( h \):

- Evaluate \( g \left( \frac{a}{2} \right) \):
[tex]\[ g \left( \frac{a}{2} \right) = 3 \left( \frac{a}{2} \right) + 4 \][/tex]
Simplifying this,
[tex]\[ g \left( \frac{a}{2} \right) = \frac{3a}{2} + 4 \][/tex]
Substituting \( a = 1 \),
[tex]\[ g \left( \frac{1}{2} \right) = \frac{3 \cdot 0.5}{2} + 4 = 1.5 + 4 = 5.5 \][/tex]

- Evaluate \( h \left( \frac{a}{2} \right) \):
[tex]\[ h \left( \frac{a}{2} \right) = \left( \frac{a}{2} \right)^2 + \frac{a}{2} \][/tex]
Simplifying this,
[tex]\[ h \left( \frac{a}{2} \right) = \left( \frac{a^2}{4} \right) + \left( \frac{a}{2} \right) \][/tex]
Substituting \( a = 1 \),
[tex]\[ h \left( \frac{1}{2} \right) = \left( \frac{0.5^2}{4} \right) + \left( \frac{0.5}{2} \right) = 0.25 + 0.5 = 0.75 \][/tex]

3. Calculate \(\left( \frac{g}{h} \right) \left( \frac{a}{2} \right)\):
[tex]\[ \left( \frac{g}{h} \right) \left( \frac{a}{2} \right) = \frac{g \left( \frac{a}{2} \right)}{h \left( \frac{a}{2} \right)} \][/tex]
Substituting the values we found,
[tex]\[ \left( \frac{g}{h} \right) \left( \frac{1}{2} \right) = \frac{5.5}{0.75} = 7.\overline{3} \][/tex]

Therefore, the result is:
[tex]\[(5.5, 0.75, 7.333333333333333)\][/tex]