Question

Evaluate the following expressions 3a/2b+z if a=11and b=1

1-x⁵+2x+7x if x=1/2
x³+3x²+2 if x=1

Answer 1

Let's evaluate each expression step by step.

1. For the expression \( \frac{3a}{2b} + z \) with \( a = 11 \) and \( b = 1 \):

\[
\frac{3(11)}{2(1)} + z = \frac{33}{2} + z = 16.5 + z
\]

So, the expression evaluates to \( 16.5 + z \).

---

2. For the expression \( 1 - x^5 + 2x + 7x \) if \( x = \frac{1}{2} \):

First, simplify the expression:
\[
1 - x^5 + 2x + 7x = 1 - x^5 + 9x
\]
Now substitute \( x = \frac{1}{2} \):
\[
1 - \left(\frac{1}{2}\right)^5 + 9\left(\frac{1}{2}\right)
\]
Calculating \( \left(\frac{1}{2}\right)^5 = \frac{1}{32} \):
\[
1 - \frac{1}{32} + \frac{9}{2}
\]
Convert \( 1 \) to a fraction with a denominator of 32:
\[
1 = \frac{32}{32}
\]
Now we can rewrite the expression:
\[
\frac{32}{32} - \frac{1}{32} + \frac{9}{2} = \frac{32 - 1}{32} + \frac{9 \cdot 16}{32} = \frac{31}{32} + \frac{144}{32} = \frac{31 + 144}{32} = \frac{175}{32}
\]

So, the expression evaluates to \( \frac{175}{32} \).

---

3. For the expression \( x^3 + 3x^2 + 2 \) if \( x = 1 \):

Substituting \( x = 1 \):
\[
1^3 + 3 \cdot 1^2 + 2 = 1 + 3 + 2 = 6
\]

So, the expression evaluates to \( 6 \).

---

### Summary of Results

1. \( \frac{3a}{2b} + z = 16.5 + z \)
2. \( 1 - x^5 + 2x + 7x = \frac{175}{32} \)
3. \( x^3 + 3x^2 + 2 = 6 \)