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

1 answer

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 \( 16.5 + z \).


  1. 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 \( \frac{175}{32} \).


  1. 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. \( x^3 + 3x^2 + 2 = 6 \)