https://file.aiquickdraw.com/mathbot/file/17273231893549hdqlyt8.jpg

Let's solve the indicated operations one by one based on the given functions:

### Functions:
- \( a(x) = 2x + 2 \)
- \( b(x) = x - 8 \)
- \( c(x) = 2x^2 + 24x + 20 \)
- \( d(x) = x^2 + 2x - 80 \)

### Now, let's compute the operations step by step.

#### 1-2. \( c(0) + d(3) \)
- \( c(0) = 2(0)^2 + 24(0) + 20 = 20 \)
- \( d(3) = (3)^2 + 2(3) - 80 = 9 + 6 - 80 = -65 \)
Thus, \( c(0) + d(3) = 20 + (-65) = -45 \).

#### 3-4. \( a(x) + (d - c)(x) \)
- \( d(x) - c(x) = (x^2 + 2x - 80) - (2x^2 + 24x + 20) \)
  Simplifying this:
  \[
  d(x) - c(x) = x^2 + 2x - 80 - 2x^2 - 24x - 20 = -x^2 - 22x - 100
  \]
- Now, \( a(x) + (d(x) - c(x)) = (2x + 2) + (-x^2 - 22x - 100) \)
  \[
  = -x^2 - 20x - 98
  \]

#### 5-6. \( (a - b)(x) \)
- \( a(x) - b(x) = (2x + 2) - (x - 8) = 2x + 2 - x + 8 = x + 10 \)

#### 7-8. \( b(5) \cdot (c(-1)) \)
- \( b(5) = 5 - 8 = -3 \)
- \( c(-1) = 2(-1)^2 + 24(-1) + 20 = 2(1) - 24 + 20 = -2 \)
Thus, \( b(5) \cdot c(-1) = (-3) \cdot (-2) = 6 \).

#### 9-10. \( \left(\frac{c}{d}\right)(4) \)
- \( c(4) = 2(4)^2 + 24(4) + 20 = 2(16) + 96 + 20 = 32 + 96 + 20 = 148 \)
- \( d(4) = (4)^2 + 2(4) - 80 = 16 + 8 - 80 = -56 \)
Thus, \( \left(\frac{c}{d}\right)(4) = \frac{148}{-56} = -\frac{37}{14} \).

#### 11-12. \( (b \circ d)(x) \)
- \( (b \circ d)(x) = b(d(x)) = b(x^2 + 2x - 80) \)
- Using \( b(x) = x - 8 \), we get:
  \[
  b(d(x)) = (x^2 + 2x - 80) - 8 = x^2 + 2x - 88
  \]

#### 13-15. \( (c \circ a)(x) \)
- \( (c \circ a)(x) = c(a(x)) = c(2x + 2) \)
- Using \( c(x) = 2x^2 + 24x + 20 \), we substitute \( x = 2x + 2 \):
  \[
  c(2x + 2) = 2(2x + 2)^2 + 24(2x + 2) + 20
  \]
  First, calculate \( (2x + 2)^2 = 4x^2 + 8x + 4 \), so:
  \[
  c(2x + 2) = 2(4x^2 + 8x + 4) + 24(2x + 2) + 20 = 8x^2 + 16x + 8 + 48x + 48 + 20
  \]
  Simplifying:
  \[
  c(2x + 2) = 8x^2 + 64x + 76
  \]

#### 16-17. \( c \circ d(10) \)
- \( d(10) = 10^2 + 2(10) - 80 = 100 + 20 - 80 = 40 \)
- \( c(40) = 2(40)^2 + 24(40) + 20 = 2(1600) + 960 + 20 = 3200 + 960 + 20 = 4180 \)

#### 18-20. \( a \circ (c - d)(5) \)
- First, calculate \( (c - d)(5) \):
  \[
  c(5) = 2(5)^2 + 24(5) + 20 = 2(25) + 120 + 20 = 50 + 120 + 20 = 190
  \]
  \[
  d(5) = (5)^2 + 2(5) - 80 = 25 + 10 - 80 = -45
  \]
  So, \( (c - d)(5) = 190 - (-45) = 190 + 45 = 235 \).
- Now, \( a(235) = 2(235) + 2 = 470 + 2 = 472 \).

### Summary of Results:
1-2: \( -45 \)  
3-4: \( -x^2 - 20x - 98 \)  
5-6: \( x + 10 \)  
7-8: \( 6 \)  
9-10: \( -\frac{37}{14} \)  
11-12: \( x^2 + 2x - 88 \)  
13-15: \( 8x^2 + 64x + 76 \)  
16-17: \( 4180 \)  
18-20: \( 472 \)

Would you like any further details on any of these steps or explanations?

### Related Questions:
1. How do you evaluate composite functions like \( f(g(x)) \)?
2. Can you explain how to simplify polynomial subtraction?
3. What are some tips for factoring quadratics effectively?
4. How do you find the quotient when dividing two functions?
5. What is the process for computing function composition with specific values?

**Tip:** When working with composite functions, always simplify the inner function first before applying the outer function.

Given the following functions: a(x) = 2x + 2, b(x) = x - 8, c(x) = 2x^2 + 24x + 20, d(x) = x^2 + 2x - 80. Perform the indicated operations for 1-20.

Solve a variety of function operations, including addition, subtraction, multiplication, and composition, using polynomial functions such as a(x) = 2x + 2 and c(x) = 2x^2 + 24x + 20. Suitable for Grades 10-12.

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