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Write the given function for the drone's height.

Computer Science

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Here are the solutions to the questions:

  1. a)
  1. Write the given function for the drone's height.

    h(t)=t2−6t+8h(t) = t^2 - 6t + 8
  2. Substitute t=4t = 4 seconds into the function.

    h(4)=(4)2−6(4)+8h(4) = (4)^2 - 6(4) + 8 h(4)=16−24+8h(4) = 16 - 24 + 8 h(4)=−8+8h(4) = -8 + 8 h(4)=0h(4) = 0

    The height of the drone at t=4t = 4 seconds is 0 units\boxed{0 \text{ units}}.

    b)

  3. To find the minimum height, complete the square for the function h(t)=t2−6t+8h(t) = t^2 - 6t + 8.

    Take half of the coefficient of tt and square it: (−62)2=(−3)2=9\left(\frac{-6}{2}\right)^2 = (-3)^2 = 9.

  4. Add and subtract this value to the expression.

    h(t)=(t2−6t+9)−9+8h(t) = (t^2 - 6t + 9) - 9 + 8
  5. Factor the perfect square trinomial and simplify the constants.

    h(t)=(t−3)2−1h(t) = (t - 3)^2 - 1

    The minimum value of (t−3)2(t-3)^2 is 0 when t=3t=3.
    Therefore, the minimum height is −1-1.
    The minimum height of the drone is −1 units\boxed{-1 \text{ units}}.

    1. a)
  6. Write the quadratic equation.

    x2−2x−8=0x^2 - 2x - 8 = 0
  7. Find two numbers that multiply to −8-8 and add to −2-2.

    These numbers are −4-4 and 2.

  8. Rewrite the middle term using these numbers and factor by grouping.

    x2−4x+2x−8=0x^2 - 4x + 2x - 8 = 0 x(x−4)+2(x−4)=0x(x - 4) + 2(x - 4) = 0 (x−4)(x+2)=0(x - 4)(x + 2) = 0
  9. Set each factor to zero and solve for xx.

    x−4=0orx+2=0x - 4 = 0 \quad \text{or} \quad x + 2 = 0 x=4orx=−2x = 4 \quad \text{or} \quad x = -2

    The solutions to the equation are ∗x=4 and x=−2∗\boxed*{x = 4 \text{ and } x = -2*}.

    b)

  10. The dimensions of a metal sheet must be positive.

    From part (a), the solutions are x=4x = 4 and x=−2x = -2.

  11. Discard the negative solution as a dimension cannot be negative.

    The two possible dimensions are ∗4 units and 2 units∗\boxed*{4 \text{ units and } 2 \text{ units}*}. (Assuming the dimensions are xx and x−2x-2 or similar, or simply the values of xx that represent dimensions). Given the context of dimensions, we usually take the positive values. If the dimensions are xx and x−2x-2, then x=4x=4 gives dimensions 4 and 2. If x=−2x=-2, it would give −2-2 and −4-4, which are not valid. So, the dimensions are 4 and 2.

    1. a)
  12. Write the expression for the robot's wheel rotation.

    w=3x3y2×4x5y4w = 3x^3y^2 \times 4x^5y^4
  13. Multiply the numerical coefficients.

    3×4=123 \times 4 = 12
  14. Apply the product rule for exponents (am×an=am+na^m \times a^n = a^{m+n}) for the xx terms and yy terms.

    For xx: x3×x5=x3+5=x8x^3 \times x^5 = x^{3+5} = x^8
    For yy: y2×y4=y2+4=y6y^2 \times y^4 = y^{2+4} = y^6

  15. Combine the results.

    w=12x8y6w = 12x^8y^6

    The simplified expression is 12x8y6\boxed{12x^8y^6}.

    b)

  16. Substitute x=2x = 2 and y=3y = 3 into the simplified expression from part (a).

    w=12(2)8(3)6w = 12(2)^8(3)^6
  17. Calculate the powers.

    28=2562^8 = 256 36=7293^6 = 729
  18. Multiply the values.

    w=12×256×729w = 12 \times 256 \times 729 w=3072×729w = 3072 \times 729 w=2239488w = 2239488

    The robot's wheel rotation value is 2239488\boxed{2239488}.

    1. a) A composite number is a positive integer that has at least one divisor other than 1 and itself. Numbers between 20 and 40: 21 (divisors: 3, 7) 22 (divisors: 2, 11) 24 (divisors: 2, 3, 4, 6, 8, 12) 25 (divisors: 5) 26 (divisors: 2, 13) 27 (divisors: 3, 9) 28 (divisors: 2, 4, 7, 14) 30 (divisors: 2, 3, 5, 6, 10, 15) 32 (divisors: 2, 4, 8, 16) 33 (divisors: 3, 11) 34 (divisors: 2, 17) 35 (divisors: 5, 7) 36 (divisors: 2, 3, 4, 6, 9, 12, 18) 38 (divisors: 2, 19) 39 (divisors: 3, 13) Any four composite numbers between 20 and 40 are: ∗21,22,24,25∗\boxed*{21, 22, 24, 25*}. (Other valid answers include 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39).

    b)

  19. Choose one of the numbers from part (a).

    Let's choose 24.

  20. Find its prime factorization.

    24=2×1224 = 2 \times 12 12=2×612 = 2 \times 6 6=2×36 = 2 \times 3

    So, 24=2×2×2×3=23×324 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3.
    The prime factorization of 24 is 23×3\boxed{2^3 \times 3}. (If a different number was chosen, the prime factorization would be different, e.g., for 21: 3×73 \times 7; for 22: 2×112 \times 11; for 25: 525^2).

    1. a)
  21. Write the expression for profit.

    P=5x2−20xP = 5x^2 - 20x
  22. Find the greatest common factor (GCF) of the terms 5x25x^2 and −20x-20x.

    The GCF of 5 and 20 is 5.
    The GCF of x2x^2 and xx is xx.
    So, the GCF of 5x25x^2 and −20x-20x is 5x5x.

  23. Factor out the GCF.

    P=5x(x−4)P = 5x(x - 4)

    The factorized expression is 5x(x−4)\boxed{5x(x - 4)} .

    b)

  24. Substitute x=7x = 7 into the profit equation.

    P=5x2−20xP = 5x^2 - 20x P=5(7)2−20(7)P = 5(7)^2 - 20(7)
  25. Calculate the values.

    P=5(49)−140P = 5(49) - 140 P=245−140P = 245 - 140 P=105P = 105

    The profit when x=7x = 7 (representing 7 thousand items) is 105 units of profit\boxed{105 \text{ units of profit}}. (The units of profit are not specified, but it's a numerical value).

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