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32 Factorize: 3p2 + 16 33 A straight line AB passes through points A(3,2) and B(5,6). Calculate the gradient of the line. 34 A straight line whose gradient is rac13 passes through points R(8,4) and P(

Mathematics

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2
3

8≤h<128 \le h < 12

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32 Factorize: 3p2+163p^2 + 16

This expression is a sum of two terms, where both terms are positive for any real value of pp. It cannot be factorized into simpler expressions with real coefficients.

33 A straight line AB passes through points A(3,2) and B(5,6). Calculate the gradient of the line.

  1. Identify the coordinates of the points.

    Let (x1,y1)=(3,2)(x_1, y_1) = (3,2) and (x2,y2)=(5,6)(x_2, y_2) = (5,6).

  2. Use the gradient formula.

    The gradient mm is given by the formula:

    m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}
  3. Substitute the values and calculate.

    m=6−25−3m = \frac{6 - 2}{5 - 3} m=42m = \frac{4}{2} m=2m = 2

    The gradient of the line is 2\boxed{2}.

    34 A straight line whose gradient is 13\frac{1}{3} passes through points R(8,4) and P(5,k). Find the value of k.

  4. Identify the given gradient and points.

    Given gradient m=13m = \frac{1}{3}.
    Let (x1,y1)=(8,4)(x_1, y_1) = (8,4) and (x2,y2)=(5,k)(x_2, y_2) = (5,k).

  5. Use the gradient formula.

    m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}
  6. Substitute the known values into the formula.

    13=k−45−8\frac{1}{3} = \frac{k - 4}{5 - 8} 13=k−4−3\frac{1}{3} = \frac{k - 4}{-3}
  7. Solve for kk.

    Multiply both sides by −3-3:

    −3×13=k−4-3 \times \frac{1}{3} = k - 4 −1=k−4-1 = k - 4

    Add 4 to both sides:

    −1+4=k-1 + 4 = k 3=k3 = k

    The value of kk is 3\boxed{3}.

    35 The height of a flag post must be 8m or more but shorter than 12m. The flag post at Umoja Junior School is 2m tall. Form an inequality to represent the information.

    Let hh be the height of the flag post in meters.
    "8m or more" means h≥8h \ge 8.
    "shorter than 12m" means h<12h < 12.

    Combining these two conditions, the inequality is:

    8≤h<12\boxed{8 \le h < 12}

    36 The equation of a straight line is given as: 2x+3y=92x + 3y = 9. Determine the gradient of the straight line.

  8. Rearrange the equation into the slope-intercept form y=mx+cy = mx + c.

    2x+3y=92x + 3y = 9

    Subtract 2x2x from both sides:

    3y=−2x+93y = -2x + 9

    Divide all terms by 3:

    y=−2x3+93y = \frac{-2x}{3} + \frac{9}{3} y=−23x+3y = -\frac{2}{3}x + 3
  9. Identify the gradient.

    In the form y=mx+cy = mx + c, mm is the gradient.
    The gradient of the straight line is ∗−23∗\boxed*{-\frac{2}{3}}*.

    37 A straight line passes through points N(0,2) and M(8,5). Determine its equation in the form y=mx+cy = mx + c.

  10. Calculate the gradient mm.

    Let (x1,y1)=(0,2)(x_1, y_1) = (0,2) and (x2,y2)=(8,5)(x_2, y_2) = (8,5).

    m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} m=5−28−0m = \frac{5 - 2}{8 - 0} m=38m = \frac{3}{8}
  11. Determine the y-intercept cc.

    Since point N is (0,2)(0,2), this means that when x=0x=0, y=2y=2. In the equation y=mx+cy = mx + c, when x=0x=0, y=cy=c. Therefore, c=2c=2.

  12. Write the equation of the line.

    Substitute the gradient m=38m = \frac{3}{8} and the y-intercept c=2c = 2 into the form y=mx+cy = mx + c.

    y=38x+2y = \frac{3}{8}x + 2

    The equation of the line is ∗y=38x+2∗\boxed*{y = \frac{3}{8}x + 2}*.

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