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Here are the solutions to the questions. 1) Express: (x+1)/3 - (2x+4)/5 as a single fraction in its simplest form. 2) Calculate the value of 0.45 + 0.005. 3) Make y the subject: p = (2y−c)/m. 4) The r

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2 parts
a

070∘070^\circ

b

250∘250^\circ

y=23x−3y = \frac{2}{3}x - 3

3y=2x−93y = 2x - 9

Step-by-step solution

12 steps

Here are the solutions to the questions.

QUESTION ONE

1) Express: x+13−2x+45\frac{x+1}{3} - \frac{2x+4}{5} as a single fraction in its simplest form.

Step 1: Find the least common multiple (LCM) of the denominators 3 and 5, which is 15.

Step 2: Rewrite each fraction with the common denominator.

5(x+1)15−3(2x+4)15\frac{5(x+1)}{15} - \frac{3(2x+4)}{15}

Step 3: Combine the fractions over the common denominator.

5(x+1)−3(2x+4)15\frac{5(x+1) - 3(2x+4)}{15}

Step 4: Expand the terms in the numerator.

5x+5−(6x+12)15\frac{5x+5 - (6x+12)}{15} 5x+5−6x−1215\frac{5x+5 - 6x-12}{15}

Step 5: Simplify the numerator by combining like terms.

(5x−6x)+(5−12)15\frac{(5x-6x) + (5-12)}{15} −x−715\frac{-x-7}{15}

The single fraction in its simplest form is −x−715\boxed{\frac{-x-7}{15}}.

2) Calculate the value of 0.45+0.0050.45 + 0.005.

Step 1: Align the decimal points and add the numbers.

0.4500.450 +0.005+ 0.005

−−−−−-----

0.4550.455

The value is 0.455\boxed{0.455}.

3) Make yy the subject: p=2y−cmp = \frac{2y-c}{m}

Step 1: Multiply both sides of the equation by mm.

pm=2y−cpm = 2y-c

Step 2: Add cc to both sides of the equation.

pm+c=2ypm+c = 2y

Step 3: Divide both sides by 2.

y=pm+c2y = \frac{pm+c}{2}

Making yy the subject gives y=pm+c2\boxed{y = \frac{pm+c}{2}}.

4) The ratio of female to male student at NACE is 5:4. If the number of male students is 1000. Find the total number of students at NACE.

Step 1: Set up the ratio of female (F) to male (M) students.

FM=54\frac{F}{M} = \frac{5}{4}

Step 2: Substitute the given number of male students, M=1000M=1000.

F1000=54\frac{F}{1000} = \frac{5}{4}

Step 3: Solve for the number of female students, FF.

F=54×1000F = \frac{5}{4} \times 1000 F=5×250F = 5 \times 250 F=1250F = 1250

Step 4: Calculate the total number of students.

Total students=F+M=1250+1000=2250\text{Total students} = F + M = 1250 + 1000 = 2250

The total number of students at NACE is 2250\boxed{2250}.

5) The diagram below shows the position of town P and town Q.

Part a· 3 steps

Find the bearing of town P from Q.

  1. Identify the North line at Q.

  2. Bearings are measured clockwise from the North line.

  3. The angle given in the diagram from the North line at Q to the line segment QP is 70∘70^\circ.

    The bearing of town P from Q is 070∘\boxed{070^\circ}.

Part b

250∘250^\circ

Find the bearing of town Q from P.

  1. Draw a North line at town P, parallel to the North line at town Q.

  2. The angle between the North line at Q and the line QP is 70∘70^\circ.

  3. Due to parallel North lines, the alternate interior angle formed by the line PQ and the North line at P (if extended to the left) would be 70∘70^\circ.

  4. The angle between the line PQ and the South line at P (which is 180∘180^\circ from North) is also 70∘70^\circ (alternate interior angles with the angle NQPNQP).

  5. The bearing of Q from P is measured clockwise from the North line at P.

    This angle is 180∘180^\circ (to the South line) plus the 70∘70^\circ angle.

    Bearing=180∘+70∘=250∘\text{Bearing} = 180^\circ + 70^\circ = 250^\circ

    The bearing of town Q from P is 250∘\boxed{250^\circ}.

    QUESTION TWO

    1) Find the equation of the straight line passing through the points P (6,1) and Q (0,-3).

  6. Calculate the gradient (mm) of the line using the formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}.

    Let (x1,y1)=(6,1)(x_1, y_1) = (6,1) and (x2,y2)=(0,−3)(x_2, y_2) = (0,-3).

    m=−3−10−6=−4−6=23m = \frac{-3 - 1}{0 - 6} = \frac{-4}{-6} = \frac{2}{3}

  7. Use the point-slope form of a linear equation, y−y1=m(x−x1)y - y_1 = m(x - x_1), with one of the points (e.g., P(6,1)) and the gradient.

    y−1=23(x−6)y - 1 = \frac{2}{3}(x - 6)

  8. Multiply both sides by 3 to eliminate the fraction.

    3(y−1)=2(x−6)3(y - 1) = 2(x - 6) 3y−3=2x−123y - 3 = 2x - 12

  9. Rearrange the equation into the form y=mx+cy = mx+c or Ax+By+C=0Ax+By+C=0.

    3y=2x−12+33y = 2x - 12 + 3 3y=2x−93y = 2x - 9

    Or, in y=mx+cy = mx+c form:

    y=23x−93y = \frac{2}{3}x - \frac{9}{3} y=23x−3y = \frac{2}{3}x - 3

    The equation of the straight line is y=23x−3\boxed{y = \frac{2}{3}x - 3} or 3y=2x−9\boxed{3y = 2x - 9}.

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