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1.1 The picture below shows a rectangular advertising board that consists of a frame and a rectangular area for placing an advertisement poster - NSC Technical Mathematics - Question 1 - 2020 - Paper 1

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1.1-The-picture-below-shows-a-rectangular-advertising-board-that-consists-of-a-frame-and-a-rectangular-area-for-placing-an-advertisement-poster-NSC Technical Mathematics-Question 1-2020-Paper 1.png

1.1 The picture below shows a rectangular advertising board that consists of a frame and a rectangular area for placing an advertisement poster. (Note that the board... show full transcript

Worked Solution & Example Answer:1.1 The picture below shows a rectangular advertising board that consists of a frame and a rectangular area for placing an advertisement poster - NSC Technical Mathematics - Question 1 - 2020 - Paper 1

Step 1

1.1 (a) The outside length of the rectangular board

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Answer

The outside length of the rectangular board is given by the sum of the length of the poster (12 meters) and twice the width of the frame (2x meters). Thus, the formula for the outside length is:

L=12+2xL = 12 + 2x

Step 2

1.1 (b) The outside breadth of the rectangular board

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Answer

The outside breadth of the rectangular board is determined similarly by adding the height of the poster (3 meters) and twice the width of the frame (2x meters). Thus, the formula for the outside breadth is:

B=3+2xB = 3 + 2x

Step 3

1.1.2 Show that A = 4x² + 30x + 36

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To find the total area A of the rectangular board, we multiply the outside length and outside breadth:

A=(12+2x)(3+2x)A = (12 + 2x)(3 + 2x)

Expanding this, we have:

A=123+122x+2x3+2x2xA = 12 \cdot 3 + 12 \cdot 2x + 2x \cdot 3 + 2x \cdot 2x

This simplifies to:

A=36+24x+6x+4x2=4x2+30x+36A = 36 + 24x + 6x + 4x^2 = 4x^2 + 30x + 36

Step 4

1.1.3 Hence, determine the outside length (in metres) of the rectangular board.

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To find the outside length of the rectangular board, we set the equation obtained for A equal to 52 and solve:

4x2+30x+36=524x^2 + 30x + 36 = 52

This simplifies to:

4x2+30x16=04x^2 + 30x - 16 = 0

Using the quadratic formula, we find:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting the coefficients a = 4, b = 30, and c = -16 gives the values for x. The corresponding length is:

L=12+2xL = 12 + 2x

Step 5

1.2.1 Solve for x ∈ {Real Numbers}

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Answer

Starting with the equation:

3x=7x5\frac{3}{x} = 7x - 5

Multiply both sides by x (where x ≠ 0) to eliminate the fraction:

3=7x25x3 = 7x^2 - 5x

Rearranging gives:

7x25x3=07x^2 - 5x - 3 = 0

Using the quadratic formula:

x=5±(5)247(3)27x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 7 \cdot (-3)}}{2 \cdot 7}

Calculating gives two possible roots to 2 decimal places.

Step 6

1.2.2 Solve x² + 4 > 0

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The inequality (x^2 + 4 > 0) is always true since (x^2) is non-negative and the minimum value of (x^2) is 0. Therefore, the solution set is:

xRx \in \mathbb{R}

Step 7

1.3 Solve for x and y if y - x = 3 and 3x² + xy - y² = -3

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From the first equation, rearranging gives:

y=x+3y = x + 3 Substituting into the second equation:

3x2+x(x+3)(x+3)2=33x^2 + x(x + 3) - (x + 3)^2 = -3

Expanding and rearranging leads to a quadratic in terms of x, which can be solved accordingly.

Step 8

1.4.1 Express f as the subject of the formula

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Answer

Starting with:

Xc=12πfCX_c = \frac{1}{2\pi f C}

Rearranging gives:

f=12πXcCf = \frac{1}{2\pi X_c C}

Step 9

1.4.2 Hence, determine, to the nearest integer, the numerical value of f

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Substituting the values for (X_c = 63.66 \text{ ohms}) and (C = 50 \times 10^{-6} \text{ farads}):

f=12π(63.66)(50×106)f = \frac{1}{2\pi (63.66)(50 \times 10^{-6})}

Calculating gives approximately 50 hertz.

Step 10

1.5.1 Determine the sum (in binary form) of the TWO binary numbers

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Answer

To sum the binary numbers 110011 and 111010:

  110011
+ 111010
 --------
 1101101

Step 11

1.5.2 Convert the sum to its equivalent decimal number notation

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The binary sum 1101101 can be converted to decimal as follows:

126+125+024+123+122+021+120=64+32+0+8+4+0+1=109.1*2^6 + 1*2^5 + 0*2^4 + 1*2^3 + 1*2^2 + 0*2^1 + 1*2^0 = 64 + 32 + 0 + 8 + 4 + 0 + 1 = 109.

Thus, the equivalent decimal number notation is 109.

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