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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

1.1.1 (a) The outside length of the rectangular board

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Answer

The outside length of the rectangular board can be expressed in terms of x as:

12+2x12 + 2x

This accounts for the length of the advertisement poster (12m) plus twice the width of the frame.

Step 2

1.1.1 (b) The outside breadth of the rectangular board

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Answer

The outside breadth of the rectangular board is given by:

3+2x3 + 2x

This includes the width of the advertisement poster (3m) plus twice the width of the frame.

Step 3

1.1.2 Show that the total front area (A) of the rectangular board can be expressed as A = 4x² + 30x + 36

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Answer

The total front area A can be calculated using the formula for the area of a rectangle:

egin{align*} A &= (12 + 2x)(3 + 2x)
&= 12 \cdot 3 + 12 \cdot 2x + 3 \cdot 2x + 2x \cdot 2x
&= 36 + 24x + 6x + 4x^2
&= 4x^2 + 30x + 36 \end{align*}

Thus, the expression for A is confirmed.

Step 4

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

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Answer

To find the outside length, we must first solve the equation for the area:

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

which simplifies to:

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

Using the quadratic formula, we get:

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

Substituting values gives:

x=30±30244(16)24x = \frac{-30 \pm \sqrt{30^2 - 4 \cdot 4 \cdot (-16)}}{2 \cdot 4}

which leads to:

After finding the values of x, plug the valid value of x into the expression for the outside length:

Outside length = 12+2x12 + 2x

Step 5

1.2.1 Solve \( \frac{3}{x} = 7x - 5, \; x ≠ 0 \) (correct to TWO decimal places)

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Answer

To solve this equation:

3=x(7x5)3 = x(7x - 5)

which simplifies to:

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}

Calculate to find the two possible values of x, rounding to two decimal places.

Step 6

1.2.2 Solve \( x^2 + 4 > 0 \)

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Answer

The inequality ( x^2 + 4 > 0 ) is satisfied for all x since

yields no real roots. Hence, it is true for all real numbers.

Step 7

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

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Answer

From the first equation, we can express y as:

y=x+3y = x + 3

Substituting this into the second equation gives:

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

Expanding and rearranging leads to a quadratic equation in terms of x. Solve this quadratic equation to find the values of x and use them to find y.

Step 8

1.4.1 Express f as the subject of the formula.

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Answer

Starting from the formula:

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 if \( X_c = 63.66 \; ohms \) and \( C = 50 \times 10^{-6} \; farads \)

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Answer

Using the previous result for f:

f=12π63.6650×106f = \frac{1}{2 \pi \cdot 63.66 \cdot 50 \times 10^{-6}}

Calculating this value gives the frequency, which is rounded to the nearest integer.

Step 10

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

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Answer

Adding the two binary numbers 1100111 and 1111010:

   1100111
 + 1111010
 ---------
  11010001

This gives a binary sum of 11010001.

Step 11

1.5.2 Convert, clearly showing all calculations, the sum obtained in QUESTION 1.5 to its equivalent decimal number notation.

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Answer

To convert the binary number 11010001 to decimal, compute:

127+126+025+124+023+022+021+120=128+64+16+1=2091 \cdot 2^7 + 1 \cdot 2^6 + 0 \cdot 2^5 + 1 \cdot 2^4 + 0 \cdot 2^3 + 0 \cdot 2^2 + 0 \cdot 2^1 + 1 \cdot 2^0 = 128 + 64 + 16 + 1 = 209

Thus, the equivalent decimal number is 209.

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