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1.1 Los op vir $x$: 1.1.1 $x (7 + x) = 0$ 1.1.2 $4x^2 - 5x - 4 = 0$ (korrek tot TWEE desimale plekke) 1.1.3 $2x^2 - 8 > 0$ 1.2 Los op vir $x$ en $y$ indien: $y = 5x - 2$ en $y = x^3 + 4x - 8$ 1.3 Die diagram hieronder toon die beweging van 'n suier binne die motor-ein se silinder - NSC Technical Mathematics - Question 1 - 2022 - Paper 1

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1.1-Los-op-vir-$x$:--1.1.1-$x-(7-+-x)-=-0$--1.1.2-$4x^2---5x---4-=-0$-(korrek-tot-TWEE-desimale-plekke)--1.1.3-$2x^2---8->-0$--1.2-Los-op-vir-$x$-en-$y$-indien:--$y-=-5x---2$-en-$y-=-x^3-+-4x---8$--1.3-Die-diagram-hieronder-toon-die-beweging-van-'n-suier-binne-die-motor-ein-se-silinder-NSC Technical Mathematics-Question 1-2022-Paper 1.png

1.1 Los op vir $x$: 1.1.1 $x (7 + x) = 0$ 1.1.2 $4x^2 - 5x - 4 = 0$ (korrek tot TWEE desimale plekke) 1.1.3 $2x^2 - 8 > 0$ 1.2 Los op vir $x$ en $y$ indien: $y ... show full transcript

Worked Solution & Example Answer:1.1 Los op vir $x$: 1.1.1 $x (7 + x) = 0$ 1.1.2 $4x^2 - 5x - 4 = 0$ (korrek tot TWEE desimale plekke) 1.1.3 $2x^2 - 8 > 0$ 1.2 Los op vir $x$ en $y$ indien: $y = 5x - 2$ en $y = x^3 + 4x - 8$ 1.3 Die diagram hieronder toon die beweging van 'n suier binne die motor-ein se silinder - NSC Technical Mathematics - Question 1 - 2022 - Paper 1

Step 1

1.1.1 $x (7 + x) = 0$

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Answer

To solve the equation x(7+x)=0x (7 + x) = 0, we can set each factor equal to zero:

  • x=0x = 0
  • 7+x=0    x=77 + x = 0 \implies x = -7

Thus, the solutions are x=0x = 0 or x=7x = -7.

Step 2

1.1.2 $4x^2 - 5x - 4 = 0$ (korrek tot TWEE desimale plekke)

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Answer

Using the quadratic formula:

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

where a=4a = 4, b=5b = -5, and c=4c = -4:

x=5±(5)24(4)(4)2(4)=5±25+648=5±898x = \frac{5 \pm \sqrt{(-5)^2 - 4(4)(-4)}}{2(4)} = \frac{5 \pm \sqrt{25 + 64}}{8} = \frac{5 \pm \sqrt{89}}{8}

Calculating the roots:

  • x11.80x_1 \approx 1.80
  • x20.55x_2 \approx -0.55

Thus, the solutions are x1.80x \approx 1.80 and x0.55x \approx -0.55.

Step 3

1.1.3 $2x^2 - 8 > 0$

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Answer

To find the critical points, we set 2x28=02x^2 - 8 = 0:

2(x24)=0    (x2)(x+2)=02(x^2 - 4) = 0 \implies (x - 2)(x + 2) = 0

Thus, x=2x = 2 or x=2x = -2. We analyze the intervals to determine the solution set:

  • For x<2x < -2, the expression is positive.
  • For 2<x<2-2 < x < 2, the expression is negative.
  • For x>2x > 2, the expression is positive.

Thus, the solution set is (,2)(2,)(-\infty, -2) \cup (2, \infty).

Step 4

1.2 Los op vir $x$ en $y$ indien: $y = 5x - 2$ en $y = x^3 + 4x - 8$

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Answer

To find xx and yy, we set the equations equal to each other:

5x2=x3+4x85x - 2 = x^3 + 4x - 8

Rearranging leads to:

x3x+6=0x^3 - x + 6 = 0

By testing potential rational roots, we find:

  • At x=2x = -2: (2)3(2)+6=0(-2)^3 - (-2) + 6 = 0

Thus, x=2x = -2. We can use synthetic division to factor the cubic and find other roots as x=3x = 3 and x=2x = -2:

So y=5(2)2=12y = 5(-2) - 2 = -12 and y=5(3)2=13y = 5(3) - 2 = 13. Therefore, the solutions are:

  • x=2x = -2 leads to y=12y = -12
  • x=3x = 3 leads to y=13y = 13.

Step 5

1.3.1 Maak $L$ die onderwerp van die formule.

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Answer

Starting with the formula for the slagvolume:

SV=πd2×L4SV = \frac{\text{π} d^2 \times L}{4}

To isolate LL, we rearrange the equation:

L=4SVπd2L = \frac{4SV}{\text{π}d^2}

Step 6

1.3.2 Bereken vervolgens (afgerond tot die naaste cm), die numeriese waarde van $L$ as $SV = 1 020,5$ cm³ en die middellyn $d = 10$ cm.

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Answer

Using the formula derived above:

Substituting SV=1020,5SV = 1 020,5 and d=10d = 10:

L=4(1020,5)π(10)24082100π13cmL = \frac{4(1 020,5)}{\text{π}(10)^2} \approx \frac{4 082}{100\text{π}} \approx 13 \, \text{cm}

Step 7

1.3.3 Gee die sisteem eienskappe: $P = 1 010$, en $Q = 10 000$.

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Answer

The values given are:

  • P=1010P = 1 010
  • Q=10000Q = 10 000

These could represent various properties like pressure and volume, expressed in appropriate units depending on the context.

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