1.1 Los op vir $x$:
1.1.1 $2x(x + 3) = 0$
1.1.2 $x + 9 = 12$ (korrekt tot TWEE desimale plakke)
1.1.3 $x(6 - x) ext{ ge } 0$ en stel dan die oplossing op $n$ getallelig voor
1.2 Los op vir $x$ en $y$ indien:
$x = 1 - 2y$ en $3x^2 = 3 + x + y$
1.3 Die diagram hieronder toon ʼn eenvoudige pendulum wat van punt A na punt C swaai - NSC Technical Mathematics - Question 1 - 2021 - Paper 1
Question 1
1.1 Los op vir $x$:
1.1.1 $2x(x + 3) = 0$
1.1.2 $x + 9 = 12$ (korrekt tot TWEE desimale plakke)
1.1.3 $x(6 - x) ext{ ge } 0$ en stel dan die oplossing op $n$... show full transcript
Worked Solution & Example Answer:1.1 Los op vir $x$:
1.1.1 $2x(x + 3) = 0$
1.1.2 $x + 9 = 12$ (korrekt tot TWEE desimale plakke)
1.1.3 $x(6 - x) ext{ ge } 0$ en stel dan die oplossing op $n$ getallelig voor
1.2 Los op vir $x$ en $y$ indien:
$x = 1 - 2y$ en $3x^2 = 3 + x + y$
1.3 Die diagram hieronder toon ʼn eenvoudige pendulum wat van punt A na punt C swaai - NSC Technical Mathematics - Question 1 - 2021 - Paper 1
Step 1
1.1.1 $2x(x + 3) = 0$
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Answer
Om die vergelyking op te los, stel ons elk van die faktore gelyk aan nul:
2x=0
Dit gee x=0.
x+3=0
Dit gee x=−3.
So, die oplosings is x=0 of x=−3.
Step 2
1.1.2 $x + 9 = 12$ (korrekt tot TWEE desimale plakke)
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Answer
Oplossing:
x+9=12 x=12−9 x=3
Die antwoord is 3.
Step 3
1.1.3 $x(6 - x) ext{ ge } 0$ en stel dan die oplossing op $n$ getallelig voor
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Answer
Hierdie vergelyking kan herskryf word as:
x(6−x)extge0
Hier kan ons die kritieke waardes vind: x=0 of x=6.
Die oplossing is:
0ext≤xext≤6.
Step 4
1.2 Los op vir $x$ en $y$ indien: $x = 1 - 2y$ en $3x^2 = 3 + x + y$
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Answer
Begin met x=1−2y.
Substitueer in die tweede vergelyking:
egin{align*}
3(1 - 2y)^2 &= 3 + (1 - 2y) + y \
3(1 - 4y + 4y^2) &= 3 + 1 - 2y + y \
3 - 12y + 12y^2 &= 4 - y \
12y^2 - 11y - 1 &= 0\
ext{Gebruik die kwadratiese formule:}\
y = rac{-b ext{ ± } ext{√}(b^2 - 4ac)}{2a} = rac{-(-11) ext{ ± } ext{√}(121 + 48)}{24} = rac{11 ext{ ± } ext{√}(169)}{12} = rac{11 ext{ ± } 13}{12}
ext{Dus: } ext{die antwoord is } y = 2 ext{ of } y = - rac{1}{6}.
ext{Nou, substitueer y terug om x te kry.}
ext{As } y=2: x = 1 - 2(2) = -3. ext{ As } y=- rac{1}{6}: x = 1 - 2(- rac{1}{6}) = rac{4}{3}.
ext{ Oplossingspar: } (-3, 2) ext{ en } ( rac{4}{3}, - rac{1}{6}).
Step 5
1.3.1 Maak $L$ die onderwerp van die formule.
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Answer
Begin met die formule:
T = 2ar{ ext{π}}igg( rac{L}{g}igg)^{ rac{1}{2}}
Kwadreer beide kante:
T^2 = 4ar{ ext{π}}^2 rac{L}{g}.
Los op vir L:
L = rac{g T^2}{4ar{ ext{π}}^2}
Step 6
1.3.2 Bereken vervolgens die numeriese waarde van $L$ as $g = 9,8$ m/s² en die tyd geneem $T = 1,74$ s.
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Answer
Vervang die waardes in die formule:
L = rac{9.8 imes (1.74)^2}{4 ar{ ext{π}}^2}
Bereken die waarde:
Lextapproximatelyequals0.75extm
Step 7
1.4 Gegee die binêre getalle: A: $1101100_2$ en B: $11100_2$
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Answer
Evalueer A – B:
egin{align*}
A: & ext{ } 1101100_2 = 100 \
B: & ext{ } 11100_2 = 28 \
ext{Dus,} \
A - B = 100 - 28 = 72_{10} \ ext{Gegee in binêre vorm: } 1001000_2.
ext{Daarom is die antwoord } 1001000_2.
ext{ }
ext{Dus die finale antwoord is } 1001000_2.
