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South Africa
GRADE 11 3. Given that: 5tan⁡θ=6cos⁡θ 3.3.1. show that the equation can be written as 6sin2⁡θ+5sin⁡θ−6=0
Trigonometry Mar 25, 2025
Find the general solution for the following equation. 1,2=sin⁡2xsin⁡x+0,2
Trigonometry Mar 25, 2025
Determine the general solution of: cos⁡2x−19cos⁡x+10=0 Answer: x=◻∘+k⋅◻∘+k⋅◻∘, or x=◻∘, where k is any Select
Trigonometry Mar 25, 2025
Simplify the following expression completely, given that 0∘<A<90∘ 1−cos2⁡(2A)cos⁡(−A)cos⁡(90∘−A)
Trigonometry Mar 25, 2025
Simplify the following expressions: a) 3x(x+5)−4x(3x−7)
Algebra Mar 25, 2025
4 Show that the maximum value of −x2+4x+12 is 4.
Pre Calculus Mar 25, 2025
.3 The roots of the equation f(x)=0 are x=4±16−4m(−m+5)2m Determine the values of m for which the roots will be non-real.
Algebra Mar 25, 2025
Evaluate the following limits. If the limit does not exist, explain why.  (a) limt→−2(2−|t|2+t) (b) limx→2−6−3x|x−2| (c) limx→31|3−x| (c) limx→0(x2cos⁡(2x+4)) (d) limx→0x2(1+sin⁡πx) (e) limx→2+x−2cos⁡(1x−2) (f) limx→∞sin⁡(2tan−1⁡(x)) (g) limx→∞arctan⁡ex (h) limx→∞arctan⁡e−x
Calculus Mar 25, 2025
Calculate the volume of water that would be collected from the roof each month, if the average rainfall actually fell that month (in kilolitres). Fill these values into your copy of the table. (For example, for June the rainfall is 84 mm , which is 0,084 m. The area of the roof is 65 m2, so the volume is 65×0,084= 5,46 m3, which is the same as 5,46 kilolitres.)
Arithmetic Mar 25, 2025
Evaluate the following limits. If the limit does not exist, explain why.  (a) limt→−2(2−|t|2+t) (b) limx→2−6−3x|x−2| (c) limx→31|3−x| (c) limx→0(x2cos⁡(2x+4)) (d) limx→0x2(1+sin⁡πx) (e) limx→2+x−2cos⁡(1x−2) (f) limx→∞sin⁡(2tan−1⁡(x)) (g) limx→∞arctan⁡ex (h) limx→∞arctan⁡e−x
Calculus Mar 25, 2025
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