Evaluate Cos 45o Cos 30o - Sin 45o Sin 30o leaving the answer in surd form
\(\frac{\sqrt{2}-1}{2}\)
\(\frac{\sqrt{3}-\sqrt{2}}{4}\)
\(\frac{\sqrt{6}-\sqrt{2}}{2}\)
\(\frac{\sqrt{6}-\sqrt{2}}{4}\)
Correct answer is D
\(cos45^o \times cos30^o - sin45^o \times sin30^o\\
\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\times \frac{1}{2}\\
\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}}; = \frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}\)
Find the equation whose roots are \(-\frac{2}{3}\) and 3
3x2+11x-6=0
3x2+7x+6=0
3x2-11x-6=0
3x2-7x-6=0
Correct answer is D
\(x = -\frac{2}{3} \implies x + \frac{2}{3} = 0\)
\(x = 3 \implies x - 3 = 0\)
\(\implies (x - 3)(x + \frac{2}{3}) = 0\)
\(x^2 - 3x + \frac{2}{3}x - 2 = 0\)
\(x^2 - \frac{7}{3}x - 2 = 0\)
\(3x^2 - 7x - 6 = 0\)
The locus of points equidistant from two intersecting straight lines PQ and PR is
a circle centre P radius Q.
a circle centre P radius PR
the point of intersection of the perpendicular bisectors of PQ and PR
the bisector of angle QPR
Correct answer is C
No explanation has been provided for this answer.
1-(x-y)+xy
1-(x+y)-xy
1-(x+y)+xy
1 - (x - y) + x
Correct answer is C
Prob (passing English) = x
Prob (passing Maths) = Y
Prob (failing English) = 1 - x
Prob (failing Maths) = 1 - y
Prob (failing both test) = Prob(failing English) and Prob(failing Maths) = (1 - x)(1 - y)
=1 - y - x + xy
=1 - (y + x) + xy
p varies as the square of r
p varies as the square root of r
p varies inversely as the square of r
p varies inversely as r
Correct answer is D
\(p \propto q^2\)
\(q\propto\frac{1}{\sqrt{r}}\)
\(p = kq^2\)
\(q = \frac{c}{\sqrt{r}}\)
where c and k are constants.
\(q^2 = \frac{c^2}{r}\)
\(p = \frac{kc^2}{r}\)
If k and c are constants, then kc\(^2\) is also a constant, say z.
\(p = \frac{z}{r}\)
p varies inversely as r.