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.
The square root of a number is 2k. What is half of the number
\(\sqrt{\frac{k}{2}}\)
\(\sqrt{k}\)
\(\frac{1}{2}k^2\)
2k2
Correct answer is D
Let the number be x.
\(\sqrt{x} = 2k \implies x = (2k)^2\)
= \(4k^2\)
\(\frac{1}{2} \times 4k^2 = 2k^2\)
From the Venn Diagram below, find Q' ∩ R.
(e)
(c, h)
(c, g, h)
(c, e, g, h)
Correct answer is C
Q' ∩ R Q' = U - Q Q' = {a, b, c, d, g, h, i} R = {c, e, h, g} Q' ∩ R = {c, h, g}
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