If →PQ = -2i + 5j and →RQ = -i - 7j, find →PR
-3i + 12j
-3i - 12j
-i + 12j
i - 12j
Correct answer is C
→PQ = →PR + →RQ
→PR = →PQ - →RQ
→PR = -2i + 5j - (-i - 7j)
→PR = -2i + 5j + i + 7j
→PR = -i + 12j.
(10 units, 053º)
(9 units, 049º)
(10 units, 037º)
(9 units, 027º)
Correct answer is A
|→PQ| = √[(2-(-4))\(^2\) + (3-(-5))\(^2\)]
|→PQ| = √\(6^2 + 8^2\)
|→PQ| = √100
|→PQ| = 10units
tanθ = \(\frac{3--5}{2--4}\)
tanθ = \(\frac{4}{3}\)
θ = \(tan^{-1}\frac{4}{3}\)
= 53º
Solve: \(3^{2x-2} - 28(3^{x-2}) + 3 = 0\)
x = -2 or x = 1
x = 0 or x = -3
x = 2 or x = 1
x = 0 or x = 3
Correct answer is D
\(3^{2x-2} - 28(3^{x-2}) + 3 = 0\)
\(\frac{3^{2x}}{3^2} - \frac{28.3^x}{3^2} + 3 = 0\)
\(\frac{3^{2x}}{9} - \frac{28.3^x}{9} + 3 = 0\)
let p = 3\(^x\)
\(\frac{p^{2}}{9} - \frac{{28p}}{9} + 3 = 0\)
multiply through by 9
p\(^2\) - 28p + 27 = 0
p\(^2\) - p - 27p + 27 = 0
p (p - 1) - 27(p - 1) = 0
(p-1)(p-27) = 0
p = 1 or 27
when p = 1
p = 3\(^x\)
3\(^x\) = 1
3\(^x\) = 3\(^0\)
x = 0
when p = 27
3x = 27
3x = 33
x = 3
In how many ways can six persons be paired?
5
10
15
20
Correct answer is C
6C\(_2 = \frac{6!}{[6-2]![2!]}\)
\(\frac{6*5*4!}{4!*2!}\)
= \(\frac{6*5}{2}\)
= 15
If Un = kn\(^2\) + pn, U\(_1\) = -1, U\(_5\) = 15, find the values of k and p.
k = -1, p = 2
k = -1, p = -2
k = 1, p = -2
k = 1, p = 2
Correct answer is C
Un = kn\(^2\) + pn,
U\(_1\) = -1,
U\(_5\) = 15,
when n = 1
U1 = k(1)\(^2\)+ p(1) = -1
k + p = -1 --------eqn1
when n = 5
U\(_5\) = k(5)\(^2\)+ p(5) = 15
25k + 5p = 15 --------eqn2
multiply eqn1 by 5 and eqn2 by 1
5k + 5p = -5 -------eqn3
25k + 5p = 15 -------eqn4
eqn4 - eqn3
20k = 20
k = 1
sub for k in eqn1
1 + p = -1
p = -1 -1 = -2