The gradient of the straight line joining the points P(5, -7) and Q(-2, -3) is
\(\frac{1}{2}\)
\(\frac{2}{5}\)
\(-\frac{4}{7}\)
\(-\frac{2}{3}\)
Correct answer is C
PQ = \(\frac{y_1 - y_0}{x_1 - x_0}\) = \(\frac{-3 - (-7)}{-2 - 5}\) = \(\frac{-3 + 7}{-2 - 5}\) = \(\frac{4}{-7}\)
Calculate the volume of a cuboid of length 0.76cm, breadth 2.6cm and height 0.82cm.
3.92cm3
2.13cm3
1.97cm3
1.62cm3
Correct answer is D
Volume of cuboid = L x b x h
= 0.76cm x 2.6cm x 0.82cm
= 1.62cm3
The angles of a polygon are given by x, 2x, 3x, 4x and 5x respectively. Find the value of x.
24o
30o
33o
36o
Correct answer is D
Since there are 5 angles given, the polygon is a pentagon.
Sum of interior angles of a pentagon = (2(5) - 4) x 90° = 540°
\(\therefore\) x + 2x + 3x + 4x + 5x = 15x
15x = 540°
\(x = \frac{540}{15} = 36°\)
Given that I3 is a unit matrix of order 3, find |I3|
-1
0
1
2
Correct answer is C
Recall that a unit matrice is a diagonal matrix in which the elements in the leading diagonal is unity. Therefore,
I3 = \(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\)
I3 = \(+1\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} - 0\begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix} + 0 \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix} \)
I3 = +1(1 - 0) - 0(0 - 0) + 0(0 - 0)
= 1(1)
= 1
3
4
5
7
Correct answer is C
\(\begin{vmatrix} 5 & 3 \\ x & 2 \end{vmatrix}\) = \(\begin{vmatrix} 3 & 5 \\ 4 & 5 \end{vmatrix}\)
10 - 3x = 15 - 20
-3x = 15 - 20 - 10
-3x = -15
x = 5