If y = 4x3 - 2x2 + x, find \(\frac{\delta y}{\delta x}\)
8x2 - 2x + 1
8x2 - 4x + 1
12x2 - 2x + 1
12x2 - 4x + 1
Correct answer is D
If y = 4x3 - 2x2 + x, then;
\(\frac{\delta y}{\delta x}\) = 3(4x2) - 2(2x) + 1
= 12x2 - 4x + 1
If \(\sin\theta = \frac{12}{13}\), find the value of \(1 + \cos\theta\)
\(\frac{25}{13}\)
\(\frac{18}{13}\)
\(\frac{8}{13}\)
\(\frac{5}{13}\)
Correct answer is B
No explanation has been provided for this answer.
\(\begin{pmatrix} 3 & -3 \\ 8 & 2 \end{pmatrix}\)
\(\begin{pmatrix} 3 & 3 \\ 8 & 2 \end{pmatrix}\)
\(\begin{pmatrix} -2 & 2 \\ 2 & 2 \end{pmatrix}\)
\(\begin{pmatrix} -2 & 3 \\ 3 & 2 \end{pmatrix}\)
Correct answer is A
y - 4x + 3 = 0
When y = 0, 0 - 4x + 3 = 0
Then -4x = -3
x = 3/4
So the line cuts the x-axis at point (3/4, 0).
When x = 0, y - 4(0) + 3 = 0
Then y + 3 = 0
y = -3
So the line cuts the y-axis at the point (0, -3)
Hence the midpoint of the line y - 4x + 3 = 0, which lies between the x-axis and the y-axis is;
\([\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]\)
\([\frac{1}{2}(\frac{3}{4} + 0), \frac{1}{2}(0 + -3)]\)
\([\frac{1}{2}(\frac{3}{4}), \frac{1}{2}(-3)]\)
\([\frac{3}{8}, \frac{-3}{2}]\)
The gradient of a line joining (x,4) and (1,2) is \(\frac{1}{2}\). Find the value of x
5
3
-3
-5
Correct answer is A
\(\text{Gradient m} = \frac{y_2 - y_1}{x_2 - x_1}\)
\(\frac{1}{2} = \frac{2 - 4}{1 - x}\)
1 - x = 2(2 - 4)
1 - x = 4 - 8
1 - x = -4
-x = -4 - 1
x = 5
Find the mid point of S(-5, 4) and T(-3, -2)
-4, 2
4, -2
-4, 1
4, -1
Correct answer is C
Mid point of S(-5, 4) and T(-3, -2) is
\([\frac{1}{2}(-5 + -3), \frac{1}{2}(4 + 2)]\)
\([\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]\)
\([\frac{1}{2}(-8), \frac{1}{2}(2)]\)
\([-4, 1]\)