Evaluate [\(\frac{1}{0.03}\) \(\div\) \(\frac{1}{0.024}\)]-1 correct to 2 decimal places
3.76
1.25
0.94
0.75
Correct answer is B
[\(\frac{1}{0.03}\) + \(\frac{1}{0.024}\)]
= [\(\frac{1}{0.03 \times 0.024}\)]-1
= [\(\frac{0.024}{0.003}\)]-1
= \(\frac{0.03}{0.024}\)
= \(\frac{30}{24}\) = 1.25
If 10112 + x7 = 2510, solve for X.
207
14
20
24
Correct answer is A
10112 + x7 = 2510 = 10112 = 1 x 23 + 0 x 22 + 1 x 21 + 1 x 2o
= 8 + 0 + 2 + 1
= 1110
x7 = 2510 - 1110
= 1410
\(\begin{array}{c|c}
7 & 14 \\ 7 & 2 R 0 \\ & 0 R 2
\end{array}\)
X = 207
Find the roots of x\(^3\) - 2x\(^2\) - 5x + 6 = 0
1, -2, 3
1, 2, -3,
-1, -2, 3
-1, 2, -3
Correct answer is A
Equation: x\(^3\) - 2x\(^2\) - 5x + 6 = 0.
First, bring out a\(_n\) which is the coefficient of x\(^3\) = 1.
Then, a\(_0\) which is the coefficient void of x = 6.
The factors of a\(_n\) = 1; The factors of a\(_0\) = 1, 2, 3 and 6.
The numbers to test for the roots are \(\pm (\frac{a_0}{a_n})\).
= \(\pm (1, 2, 3, 6)\).
Test for +1: 1\(^3\) - 2(1\(^2\)) - 5(1) + 6 = 1 - 2 - 5 + 6 = 0.
Therefore x = 1 is a root of the equation.
Using long division method, \(\frac{x^3 - 2x^2 - 5x + 6}{x - 1}\) = x\(^2\) - x - 6.
x\(^2\) - x - 6 = (x - 3)(x + 2).
x = -2, 3.
\(\therefore\) The roots of the equation = 1, -2 and 3.
\(\begin{pmatrix} 3 & -1 & 0 \\ 4 & -3 & -1 \\ -4 & 1 & 1\end{pmatrix}\)
\(\begin{pmatrix} 3 & -1 & 0 \\ 4 & -1 & 1 \\ -4 & 1 & 1\end{pmatrix}\)
\(\begin{pmatrix} 3 & -1 & 0 \\ 4 & -3 & 1 \\ -4 & 1 & 1\end{pmatrix}\)
\(\begin{pmatrix} 3 & -1 & 0 \\ 4 & 1 & 1 \\ -4 & -1 & 1\end{pmatrix}\)
Correct answer is B
2(1) - (-1) = 3 2(2) - (0) = 4 2(-1) - (2) = -4
2(0) - (1) = -1 2(-1) - (-1) = -1 2(0) - (-1) = 1
2(1) - 2 = 0 2(0) - (-1) = 1 2(1) - (1) = 1
\(\begin{pmatrix} 3 & -1 & 0 \\ 4 & -1 & 1 \\ -4 & 1 & 1\end{pmatrix}\)
-4
45
4,2
2
Correct answer is C
2p x 1 + 8 x 2 = 24
\(\to\) 4p = 24 - 8 = 16,
p = 4
3 x 1 + -5q x 2 = -17
\(\to\) -10q = -17 - 3
-10q = -20
q = 2