Find the value of x such that the expression \(\frac{1}{x}+\frac{4}{3x}-\frac{5}{6x}+1\) equals zero
\(\frac{1}{6}\)
\(\frac{1}{4}\)
\(\frac{-3}{2}\)
\(\frac{-7}{6}\)
Correct answer is C
\(\frac{1}{x} + \frac{4}{3x} - \frac{5}{6x} + 1 = 0\)
\(\frac{6 + 8 - 5 + 6x}{6x} = 0\)
\(\frac{9 + 6x}{6x} = 0 \implies 9 + 6x = 0\)
\(6x = -9 \implies x = \frac{-3}{2}\)
Solve the equation \(3y^2 = 27y\)
y = o or 3
y = 0 or 9
y = -3 or 3
y = 3 or 9
Correct answer is B
\(3y^2 = 27y\\
3y^2 - 27y = 0\\
3y(y - 9) = 0\\
3y = 0 or y - 9 = 0\\
y = 0 or y = 9\\
y = 0 or 9\)
Make t the subject of formula \(k = m\sqrt{\frac{t-p}{r}}\)
\(\frac{rk^2 + p}{m^2}\)
\(\frac{rk^2+pm^2}{m^2}\)
\(\frac{rk^2-p}{m^2}\)
\(\frac{rk^2-p^2}{m^2}\)
Correct answer is B
\(k = m\sqrt{\frac{t - p}{r}}\)
\(\frac{k}{m} = \sqrt{\frac{t - p}{r}}\)
\((\frac{k}{m})^2 = \frac{t - p}{r}\)
\(rk^2 = m^2 (t - p)\)
\(\therefore m^2 t = rk^2 + m^2 p\)
\(t = \frac{rk^2 + m^2 p}{m^2}\)
Find the equation whose roots are -8 and 5
\(x^2 + 13x + 40=0\)
\(x^2 - 13x - 40=0\)
\(x^2 - 3x +40=0\)
\(x^2 + 3x - 40=0\)
Correct answer is D
Equation with roots -8 and 5: (x + 8)(x - 5) = 0
\(x^2 - 5x + 8x - 40 = 0\)
\(x^2 + 3x - 40 = 0\)
Form an inequality for a distance d meters which is more than 18m, but not more than 23m
18 ≤ d ≤ 23
18 < d ≤ 23
18 ≤ d < 23
d < 18 or d > 23
Correct answer is B
No explanation has been provided for this answer.