solve the inequality \((Y-3)<\frac{y}{3}\)
y > -9
y < 3
y > 4
y < 9
y > 0
Correct answer is C
\((Y-3)<\frac{y}{3}\\
3y - 9 < y\\
2y< 9\\
y < 4\frac{1}{2}\)
Find (X-Y) if 4x - 3y = 7 and 3x - 2y = 5
4
3
2
-2
-3
Correct answer is C
No explanation has been provided for this answer.
Find the equation whose roots are \(\frac{2}{3}and \frac{-1}{4}\)
\(12x^2-5x+2=0\)
\(12x^2-11x+2=0\)
\(x^2-\frac{11}{12}x+2=0\)
\(x^2+\frac{11}{12}x-2=0\)
\(12x^2+11x+2=0\)
Correct answer is A
x\(^2\) - (sum of given roots)x + (product of given roots ) = 0
x\(^2\) - (\(\frac{2}{3} + \frac{-1}{4}\))x + (\(\frac{2}{3} * \frac{-1}{4}\)) = 0
x\(^2\) - (\(\frac{8 - 3}{12}\))x + (\(\frac{- 2}{12}\))
x\(^2\) - \(\frac{5}{12}\)x + \(\frac{- 2}{12}\)
multiply through by the LCM 12
12 * x\(^2\) - 12 * \(\frac{5}{12}\)x + 12 * \(\frac{- 2}{12}\)
12x\(^2\) - 5x - 2 = 0
Make S the subject of the formula: \(V = \frac{K}{\sqrt{T-S}}\)
\(T-\frac{K^2}{V^2} = S\)
\(T+\frac{K^2}{V^2} = S\)
\(T-\frac{K^2}{V} = S\)
\(T-\frac{K}{V} = S\)
\(T-\frac{K}{V^2} = S\)
Correct answer is A
\(V = \frac{K}{\sqrt{T-S}}\)
square both sides of the equation
\(V^2 = \frac{K^2}{T-S}\)
cross multiply
\(V^2 T - S = K^2\)
\(T - S = \frac{K^2}{V^2}\)
\(T = \frac{K^2}{V^2} + S\)
\(T - = \frac{K^2}{V^2} = S\)
For what value of x is the expression
3, 1
-1, -3
-1, 3
3, -2
1, -3
Correct answer is C
x2 - 2x - 3 = 0
x2 - 3x + x - 3 = 0
x(x - 3) + (x - 3) = 0
(x + 1)(x - 3) = 0
x = -1 or 3