5n - km = 0
kn + 5m = 0
5n + km = 0
kn - 5m = 0
Correct answer is B
Two lines are parallel if and only if their slopes are equal.
\(kx - 5y + 6 = 0 \implies 5y = kx + 6\)
\(y = \frac{k}{5}x + \frac{6}{5}\)
\(Slope = \frac{k}{5}\)
\(mx + ny - 1 = 0 \implies ny = 1 - mx\)
\(y = \frac{1}{n} - \frac{m}{n}x\)
\(Slope = -\frac{m}{n}\)
\(Parallel \implies \frac{k}{5} = -\frac{m}{n}\)
\(-5m = kn \implies 5m + kn = 0\)
\(\frac{50 \cos \theta}{\sin (\theta + 45)°}\)
\(\frac{50 \cos \theta}{\cos (\theta + 45)°}\)
\(\frac{50 \sin \theta}{\cos (\theta + 45)°}\)
\(\frac{50 \sin \theta}{\sin (\theta + 45)°}\)
Correct answer is D
No explanation has been provided for this answer.
The distance between P(x, 7) and Q(6, 19) is 13 units. Find the values of x.
1 or -7
1 or 7
1 or 11
5 or -5
Correct answer is C
\(d = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}\)
\(13 = \sqrt{(x - 6)^{2} + (7 - 19)^{2}}\)
\(13^{2} = x^{2} - 12x + 36 + 144\)
\(169 = x^{2} - 12x + 180\)
\(x^{2} - 12x + 180 - 169 = 0 \implies x^{2} - 12x + 11 = 0\)
\((x - 1)(x - 11) = 0 \implies x = \text{1 or 11}\)
-4
-3
0
6
Correct answer is D
\(y = x^{2} - 6x + 11\)
\( y = a(x - h)^{2} + k\)
\(a(x - h)^{2} + k = a(x^{2} - 2hx + h^{2}) + k\)
\(ax^{2} - 2ahx + ah^{2} + k = x^{2} - 6x + 11\)
Comparing, we have
\(a = 1\)
\(2ah = 6 \implies 2h = 6; h = 3\)
\(ah^{2} + k = 11 \implies (1 \times 3^{2}) + k = 11\)
\(9 + k = 11 \implies k = 2\)
\(\therefore a + h + k = 1 + 3 + 2 = 6\)
25
15
\(3\sqrt{7}\)
\(\sqrt{10}\)
Correct answer is A
Change in momentum = m (v - u)
= \(5 \times (\begin{pmatrix} 4 \\ 7 \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \end{pmatrix})\)
= \(5 \times \begin{pmatrix} 3 \\ 4 \end{pmatrix}\)
= \(\begin{pmatrix} 15 \\ 20 \end{pmatrix}\)
\(|m(v - u)| = \sqrt{15^{2} + 20^{2}} = \sqrt{625} = 25\)
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