150o
67\(\frac{1}{2}\)o
67o
135o
68o
Correct answer is B
Representing from Asia is 150
\(\frac{150}{800}\) x 360
= \(\frac{135}{2}\)
= 67\(\frac{1}{2}\)°
If \(\frac{3e + f}{3f - e}\) = \(\frac{2}{5}\), find the value of \(\frac{e + 3f}{f - 3e}\)
\(\frac{5}{2}\)
1
\(\frac{26}{7}\)
\(\frac{1}{3}\)
Correct answer is C
\(\frac{3e + f}{3f - e}\) = \(\frac{2}{5}\)
= 3e + f
= 2 x 1
\(\frac{-e + 3f}{3e - f}\) = \(\frac{5 \times 3}{2}\)
= \(\frac{3e + 9f = 15}{10f = 17}\)
f = \(\frac{17}{10}\)
Sub. for equ. (1)
3e + \(\frac{17}{10}\) = 2
3e = 2 - \(\frac{17}{10}\)
\(\frac{3}{10}\)
e = \(\frac{3}{10}\) x \(\frac{1}{3}\)
= \(\frac{1}{10}\)
= e + 3f = \(\frac{1}{10}\) + \(\frac{3 \times}{10}\) = \(\frac{52}{10}\)
f - 3e = \(\frac{17}{10}\) - 3 x \(\frac{1}{10}\)
= \(\frac{14}{10}\)
= \(\frac{52}{10}\) x \(\frac{10}{14}\)
= \(\frac{26}{7}\)
y = Q + \(\frac{P}{x^2}\)
y = Q + px
y = \(\frac{PQ}{x^2}\)
y = Q - \(\frac{P}{x^2}\)
Correct answer is A
Given the above statement,
\(y = Q + \frac{P}{x^{2}}\)
Express 130 kilometers per second in meters per hour
7.8 x 10-5
4.68 x 106
7,800,000
4.68 x 108
7.80 x 105
Correct answer is D
1km = 1000m
60sec. = 1 mins
60 mins. = 1 hr
130000m per sec = 130000 x 3600
= 468000000m/hr
= 468 x 108 m/hr
It does not rise
2\(\sqrt{3}\) meters
3 meters
1 meter
3
Correct answer is D
Cos 60o = \(\frac{PD}{2}\)
Cos 60o x 2 = 0.5 x 2
= 1m