If y = 243(4x + 5)-2, find \(\frac{dy}{dx}\) when x = 1
\(\frac{-8}{3}\)
\(\frac{3}{8}\)
\(\frac{9}{8}\)
-\(\frac{8}{9}\)
Correct answer is A
y = 243(4x + 5)-2, find \(\frac{dy}{dx}\)
= -1944(4x + 5)-3
= 1944(9)-3
\(\frac{dy}{dx}\) when x = 1
= -\(\frac{1944}{9^3}\)
= -\(\frac{1944}{729}\)
= \(\frac{-8}{3}\)
150(1 + \(\sqrt{3}\))m
50( \(\sqrt{3}\) - \(\sqrt{3}\))m
150 \(\sqrt{3}\)m
\(\frac{50}{\sqrt{3}}\)m
Correct answer is B
\(\frac{150}{Z}\) = tan 60o,
Z = \(\frac{150}{tan 60^o}\)
= \(\frac{150}{3}\)
= 50\(\sqrt{3}\)cm
\(\frac{150}{X x Z}\) = tan45o = 1
X + Z = 150
X = 150 - Z
= 150 - 50\(\sqrt{3}\)
= 50( \(\sqrt{3}\) - \(\sqrt{3}\))m
\(\frac{\pi}{2}\), \(\frac{3\pi}{2}\)
\(\frac{\pi}{3}\), \(\frac{2\pi}{3}\)
0, \(\frac{\pi}{3}\)
0, \(\pi\)
Correct answer is D
cos x + sin x \(\frac{1}{cos x - sinx}\)
= (cosx + sinx)(cosx - sinx) = 1
= cos2x + sin2x = 1
cos2x - (1 - cos2x) = 1
= 2cos2x = 2
cos2x = 1
= cosx = \(\pm\)1 = x
= cos-1x (\(\pm\), 1)
= 0, \(\pi\) \(\frac{3}{2}\pi\), 2\(\pi\)
(possible solution)
(\(\frac{3}{2}\), \(\frac{3}{2}\))
(\(\frac{2}{3}\), \(\frac{3}{2}\))
(\(\frac{3}{8}\), \(\frac{3}{2}\))
(-\(\frac{3}{8}\), \(\frac{3}{2}\))
Correct answer is D
y = 4x + 3
when x = 0, y = 3 \(\to\) (0, 3)
when y = 0, x = -\(\frac{3}{4}\) \(\to\) (\(\frac{3}{4}\), 0)
mid-point \(\frac{0 + (-{\frac{3}{4}})}{2}\), \(\frac{3 + 0}{4}\)
-\(\frac{3}{8}\), \(\frac{3}{2}\)
If the distance between the points (x, 3) and (-x, 2) is 5. Find x
6.0
2.5
\(\sqrt{6}\)
\(\sqrt{3}\)
Correct answer is C
d2 = (y - y)2 + (x - x)2
5 = 4x2 + 1 = 25= 4x2 + 1
= 4x2 = 25 - 1= 24
x2 = \(\frac{24}{4}\)
x = \(\sqrt{6}\)