Calculate the distance travelled in the third second.
5.5m
14.5m
26.0m
30.0m
Correct answer is A
From S = 12t + \(\frac{5}{2t^2}\) - t\(^3\);
Distance traveled in 2 seconds
S = 12[2] + \(\frac{5}{2[2]^2}\) - 2\(^3\).
= 24 + 10 - 8 = 26m
Distance traveled in 3 seconds
S = 12[3] + \(\frac{5}{2[3]^2}\) - 3\(^3\)
= 36 + 22.5 - 27 = 31.5m
Distance traveled in the 3rd second
= 31.5m - 26m
: S = 5.5m
Calculate the maximum height reached.
418.5m
56.0m
31.5m
30.0m
Correct answer is C
S = 12t + \(\frac{5}{2t^2}\) - t\(^3\);
ds/dt = 12 + 5t - 3t\(^2\)
At max height ds/dt = 0
i.e 12 + 5t - 3t\(^2\)
(3t + 4)(t -3) = 0;
t = -4/3 or 3
Hmax = 12[3] + \(\frac{5}{2[3]^2}\) - 3\(^3\)
= 36 + 45/2 - 27
= 31.5m
In △PQR, \(\overline{PQ}\) = 5i - 2j and \(\overline{QR}\) = 4i + 3j. Find \(\overline{RP}\)
-i - 5j
-9 - j
i + 5j
-9i + j
Correct answer is A
\(\overline{PQ}\) = 5i - 2j; \(\overline{QR}\) = 4i + 3j
\(\overline{RP}\) = \(\overline{PQ}\) - \(\overline{QR}\)
= 5i - 2j - [4i + 3j] = 5i - 2j - 4i - 3j
= i - 5j
5√3
6√2
6√3
9√3
Correct answer is C
No explanation has been provided for this answer.
If 2y\(^2\) + 7 = 3y - xy, find \(\frac{dy}{dx}\)
\(\frac{-y}{4x+y-3}\)
\(\frac{y}{4x+y-3}\)
\(\frac{-y}{4x+x-3}\)
\(\frac{y}{4x+x-3}\)
Correct answer is D
2y\(^2\) + 7 = 3y - xy, find \(\frac{dy}{dx}\)
4y \(\frac{dy}{dx}\) + 0 = 3 \(\frac{dy}{dx}\) - x \(\frac{dy}{dx}\) + y
{4x+x-3} \(\frac{dy}{dx}\) = y
= \(\frac{y}{4x+x-3}\)
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