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Verizon [17]
4 years ago
5

Help on this problem?? please?

Mathematics
1 answer:
Reptile [31]4 years ago
5 0

MrBillDoesMath!


Answer to #4:  81/256 * s^8 * t^ 12


Comments:

(7x^3) ^ (1/2)   =  7 ^ (1/2)  *  x^(3/2)   where  ^(1/2) means the square root of a quantity. The answer written (7x^3) is NOT correct.

---------------------

(1)             (27s^7t^11)^ (4/3)  

               = 27^(4/3) * (s^7)^(4/3) * (t^11)^ (4/3)


As 27 = 3^3, 27 ^(4/3) = 3^4 = 81

               

(2)  (-64st^2)^ (4/3)  =     (-64)^(4/3) * (s^4/3) * t(^8/3)


As 64 = (-4)^3,  (-64)^(4/3) = (-4)^4 = +256                                        


So (1)/(2) =

81 * s^(28/3)* t^(44/3)

-------------------------------     =

256 s^(4/3) * t^((8/3)


81/256 *  s ^ (28/3 - 4/3) * t^(44/3 - 8/3) =


81/256 * s^(24/3) * t (36/3) =

81/256 * s^8        * t^ 12



MrB

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cestrela7 [59]
(i):
x = Vtcos\theta
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V^{2} = \frac{x^{2}}{t^{2}} + \frac{y^{2} + gt^{2}y + \frac{1}{4}g^{2}t^{4}}{t^{2}}
t^{2}V^{2} = x^{2} + y^{2} + gt^{2}y + \frac{1}{4}g^{2}t^{4}
4t^{2}V^{2} = 4x^{2} + 4y^{2} + 4gt^{2}y + g^{2}t^{4}
4x^{2} + 4y^{2} + 4gt^{2}y + g^{2}t^{4} - 4t^{2}V^{2} = 0
4y^{2} + 4gt^{2}y + (gt^{2}t^{4} + 4x^{2} - 4t^{2}V^{2}) = 0

(ii): Impact is when x = d.
\text{Impact: } d = Vtcos\theta
t = \frac{d}{Vcos\theta}

First impact occurs when t is minimised.
This means that Vcos theta is maximised, which means cos theta = 1, and theta = 0
\therefore \text{First impact occurs at } \theta = 0\text{: }t = \frac{d}{V(1)} = \frac{d}{V}

i'll do the rest later.
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