Can someone please help me on this one!!!!
1 answer:
This appears to be about rules of exponents as much as anything. The applicable "definitions, identities, and properties" are
i^0 = 1 . . . . . as is true for any non-zero value to the zero power
i^1 = i . . . . . . as is true for any value to the first power
i^2 = -1 . . . . . from the definition of i
i^3 = -i . . . . . = (i^2)·(i^1) = -1·i = -i
i^n = i^(n mod 4) . . . . . where "n mod 4" is the remainder after division by 4
1. = -3^4·i^(3·2+0+2·4) = -81·i^14 = 81
2. = i^((3-5)·2+0 = i^-4 = 1
3. = -2^2·i^(4+2+2+(-1+1+5)·3+0) = -4·i^23 = 4i
4. = i^(3+(2+3+4+0+2+5)·2) = i^35 = -i
i^0 = 1 . . . . . as is true for any non-zero value to the zero power
i^1 = i . . . . . . as is true for any value to the first power
i^2 = -1 . . . . . from the definition of i
i^3 = -i . . . . . = (i^2)·(i^1) = -1·i = -i
i^n = i^(n mod 4) . . . . . where "n mod 4" is the remainder after division by 4
1. = -3^4·i^(3·2+0+2·4) = -81·i^14 = 81
2. = i^((3-5)·2+0 = i^-4 = 1
3. = -2^2·i^(4+2+2+(-1+1+5)·3+0) = -4·i^23 = 4i
4. = i^(3+(2+3+4+0+2+5)·2) = i^35 = -i
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Answer:
- 168.25 ft
- 3.125 s
Step-by-step explanation:
It is almost too easy to let a graphing calculator show you the solution. (See attached)
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You may be expected to use some actual algebra (or even calculus) to find the solution.
If you've spent any time with quadratics, you know the vertex of ax²+bx+c is located at x=-b/(2a). Here, that means the time to get to the highest point is ...
... t = -100/(2·(-16)) = 100/32 = 3.125 . . . . seconds
Then the highest point is ...
... h(3.125) = -16·3.125² +100·3.125 +12 = 168.25 . . . . feet
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Or, you can put the equation into vertex form.
... h(t) = -16t² +100t +12
... = -16(t² +6.25t) +12
... = -16(t² +6.25t +3.125²) +12 +16·3.125² . . . . . add the square of half the t coefficient inside parentheses; add the opposite of the same amount outside parentheses
... = -16(t +3.125)² +168.25 . . . . . simplify
This tells us the peak of the travel of the bullet is at 168.25 ft after 3.125 seconds.
