To find a fraction equivalent to 3/5 multiply both top and bottom numbers by the same number .Example of answer 6/10.
Answer:
A
Step-by-step explanation:
how long is the ball in the air ?
that is the same as asking : after how many seconds will the ball hit the ground (= reach the height of 0) ?
so, that means we need to find the zero solution of h(t).
at what t is h(t) = 0 ?
when at least one of the factors is 0 :
2(-2 - 4t)(2t - 5)
we have 3 factors
2 : can never be 0.
(-2 -4t) : can only be 0 for negative t, which does not make sense in our scenario (we cannot go back in time, only forward).
(2t - 5) : is 0 when 2t = 5 or t = 2.5
so, A is the right answer.
FYI : the starting height (on the hill) is given by t = 0 :
2(-2 - 0)(0 - 5) = 2×-2×-5 = 20 ft
Answer:it would be 15cm^2 I think
Step-by-step explanation:
The function is (-x+3)/ (3x-2) and we get f(1)=1 and differentiation is f'(x)=-7/ (9x²- 12x+4).
Given that,
The function is (-x+3)/ (3x-2)
We have to find f(1) and f'(x).
Take the function expression
f(x)= (-x+3)/ (3x-2)
Taking x as 1 value
f(1)= (-1+3)/(3(1)-2)
f(1)=2/1
f(1)=1
Now, to get f'(x)
With regard to x, we must differentiate.
f(x) is in u/v
We know
u/v=(vu'-uv')/ v² (formula)
f'(x)= ((3x-2)(-1)- (-x+3)(3))/ (3x-2)²
f'(x)= ((-3x+2)-(-3x+9))/ 9x²- 12x+4
f'(x)=(-3x+2+3x-9)/ 9x²- 12x+4
f'(x)=2-9/ (9x²- 12x+4)
f'(x)=-7/ (9x²- 12x+4)
Therefore, The function is (-x+3)/ (3x-2) and we get f(1)=1 and differentiation is f'(x)=-7/ (9x²- 12x+4).
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