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pickupchik [31]
4 years ago
12

The distance s (in feet) covered by a car t sec after starting is given by the following function. s = −t3 + 11t2 + 20t (0 ≤ t ≤

6) Find a general expression for the car's acceleration at any time t (0 ≤ t ≤ 6). s ''(t) = Correct: Your answer is correct. At what time t does the car begin to decelerate? (Round your answer to one decimal place.) t = sec
Mathematics
1 answer:
VladimirAG [237]4 years ago
6 0

8.94

Answer:

Step-by-step explanation:

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3 years ago
What is y+5=-3(x-4)​
schepotkina [342]

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8 0
3 years ago
How to solve and respond.
son4ous [18]

Step-by-step explanation:

the error:

it is stated that

\frac{5x}{x+7} +\frac{7}{x} = \frac{5x}{x+7} +\frac{7(7)}{x+7}

subtract (5x)/(x+7) from both sides

\frac{7}{x}  = \frac{7(7)}{x+7}

multiply both sides by x + 7

7(x+7) = 49x

7x + 49 = 49x

subtract 7x from both sides to isolate x and its coefficient

49 = 42x

thus, this is only true when 49 = 42x. in order for these two equations to be equal, they must <em>always </em>be true, so this is wrong

the solution:

we want to express 7/x as (something) / (x+7). to do this, we can multiply 7/x  by 1.

anything divided by itself = 1. thus, if we multiply both the numerator and the denominator by something that turns x into (x+7), we can do what we want to do.

(x+7)/x * x turns x into (x+7), so we multiply both the numerator and denominator by (x+7)/x to get

\frac{7}{x} = \frac{7(x+7)/x}{x+7}

substitute this for 7/x in our original problem

\frac{5x}{x+7} +\frac{7(x+7)/x}{x+7} =  \frac{5x}{x+7} +\frac{(7x+49)/x}{x+7} = \frac{5x}{x+7} +\frac{7+49/x}{x+7} = \frac{5x+7+49/x}{x+7}

3 0
3 years ago
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