the position of an object moving in a straight line is given by s=tsquared-8t, where s is in feet and t is the time in seconds the object has been in motion. how long(to the nearest tenth)will it take the object to move 10 feet?

Umm I saw you critique someone else's paper for cheating...nice! Hypocritical are you not?

9.09

To find the time it takes for the object to move 10 feet, we need to solve the equation s = 10.

Given that the position of the object is given by s = t^2 - 8t, we can substitute s with 10:

10 = t^2 - 8t

Now, let's rearrange the equation to get it in the form of a quadratic equation:

t^2 - 8t - 10 = 0

We can solve this quadratic equation using the quadratic formula, which is given by:

t = (-b ± sqrt(b^2 - 4ac)) / 2a

In this case, the coefficients of the quadratic equation are a = 1, b = -8, and c = -10. Substituting these values into the quadratic formula, we get:

t = (-(-8) ± sqrt((-8)^2 - 4(1)(-10))) / (2(1))
t = (8 ± sqrt(64 + 40)) / 2
t = (8 ± sqrt(104)) / 2
t = (8 ± √104) / 2
t = (8 ± 2√26) / 2

Now, we can simplify this expression by canceling out the common factor of 2:

t = 4 ± √26

So, the two possible values for t are t = 4 + √26 and t = 4 - √26.

Therefore, the object will take approximately 4 + √26 seconds and 4 - √26 seconds to move 10 feet, depending on the direction of motion.