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pishuonlain [190]

3 years ago

11

kaitlin is on a long car trip every time she stops she loses 15 minutes of travel time is she stops 5 times how late will she be

to her destination

2 answers:

yanalaym [24]3 years ago

8 0

5(15)= 75 minutes or an hour and 15 minutes lost

vodomira [7]3 years ago

7 0

5 times 15 equals 75 so she will be 75 minutes late

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Consider the provided cubic function.

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Substitute x=-2 in $f(x) = x^3 + 2x^2 + 2x - 4$ .

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Therefore, the option is incorrect.

Substitute x=-2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (-2)^3+2(-2)^2 - 2(-2) - 4\\f(x) =-8+8+4-4\\f(x) =0$

Now check for other roots as well.

Substitute x=√2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (\sqrt{2})^3+2(\sqrt{2})^2 - 2(\sqrt{2}) - 4\\f(x) =2\sqrt{2}+4-2\sqrt{2}-4\\f(x) =0$

Substitute x=-√2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (-\sqrt{2})^3+2(-\sqrt{2})^2 - 2(-\sqrt{2}) - 4\\f(x) =-2\sqrt{2}+4+2\sqrt{2}-4\\f(x) =0$

Therefore, the option is correct.

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