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VashaNatasha [74]

3 years ago

7

3. WHAT IS THE END BEHAVIOR?SHOW WORK. - 10x -x+ 2

1 answer:

nekit [7.7K]3 years ago

3 0

Answer:

Step-by-step explanatio

−10x−x+2

=−10x+−x+2

=−10x+−x+2

=(−10x+−x)+(2)

=−11x+2

Answer:

=−11x+2

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Answer:

Vertex form: $f(x) =(x +\frac{7}{2})^2 -\frac{53}{4}$

The vertex is $(-\frac{7}{2},-\frac{53}{4})$

Step-by-step explanation:

For a general quadratic function the form is:

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The values of the coefficients for the function are the following: $a = 1$ , $b =7$ , $c = -1$

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$\frac{b}{2}= \frac{7}{2}$

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Add and subtract $\frac{49}{4}$

$f(x) = (x ^ 2 +7x +\frac{49}{4}) -\frac{49}{4}- 1$

Write the expression of the form

$f(x) = (x+\frac{b}{2})^2 +k$

$f(x) =(x +\frac{7}{2})^2 -\frac{53}{4}$

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Here we know that the initial velocity of the car is given by:

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And the final velocity of the car is given by:

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It took 3 seconds to attain the final velocity, so we have $t=3 s$

Therefore, the acceleration can be obtained by:

$V_f-V_i=a\times t$

$a=\frac{V_f-V_i}{t}$

Plugging the values of the initial, final velocity and the time, we get:

$a= \frac{(6\hat x-3\hat y)-(3 \hat x+1 \hat y)}{3} =\frac{6\hat x-3 \hat y -3 \hat x-1 \hat y}{3} =\frac{3 \hat x-4 \hat y}{3}$

So the acceleration of the car is given by:

$a=\frac{3 \hat x}{3} -\frac{4 \hat y}{3} =1 \hat x- \frac{4}{3} \hat y$

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Here, x and y are the coefficients of the 'x' and 'y' components of the vector:

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Therefore, the direction of the average acceleration of the car is $-53.13^\circ$ .

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