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zhuklara [117]
3 years ago
9

Is my answer correct? 10 points + brainleist!

Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
8 0

Answer:

your answer is incorrect.  The correct answer is  h=-13  and k=13 .

Step-by-step explanation:

If a quadratic function is f(x)=ax^2+bx+c and a>0, then minimum value of the function at point \left(-\dfrac{b}{2a},f(-\dfrac{b}{2a})\right).

The given function is

f(x)=x^2+bx+182

Here, a=1, b=b and c=182. So.

-\dfrac{b}{2a}=-\dfrac{b}{2(1)}=-\dfrac{b}{2}

Put x=-\dfrac{b}{2} in the given function to find the minimum value of the function.

f(-\dfrac{b}{2})=(-\dfrac{b}{2})^2+b(-\dfrac{b}{2})+182

We know that minimum value is 13. So,

13=\dfrac{b^2}{4}-\dfrac{b^2}{2}+182

13-182=-\dfrac{b^2}{4}

-169=-\dfrac{b^2}{4}

169\times 4=b^2

Taking square root on both sides.

13\times 2=b

b=26

The value of b is 26.

So, the given function is  

f(x)=x^2+26x+182

Now, add and subtract square of half of coefficient of x.

f(x)=x^2+26x+182+(13)^2-(13)^2

f(x)=(x^2+2(13)x+(13)^2)+182-169

f(x)=(x+13)^2+13

On comparing with f(x)=(x-h)^2+k, we get

h=-13

k=13

Therefore, your answer is incorrect.

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3 years ago
Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'
vodka [1.7K]

Answer:

a) v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9

a(t) = x''(t) = v'(t) =6t-12

b)  0

c) a(t) = x''(t) = v'(t) =6t-12

When the acceleration is 0 we have:

6t-12=0, t =2

And if we replace t=2 in the velocity function we got:

v(t) = 3(2)^2 -12(2) +9=-3

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9

a(t) = x''(t) = v'(t) =6t-12

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

v(t) = 3t^2 -12t +9

We can factorize this function as v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is 0\leq t\leq 10 we need to analyze the following intervals:

0< t

For this case if we select two values let's say 0.25 and 0.5 we see that

v(0.25) =6.1875, v(0.5)=3.75

And we see that for a=0.5 >0.25=b we have that f(b)>f(a) so then the function is decreasing on this case.  

1

We have a minimum at t=2 since at this value w ehave the vertex of the parabola :

v_x =-\frac{b}{2a}= -\frac{-12}{2*3}= -2

And at t=-2 v(2) = -3 that represent the minimum for this function, we see that if we select two values let's say 1.5 and 1.75

v(1.75) =-2.8125< -2.25= v(1.5) so then the function sis decreasing on the interval 1<t<2

2

We see that the function would be increasing.

3

For this interval we will see that for any two points a,b with a>b we have f(a)>f(b) for example let's say a=3 and b =4

f(a=3) =0 , f(b=4) =9 , f(b)>f(a)

The particle is moving to the right then the velocity is positive so then the answer for this case is: 0

Part c

a(t) = x''(t) = v'(t) =6t-12

When the acceleration is 0 we have:

6t-12=0, t =2

And if we replace t=2 in the velocity function we got:

v(t) = 3(2)^2 -12(2) +9=-3

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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