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saveliy_v [14]
3 years ago
7

Which statement correctly identifies a local minimum of the graphed function?

Mathematics
2 answers:
4vir4ik [10]3 years ago
4 0

That would be option C.

Rina8888 [55]3 years ago
3 0

Answer:

C. Over the interval [–1, 0.5], the local minimum is 1.

Step-by-step explanation:

From the graph we observe the following:

1) x intercepts are two points.

ii) y intercept = 1

f(x) = y increases from x=-infinity to -1.3

y decreases from x=-1.3 to 0

Again y increases from x=0 to end of graph.

Hence in the interval for x as (-1.3, 1) f(x) has a minimum value of (0,1)

i.e. there is a minimum value of 1 when x =0

Since [-1,0.5] interval contains the minimum value 1 we find that

Option C is right answer.

There is a local minimum of 1 in the interval [-1,0.5]



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from the edge of a 486-foot cliff, peyton shot an arrow over the ocean with an initial upward velocity of 90-feet per second
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Answer:

9 seconds

Step-by-step explanation:

The complete question is

The altitude of an object, d, can be modeled using the equation below:

d=-16t^2 +vt+h

from the edge of a 486 foot cliff, Peyton shot an arrow over the ocean with an initial upward velocity of 90 feet per second. In how many seconds will the arrow reach the water below?

Let

d ----> the altitude of an object in feet

t ---> the time in seconds

v ---> initial velocity in ft per second

h ---> initial height of an object in feet

we have

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we  know that

When the arrow reach the water the value of d is equal to zero

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substitute the values and solve for t

0=-16t^2 +(90)t+486

-16t^2+90t+486=0

Multiply by -1 both sides

16t^2-90t-486=0

The formula to solve a quadratic equation of the form ax^{2} +bx+c=0 is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

a=16\\b=-90\\c=-486

substitute in the formula

t=\frac{90(+/-)\sqrt{-90^{2}-4(16)(-486)}}{2(16)}

t=\frac{90(+/-)\sqrt{39,204}}{32}

t=\frac{90(+/-)198}{32}

t_1=\frac{90(+)198}{32}=9\ sec

t_2=\frac{90(-)198}{32}=-3.375\ sec

the solution is t=9 sec

see the attached figure to better understand the problem

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