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4 years ago

Simplify the equation

Mathematics

1 answer:

____ [38]4 years ago

6 0

Answer:

Assuming it would be -x-4

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Lonzell said the function shown in the graph is positive on the interval (−,) and negative on the interval (−,−)open negative 5

Nezavi [6.7K]

Answer:

The positive graph is on the points (-5,-4) and (2,5)

And, negative would be (-4,2)

Step-by-step explanation:

As it seems that fgraph shows positive in the case when it is over x-axis, and negative when it is below y-axis

As on the interval (-1,5)

The graph is below x-axis i.e. on (-1,2)

And, over x-axis i.e. on (2,5)

At this point Lonzell’s is not correct

As on the interval (-5,-1)

The graph is below x-axis i.e. on (-4,-1)

And, over x-axis i.e. on (-5,4)

At this point Lonzell’s is not correct

So,

The positive graph is on the points (-5,-4) and (2,5)

And, negative would be (-4,2)

5 0

4 years ago

Complete the input-output table for the function y = 3^x

Archy [21]

Answer:

x y

-2 0.111

-1 0.333

0 1

1 3

2 9

Step-by-step explanation:

3 0

3 years ago

Jessica's home town is a mid-sized city experiencing a decline in population. The following graph models the estimated populatio

Vaselesa [24]

The answer should be B=years

6 0

3 years ago

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Answer the following questions about the function whose derivative is f primef′(x)equals=x Superscript negative three fifths Ba

Elden [556K]

Answer:

(a) The critical points of f are x=0 and x=3.

(b)f is decreasing on $(0,3)$ and f is decreasing on $(3,\infty)$ .

(c) Therefore the local minimum of f is at x=3

Step-by-step explanation:

Given function is

$f'(x)= x^{-\frac35}(x-3)$

(a)

To find the critical point set f'(x)=0

$\therefore x^{-\frac35}(x-3)=0$

$\Rightarrow x=0,3$

The critical points of f are 0,3.

(b)

The interval are $(0,3)$ and $(3,\infty)$ .

To find the increasing or decreasing, taking two points one point from the interval (0,3) and another point $(3,\infty)$ .

Assume 1 and 4.

Now $f'(1)=(1)^{-\frac35}(1-3)$

and $f'(4)=(4)^{-\frac35}(4-3)>0$

Since 1∈ $(0,3)$ , f'(x)<0 and 4∈ $(3,\infty)$ , f'(x)>0

∴f is decreasing on $(0,3)$ and f is decreasing on $(3,\infty)$ .

(c)

$f'(x)= x^{-\frac35}(x-3)$

Differentiating with respect to x

$f''(x)=-\frac35x^{-\frac 85}(x-3)+x^{-\frac35}$

Now

$f''(0)=-\frac35(0)^{-\frac 85}(0-3)+(0)^{-\frac35}=0$

and

$f''(3)=-\frac35(3)^{-\frac 85}(3-3)+3^{-\frac35}$

$=0.517>0$

Since f''(x)>0 at x=3

Therefore the local minimum of f is at x=3

8 0

3 years ago

a company has 3 divisions.last year, each division earned a profit of $5×10^5 power what was the total profit the company earned

djverab [1.8K]

3 times 5× 10^5 = 3×5×10^5=15×10^5 = 1.5×10^6

6 0

3 years ago

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