Hermina cut a 10" by 15" piece of cardboard down the diagonal

Consider the function f ( x ) = − 2 x 3 + 27 x 2 − 108 x + 9 . For this function there are three important open intervals: ( − [

Darya [45]

Answer:

A=3 and B=6

Step-by-step explanation:

Increasing and Decreasing Intervals of Functions

Given f(x) as a real function and f'(x) its first derivative.

If f'(a)>0 the function is increasing in x=a

If f'(a)<0 the function is decreasing in x=a

If f'(a)=0 the function has a critical point in x=a

As we can see, the critical points may define open intervals where the function has different behaviors.

We have

$f ( x ) = - 2 x^3 + 27 x^2 - 108 x + 9$

Computing the first derivative:

$f' ( x ) = - 6 x^3 + 54 x - 108$

We find the critical points equating f'(x) to zero

$- 6 x^3 + 54 x - 108=0$

Simplifying by -6

$x^2 -9 x +18=0$

We get the critical points

$x=3,\ x=6$

They define the following intervals

$(-\infty,3),\ (3,6),\ (6,+\infty)$

Thus A=3 and B=6

3 0

3 years ago

How to solve <img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B4%7D%20x%20%7B%7D%5E%7B2%7D%20%20%3D%20x%20%2B%202" id="TexFormu

Vikentia [17]

Answer:

➩ $x=2,-\frac{2}{3}$

Step-by-step explanation:

$\sqrt{4x^2}=x+2$

➨ We can also solve by completing both squares, however. Since we can pull out the square root.

$\sqrt{4x^2}=|2x|$ ➩ Define of Absolute Value/Square Root

➩ $\sqrt{x^2}=|x|$

Thus, our new equation is ➩ $|2x|=x+2$

To solve an absolute-value equation, let there be two conditions.

➨ Where x ≥ 0

$2x=x+2\\$

Move x to another side

$2x-x=2\\x=2$

➨ Where x < 0

$-2x=x+2\\-2x-x=2\\-3x=2\\x=\frac{2}{-3}\\x=-\frac{2}{3}$

3 0

3 years ago

Can you please help me which one has the same value as 1.303

olya-2409 [2.1K]

Answer:

the answer is B

Step-by-step explanation:

6 0

2 years ago

Translate to an equation and solve the product of 2 and a number increased by 8 is -48

Marizza181 [45]

The equation would be 2x+8=-48 and x=-28

4 0

3 years ago

For a data set of weights (pounds) and highway fuel consumption amounts (mpg) of eight types of automobile, the linear correl

Jobisdone [24]

Answer:

The P value indicates that the probability of a linear correlation coefficient that is at least as extreme is 0.3% which is not significant (at α = 0.05), so there is insufficient evidence to conclude that there is a linear correlation between weight and consumption. of highway fuel in cars.

Step-by-step explanation:

We have that the correlation coefficient shows the relationship between the weights and amounts of road fuel consumption of seven types of car, now the P value establishes the importance of this relationship. If the p-value is lower than a significance level (for example, 0.05), then the relationship is said to be significant, otherwise it would not be so, this case being 0.003 not significant.

The statement would be the following:

The P value indicates that the probability of a linear correlation coefficient that is at least as extreme is 0.3% which is not significant (at α = 0.05), so there is insufficient evidence to conclude that there is a linear correlation between weight and consumption. of highway fuel in cars.

5 0

3 years ago