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3 years ago

4÷(-2) how do I figure this out

Mathematics

1 answer:

Papessa [141]3 years ago

6 0

Answer:

-2

Step-by-step explanation:

If you divide a positive by a negative, the answer will be the same a positive divided by a positive but it will be negative. So solve it normally and add a negative sign in front of it.

4/2 = 2

Add the negative

-2

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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:

Step-by-step explanation:

<u>Increasing and Decreasing Intervals of Functions</u>

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

1 p

dlinn [17]

Answer:

V = pi * R^2 * h

R = (V / (pi * h))^1/2 = (141.3 / (3.14 * 5))^1/2 = 3 cm

8 0

3 years ago

Subtract the given function and indicate the domain of the difference

maxonik [38]

Answer:

The domain is $x\in (-\infty,\infty).$

Step-by-step explanation:

Given functions $f(x)=x^2+3x+1$ and $g(x)=2x^2-4x-1$

Subtract these two functions:

$f(x)-g(x)\\ \\=(x^2+3x+1)-(2x^2-4x-1)\\ \\=x^2+3x+1-2x^2+4x+1\\ \\=(x^2-2x^2)+(3x+4x)+(1+1)\\ \\=-x^2+7x+2$

Plot these difference on the coordinate plane (see attached diagram). This function is defined for all vlues of x, so the domain is $x\in (-\infty,\infty).$