I need HELP ASAP! Rotational expressions.

Mathematics

1 answer:

Anni [7]3 years ago

4 0

So simplifie top and bottom by x-1 and will get 5/(x-1)

so choice D is right sure

hope this will help you

You might be interested in

How do you eliminate the radical sign?

vekshin1

Answer:

Step-by-step explanation:

you can eliminate the radical by squaring both sides of the equation

4 0

3 years ago

The set of all numbers less than or equal to -8 or greater than or equal to 2.

tatiyna

The set of all numbers less than or equal to -8 or greater than or equal to 2 is -8 ≥ x ≥ 2. The set is {x∈R|-8 ≥ x ≥ 2}

<h3>The set of numbers</h3>

From the question, we are to write the set of numbers l<u>ess than or equal</u> to -8 or<u> greater than or equal</u> to 2

Let x represent the set of the numbers

If a number is less than or equal to -8

Then,

-8 is greater than or equal to the number

That is,

-8 ≥ x

If a number is greater than or equal to 2

Then,

x ≥ 2

Thus, the set of all the numbers is

-8 ≥ x ≥ 2

Hence, the set of all numbers less than or equal to -8 or greater than or equal to 2 is -8 ≥ x ≥ 2. The set is {x∈R|-8 ≥ x ≥ 2}

Learn more on Set of numbers here: brainly.com/question/18667553

#SPJ1

7 0

2 years ago

x2 + 8x + 6 (x2 + 8x + ) + 6 Erin was completing the square of the quadratic function in order to find the extreme value. What i

kolezko [41]

Answer:

Step-by-step explanation:

The next step is to add (1/2 * 8)^2 = 4^2 = 16 to the 8x and subtract 16 from the + 6:

- to give (x^2 + 8x + 16) + 6 - 16 as the next step in the process.

The extreme minimum is +6 - 16 = -10.

6 0

3 years ago

For the function, f(x)=(9x+7)/(2x+4)

AVprozaik [17]

A. $f(x)=\frac{9x+7}{2x+4}\\f'(x)=\frac{(9)(2x+4)-(9x+7)(2)}{(2x+4)^2}\\f'(x)=\frac{(18x+36)-(18x+14)}{(2x+4)(2x+4)}\\f'(x)=\frac{22}{4x^2+16x+16}\\f'(x)=\frac{11}{2x^2+8x+8}\\\frac{11}{(2x+4)(x+2)}=0\\\frac{1}{(2x+4)(x+2)}=0\\x=-\infty,\infty$ - There are no critical points because the graph is neither continuous nor smooth. There is a discontinuity at x = 2.

B. $\frac{1}{(2x+4)(x+2)}=0\\x=-\infty,\infty$ - The absolute maximum is f(lim⇒-2_-) = infinity. The absolute minimum is f(lim⇒-2_+) = -infinity. This applies to the interval [-10, 7].

C. $f(x)=\frac{9x+7}{2x+4}\\f(0)=\frac{9(0)+7}{2(0)+4}\\f(0)=\frac{7}{4}\\f(0)=1.75\\f(5)=\frac{9(5)+7}{2(5)+4}\\f(5)=\frac{45+7}{10+4}\\f(5)=\frac{52}{14}\\f(5)=\frac{26}{7}\\f(5)=3.714$ - The absolute maximum is f(5) = 26/7 or 3.714. The absolute mimimum is f(0) = 1.75. This applies to the interval [0, 5]. Proof: graph f(x) at [0, 5] on a graph or graphing calculator.