How do you find the range of (x^3+x^2-6x)/(4x^2+4x-8) algebraically?

Suppose a is jointly proportional to b and c. if a=4 when b=8 and c = 9, then what is a when b=2 and c=18

asambeis [7]

From the given information, a exists jointly proportional to b and c.

then $$a \propto b c$

Therefore, the value of a exists 2.

<h3>What is the value of a?</h3>

Given, a exists jointly proportional to b and c.

$$a \propto b c$

Taking K as the constant of proportionality.

a = K b c

For b = 8 and c = 9, the value of a = 4

4 = K(8)(9)

4 = 72 K

Dividing the equation by 72, we get

$$K=\frac{1}{18}$

Using the value of K in the equation:

$$a=\frac{b c}{18}$

Substitute the values of b = 2 and c = 18, in the above equation

$$a=\frac{2 \times 18}{18} \\$

a = 2

Therefore, the value of a exists 2.

To learn more the value of x refers to:

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6 0

8 months ago

Please help!! i need to know very soon!

xeze [42]

Answer:

about 46

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Consider the following function. f(x) = 7x + 1 x (a) Find the critical numbers of f. (Enter your answers as a comma-separated li

sukhopar [10]

Answer:

a) DNE

b) The function increases for every real value of x.

c) DNE

Step-by-step explanation:

Given a function f(x), the critical points are those in which $f^{\prime}(x) = 0$ , that is, the roots of the first derivative of f(x).

Those critical points let us find the intervals in which the function increases or decreases. If the first derivative in the interval is positive, the function increases in the interval. If it is negative, the function decreases.

If the function increases before a critical point and then, as it passes the critical point, it starts to decrease, we have that the critical point $(x_{c}, f(x_{c}) is a relative maximum.If the function decreases before a critical point and then, as it passes the critical point, it starts to increase, we have that the critical point [tex](x_{c}, f(x_{c}) is a relative minimum.If the function has no critical points, it either always increases or always decreases.In this exercise, we have that:[tex]f(x) = 7x + 1$

(a) Find the critical numbers of f.

$f^{\prime}(x) = 0$

$f^{/prime}(x) = 7$

7 = 0 is false. This means that f has no critical points.

(b) Find the open intervals on which the function is increasing or decreasing.

Since there are no critical points, we know that either the function increases or decreases in the entire real interval.

We have a first order function in the following format:

$f(x) = ax + b$

In which $a > 0$ .

So the function increases for every real value of x.

(c) Apply the First Derivative Test to identify the relative extremum.

From a), we find that there are no critical numbers. So DNE

7 0

3 years ago

the perimeter of a square clock is 16 wholes and 6/9 inches.The perimeter of a rectangular clock is 16 wholes and 16/18 inches.W

sergejj [24]

6/9 and 16/18 is what we will mainly look at since the whole numbers are the same. the common denominator is 18. so you would multiply 6/9 by 2 (both the numerator and denominator multiplied by 2 ) therefore you will get 12/18. Which is bigger 12/18 or 16/18? 16/18 is bigger, therefore you would say sixteen and 6/9 is less than sixteen and 16/18. your answer should be this below. :)
16,6/9 < 16, 16/18.

4 0

3 years ago

A total of 600 advertisements were sold for a certain magazine. If 30 percent of the first 200 sold were in color, 20 percent of

solong [7]

Answer:

%35 of them were in color

Step-by-step explanation:

Since 30 percent of the first 200 sold were in color, 200x30/100 = 60 of them were in color.

Since 20 percent of the next 300 sold were in color, 300x20/100 = 60 of them were in color.

Since 90 percent of the last 100 sold were in color, 100x90/100 = 90 of them were in color.

In total, 60 + 60 + 90 = 210 out of 600 were in color.

The percentage is: $\frac{600}{100}$ = $\frac{210}{x}$ → x = $\frac{21000}{600}$ = 35

6 0

2 years ago