What is the value of x?

The graph shown below expresses a radical function that can be written in the form f(x)=a(x+k)^1/n + c. What does the graph tell

Pie

Answer: n is a positive odd number.

Step-by-step explanation:

Ok, we know that the function is something like:

f(x)=a(x+k)^1/n + c

In the graph we can see two thigns:

All the values of the graph are positive values (even for the negative values of x), but in the left side we can see that the function decreases and is different than the right side.

So this is not an even function, then n must be an odd number (n odd allows us to have negative values for y = f(x) that happen when x + k is negative).

Also, we can see that the function increases, if n was a negative number, like: n = -N

we would have:

$f(x) = \frac{a}{(x+k)^{1/N}} + c$

So in this case x is in the denominator, so as x increases, we would see that the value of y decreases, but that does not happen, so we can conclude that the value of n must be positive.

Then n is a positive odd number.

4 0

4 years ago

Read 2 more answers

What is the mean of integers -16,-27,21,-19,14,-3 -

Lelu [443]

Answer:

-5

Step-by-step explanation:

-16+(-27)+21+(-19)+14+(-3)/6

-30/6

=-5

3 0

3 years ago

Use the definition of continuity to determine whether f is continuous at a.

dmitriy555 [2]

$f(x)$ will be continuous at $x=a=7$ if
(i) $\displaystyle\lim_{x\to7}f(x)$ exists,
(ii) $f(7)$ exists, and
(iii) $\displaystyle\lim_{x\to7}f(x)=f(7)$ .

The second condition is immediate, since $f(7)=8918$ has a finite value. The other two conditions can be established by proving that the limit of the function as $x\to7$ is indeed the value of $f(7)$ . That is, we must prove that for any $\varepsilon>0$ , we can find $\delta>0$ such that

$|x-7|$

Now,

$|f(x)-f(7)|=|5x^4-9x^3+x-8925|$

Notice that when $x=7$ , we have $5x^4-9x^3+x-8925=0$ . By the polynomial remainder theorem, we know that $x-7$ is then a factor of this polynomial. Indeed, we can write

$|5x^4-9x^3+x-8925|=|(x-7)(5x^3+26x^2+182x+1275)|=|x-7||5x^3+26x^2+182x+1275|$

This is the quantity that we do not want exceeding $\varepsilon$ . Suppose we focus our attention on small values $\delta$ . For instance, say we restrict $\delta$ to be no larger than 1, i.e. $\delta\le1$ . Under this condition, we have

$|x-7|$

Now, by the triangle inequality,

$|5x^3+26x^2+182x+1275|\le|5x^3|+|26x^2|+|182x|+|1275|=5|x|^3+26|x|^2+182|x|+1275$

If $|x|$ , then this quantity is moreover bounded such that

$|5x^3+26x^2+182x+1275|\le5\cdot8^3+26\cdot8^2+182\cdot8+1275=6955$

To recap, fixing $\delta\le1$ would force $|x|$ , which makes

$|x-7||5x^3+26x^2+182x+1275|$

and we want this quantity to be smaller than $\varepsilon$ , so

$6955|x-7|$

which suggests that we could set $\delta=\dfrac{\varepsilon}{6955}$ . But if $\varepsilon$ is given such that the above inequality fails for $\delta=\dfrac{\varepsilon}{6955}$ , then we can always fall back on $\delta=1$ , for which we know the inequality will hold. Therefore, we should ultimately choose the smaller of the two, i.e. set $\delta=\min\left\{1,\dfrac{\varepsilon}{6955}\right\}$ .

You would just need to formalize this proof to complete it, but you have all the groundwork laid out above. At any rate, you would end up proving the limit above, and ultimately establish that $f(x)$ is indeed continuous at $x=7$ .

5 0

3 years ago

Plz help me with this question

german

Answer:

1 Kilogram is $1.44

Step-by-step explanation:

$2.88 divided by 2 kilograms is $1.44 for each kilogram of sand.

7 0

3 years ago

Given: △FKL, FK=a, m∠F=45°, m∠L=30° Find: FL

drek231 [11]

The best way to do this is to draw a picture of ΔFKL and include line segment KM that is perpendicular to FL. This creates ΔFKM which is a 45°-45°-90° triangle and ΔLKM which is a 30°-60°-90° triangle.

Find the lengths of FM and ML. Then, FM + ML = FL

FM

ΔFKM (45°-45°-90°): FK is the hypotenuse so FM = $\frac{a}{\sqrt{2}} = \frac{a\sqrt{2} }{2}$