Write an equation in y=mx+b form for the following data

A category within biology that deals with the physical and chemical functions.

jeyben [28]

The answer is a. Physiology. Physiology is a category within biology that deals with the chemical and physical functions. It studies the functions in living systems. Physiology studies the chemical and physical functions on different levels of biological organization, such as organisms, organ systems, organs, cells, and molecules.

4 0

3 years ago

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3 is the _________ of -3 A- opposite B- absolute value C- both

densk [106]

It’s B - Absolute Value

6 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

The ratio of people who walk home from school to people who ride the bus home is 2 : 7. The number of bus riders is how many tim

Strike441 [17]

The number of bus riders is 3.5 times the number of walkers.

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3 years ago

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Lim x→π/2 1-sinx/cot^2x any genious help please

Simora [160]

Rewrite the limand as

(1 - sin(x)) / cot²(x) = (1 - sin(x)) / (cos²(x) / sin²(x))

… = ((1 - sin(x)) sin²(x)) / cos²(x)

Recall the Pythagorean identity,

sin²(x) + cos²(x) = 1

Then

(1 - sin(x)) / cot²(x) = ((1 - sin(x)) sin²(x)) / (1 - sin²(x))

Factorize the denominator; it's a difference of squares, so

1 - sin²(x) = (1 - sin(x)) (1 + sin(x))

Cancel the common factor of 1 - sin(x) in the numerator and denominator:

(1 - sin(x)) / cot²(x) = sin²(x) / (1 + sin(x))

Now the limand is continuous at x = π/2, so

$\displaystyle\lim_{x\to\frac\pi2}\frac{1-\sin(x)}{\cot^2(x)}=\lim_{x\to\frac\pi2}\frac{\sin^2(x)}{1+\sin(x)}=\frac{\sin^2\left(\frac\pi2\right)}{1+\sin\left(\frac\pi2\right)}=\boxed{\frac12}$

4 0

3 years ago