A track team has 64 members. There are 18 more boys than girls on the team. How many girls are on the track team?

1) Let f(x)=6x+6/x. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relat

brilliants [131]

Answer:

1) increasing on (-∞,-1] ∪ [1,∞), decreasing on [-1,0) ∪ (0,1]

$x = -1$ is local maximum, $x = 1$ is local minimum

2) increasing on [1,∞), decreasing on (-∞,0) ∪ (0,1]

$x = 1$ is absolute minimum

3) increasing on (-∞,0] ∪ [8,∞), decreasing on [0,4) ∪ (4,8]

$x = 0$ is local maximum, $x = 8$ is local minimum

4) increasing on [2,∞), decreasing on (-∞,2]

$x = 2$ is absolute minimum

5) increasing on the interval (0,4/9], decreasing on the interval [4/9,∞)

$x = 0$ is local minimum, $x = 4/9$ is absolute maximum

Step-by-step explanation:

To find minima and maxima the of the function, we must take the derivative and equalize it to zero to find the roots.

1) $f(x) = 6x + 6/x$

$f\prime(x) = 6 - 6/x^2 = 0$ and $x \neq 0$

So, the roots are $x = -1$ and $x = 1$

The function is increasing on the interval (-∞,-1] ∪ [1,∞)

The function is decreasing on the interval [-1,0) ∪ (0,1]

$x = -1$ is local maximum, $x = 1$ is local minimum.

2) $f(x)=6-4/x+2/x^2$

$f\prime(x)=4/x^2-4/x^3=0$ and $x \neq 0$

So the root is $x = 1$

The function is increasing on the interval [1,∞)

The function is decreasing on the interval (-∞,0) ∪ (0,1]

$x = 1$ is absolute minimum.

3) $f(x) = 8x^2/(x-4)$

$f\prime(x) = (8x^2-64x)/(x-4)^2=0$ and $x \neq 4$

So the roots are $x = 0$ and $x = 8$

The function is increasing on the interval (-∞,0] ∪ [8,∞)

The function is decreasing on the interval [0,4) ∪ (4,8]

$x = 0$ is local maximum, $x = 8$ is local minimum.

4) $f(x)=6(x-2)^{2/3} +4=0$

$f\prime(x) = 4/(x-2)^{1/3}$ has no solution and $x = 2$ is crtitical point.

The function is increasing on the interval [2,∞)

The function is decreasing on the interval (-∞,2]

$x = 2$ is absolute minimum.

5) $f(x)=8\sqrt x - 6x$ for $x>0$

$f\prime(x) = (4/\sqrt x)-6 = 0$

So the root is $x = 4/9$

The function is increasing on the interval (0,4/9]

The function is decreasing on the interval [4/9,∞)

$x = 0$ is local minimum, $x = 4/9$ is absolute maximum.

5 0

2 years ago

Please help with this & show your working out to get brainliest. And please don’t trick me.

timurjin [86]

The lowest is the square root of 9 which is 3. After that is 3.15, 13/4 (3.25), 3 1/2 (3.50). My explanation would be: first I found out the square root of 9. Then divided 13 by 4.

7 0

2 years ago

Select either relation (if the set is a relation but not a function), function (if the set is both a relation and a function), o

galina1969 [7]

Answer: The correct options are

(A) relation

(B) function.

Step-by-step explanation: We are given to check whether the following set is a relation or a relation and a function or neither :

A = {(1, 2) (2, 2) (3, 2) (4, 2)} .

Let us consider the following two sets :

U = {1, 2, 3, 4} and V = {2}.

Then, we see that the Cartesian Product of U and V is given by

U × V = {(1, 2) (2, 2) (3, 2) (4, 2)}.

Since a relation is a subset of Cartesian product, so the given set A is a relation.

Also, we know that a relation is a function if every first element is associated with one second element.

Since the relation A satisfies this condition, so the relation A is a function.

Thus, the given set A is both a relation and a function.

Option (A) and (B) are correct.

6 0

2 years ago

En una muestra de 250 tornillos de precisión, la longitud promedio fue de 3.52 cm, con una desviación estándar de 0.52 mm. Si el

natima [27]

I cant read this. This question is not in english.

6 0

2 years ago

Jaime went to the grocery store and bought a six-pack of soda for $1.50. How much did he pay for each soda in the pack?

Ira Lisetskai [31]

Answer:

Step-by-step explanation:

1.50 / 6

0.25

4 0

3 years ago