Question

Question 1(Multiple Choice Worth 5 points)

Answer 1

Question 1:

For this case we must simplify the following expression:

$\frac {x ^ 2 + 3x + 2} {x + 1}$

We factor the numerator of the expression, looking for two numbers that when multiplied give as a result 2 and when summed give as result 3. These numbers are 2 and 1.

$2 + 1 = 3\\2 * 1 = 2$

So:

$x ^ 2 + 3x + 2 = (x + 1) (x + 2)$

Rewriting the expression we have:

$\frac {(x + 1) (x + 2)} {(x + 1)}$

Simplifying common terms in the numerator and denominator, we have that the expression is reduced to:

$\frac {x ^ 2 + 3x + 2} {x + 1} = (x + 2)$

ANswer:

Option A

Question 2:

For this case we must find the value of the variable "x" of the following expression:

$\frac {4} {7} = \frac {x-1} {12}$

Multiplying by 12 on both sides of the equation we have:

$\frac {12 * 4} {7} = x-1\\\frac {48} {7} = x-1$

We add 1 to both sides of the equation:

$\frac {48} {7} + 1 = x\\x = \frac {48 + 7} {7}\\x = \frac {55} {7}$

ANswer:

Option A

Question 3:

For this case we must find the domain of the following function:

$f (x) = \frac {x + 5} {x-2}$

The domain of a function is given by all the values ​​for which the function is defined.

The given function is not defined when the denominator is 0.

$x-2 = 0\\x = 2$

Thus, the domain is given by all real numbers except 2.

Answer:

OPTION C

Question 4:

For this case we must simplify the following expression:

$\frac {15x ^ 2-18} {6}$

We take common factor 3 from the numerator of the expression:

$3 (5x ^ 2-6)$

We can write 6 as $2 * 3.$

Rewriting the expression we have:

$\frac {3 (5x ^ 2-6)} {2 * 3} =$

Simplifying common terms of the numerator and denominator we have that the expression is reduced to:

$\frac {5x ^ 2-6} {2}$

ANswer:

Option D

Question 5:

For this case we must find the vertical asymptotes of the following function:

$f (x) = \frac {5x + 5} {x ^ 2 + x-2}$

We have by definition, that the vertical asymptotes are vertical lines to which the function is approaching indefinitely without ever cutting them.

To locate the vertical asymptotes in rational functions, we find the values ​​of "x" that annul the denominator, but not the numerator.

So:

We factor the denominator, looking for two numbers that when multiplied by -2 and when added together give 1. Thus:

$x ^ 2 + x-2 = (x + 2) (x-1)$

Now, equaling the denominator to zero, we have that the vertical asymptote occurs in areas of infinite discontinuity of:

$x = -2, x = 1$

Answer:

Option C

Question 6:

For this case we must find the discontinuities and zeros of the following function:

$f (x) = \frac {x ^ 2 + 5x + 6} {x + 2}$

We have by definition, that if a function is not continuous at a point, it is said that the function has a discontinuity at that point and that the function is discontinuous.

The function is undefined where the denominator equals 0.

$x + 2 = 0\\x = -2$

It is a discontinuity.

Now, we factor the numerator looking for two numbers that multiplied by 6 and added by 5. These numbers are 3 and 2. Then:

$(x + 3) (x + 2)$

Now we find the zeros of the funicon, for that we replace f (x) with y.

$y = \frac {(x + 3) (x + 2)} {x + 2}$

We eliminate common terms from the numerator and denominator.

$y = x + 3$

To find the roots, we do y = 0:

$0 = x + 3\\x = -3$

ANswer:

Option B

Question 7:

For this case we must find the discontinuities and zeros of the following function:

$f (x) = \frac {3x} {x ^ 2-9}$

We have by definition, that if a function is not continuous at a point, it is said that the function has a discontinuity at that point and that the function is discontinuous.

The function is undefined where the denominator equals 0. Then:

$x ^ 2-9 = 0\\x ^ 2 = 9\\x = \pm \sqrt {9}\\x = \pm3$

The function is discontinuous for $x_ {1} = 3$ and $x_ {2} = - 3$

Now factoring the denominator we have:

(x-3) (x + 3)

To find the zeros of the function we change f (x) by y.

$y = \frac {3x} {(x-3) (x + 3)}$

We make y = 0:

$0 = \frac {3x} {(x-3) (x + 3)}\\0 = 3x\\x = 0$

So, the zeros of the function are at$x = 0$

ANswer:

The function is discontinuous for $x_ {1} = 3$ and $x_ {2} = - 3$

The zeros of the function are at $x = 0$

Question 8:

For this case we have:

x: Variable representing red fish

y: Variable representing blue fish

They tell us that:

$0.6 (x + y) = x\\x = 10 + y$

Then, we substitute the second equation in the first one;

$0.6 (10 + y + y) = 10 + y\\6 + 0.6y + 0.6y = 10 + y\\6 + 1.2y = 10 + y\\1.2y-y = 10-6\\0.2y = 4\\y = \frac {4} {0.2}\\y = 20$

So, there are 20 blue fish.

Answer:

20 blue