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3 years ago

Which of the following is the quotient of the rational expressions shown here?

Mathematics

2 answers:

svetoff [14.1K]3 years ago

3 0

For this case we must find the quotient of the following expression:

$\frac {\frac {x} {x-1}} {\frac {1} {x + 1}}$

Applying double C we have the following expression:

$\frac {x (x + 1)} {x-1} =$

Applying distributive property to the terms within the parentheses of the numerator we have:

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

Thus, the quotient is given by option A

Answer:

Option A

Likurg_2 [28]3 years ago

3 0

Answer:

The answer is option A. (x^2 + x)/(x-1)

Step-by-step explanation:

To solve this problem, we must first understand how to divide fractions. When dividing fractions, the first fraction is unchanged and is multiplied by the reciprocal of the second fraction. If we apply this knowledge to this problem, we get:

x/x-1 * x+1/1

When we multiply fractions, we simply multiply both of the numerators together and both of the denominators together to create a single fraction.

In this case we get:

x(x+1)/x-1

When we simplify this single fraction by using the distributive property, we get the following:

(x^2 + x)/(x-1)

Therefore, your answer is option A.

Hope this helps!

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Let's begin by listing out the information given to us:

There are four students: n = 4

Number of students to be selected: r = 2

To calculate the combination of 2 students to be chosen, we use:

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1 year ago

Please help with any of this Im stuck and having trouble with pre calc is it basic triogmetric identities using quotient and rec

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How I was taught all of these problems is in terms of r, x, and y. Where sin = y/r, cos = x/r, tan = y/x, csc = r/y, sec = r/x, cot = x/y. That is how I will designate all of the specific pieces in each problem.

Let's start with sin here. $\frac{2\sqrt{5}}{5} = \frac{2}{\sqrt{5}}$ Therefore, because sin is y/r, r = $\sqrt{5}$ and y = +2. Moving over to cot, which is x/y, x = -1, and y = 2. We know y has to be positive because it is positive in our given value of sin. Now, to find cos, we have to do x/r.

cos = $\frac{-1}{\sqrt{5}} = \frac{-\sqrt{5}}{5}$

Let's start with secant here. Secant is r/x, where r (the length value/hypotenuse) cannot be negative. So, r = 9 and x = -7. Moving over to tan, x must still equal -7, and y = $4\sqrt{2}$ . Now, to find csc, we have to do r/y.

csc = $\frac{9}{4\sqrt{2}} = \frac{9\sqrt{2}}{8}$

The pythagorean identities are

sin^2 + cos^2 = 1,

1 + cot^2 = csc^2,

tan^2 + 1 = sec^2.

Let's take a look at the information given here. We know that cos = -3/4, and sin (the y value), must be greater than 0. To find sin, we can use the first pythagorean identity.

sin^2 + (-3/4)^2 = 1

sin^2 + 9/16 = 1

sin^2 = 7/16

sin = $\sqrt{7/16}$ = $\frac{\sqrt{7}}{4}$

Now to find tan using a pythagorean identity, we'll first need to find sec. sec is the inverse/reciprocal of cos, so therefore sec = -4/3. Now, we can use the third trigonometric identity to find tan, just as we did for sin. And, since we know that our y value is positive, and our x value is negative, tan will be negative.

tan^2 + 1 = (-4/3)^2

tan^2 + 1 = 16/9

tan^2 = 7/9

tan = - $\sqrt{7/9} = \frac{-\sqrt{7}}{3}$

Let's take a look at the information given here. If we know that csc is negative, then our y value must also be negative (r will never be negative). So, if cot must be positive, then our x value must also be negative (a negative divided by a negative makes a positive). Let's use the second pythagorean identity to solve for cot.

1 + cot^2 = $(\frac{-\sqrt{6}}{2})^{2}$

1 + cot^2 = 6/4

cot^2 = 2/4

cot = $\frac{\sqrt{2}}{2}$

tan = $\sqrt{2}$

Next, we can use the third trigonometric identity to solve for sec. Remember that we can get tan from cot, and cos from sec. And, from what we determined in the beginning, sec/cos will be negative.

$(\frac{2}{\sqrt{2}})^2$ + 1 = sec^2