I know the right answer to this I just don't know how to get it, please help.

1 answer:

7 0

$\bf \cfrac{x^2+5x+6}{x+4}\cdot \cfrac{x^2+x-12}{x^2+x-2}\implies \cfrac{\underline{(x+2)}(x+3)}{\underline{x+4}}\cdot \cfrac{\underline{(x+4)}(x-3)}{\underline{(x+2)}(x-1)}
\\\\\\
\cfrac{(x+3)(x-3)}{x-1}$

and yes, the restrictions are -4, -2 and 1.

the original expression has such restrictions because if ever "x" becomes one of those values, one of the denominators will turn to 0 making the fraction undefined.

You might be interested in

How would you describe the relationship between the real zeros and x-intercepts of the function y=log4(x-2)

notsponge [240]

Answer:

Step-by-step explanation:

First, look at y = log x. The domain is (0, infinity). The graph never touches the vertical axis, but is always to the right of it. A real zero occurs at x = 1, as log 1 = 0 => (1, 0). This point is also the x-intercept of y = log x.

Then look at y = log to the base 4 of x. The domain is (0, infinity). The graph never touches the vertical axis, but is always to the right of it. Again, a real zero occurs at x = 1, as log to the base 4 of 1 = 0 => (1, 0).

Finally, look at y=log to the base 4 of (x-2). The graph is the same as that of y = log to the base 4 of x, EXCEPT that the whole graph is translated 2 units to the right. Thus, the graph crosses the x-axis at (3, 0), which is also the x-intercept.

5 0

3 years ago

Read 2 more answers

I need help 50 points

const2013 [10]

1. m<13

2. y>2

3. p>-3

I don't know the answers for the last two

4 0

3 years ago

Read 2 more answers

Pls help I will mark brainlyesd

aev [14]

First, we have to solve for x by isolating it on one side.
6x + 10 = -8
(Subtract 10 from both sides)
6x = -18
(Divide both sides by 6)
x = -3

On option C, x is on -3, so that would be the correct answer

7 0

2 years ago

You bike 1 mile the first day of your training, 1.3 miles the second day, 1.9 miles the third day, and 3.1 miles the fourth d

JulsSmile [24]

Answer:

19.9 miles

Step-by-step explanation:

In this problem we have:

$d_1=1 mi$ is the distance travelled during the 1st day

$d_2=1.3 mi$ is the distance travelled during the 2nd day

$d_3=1.9 mi$ is the distance travelled during the 3rd day

$d_4=3.1 mi$ is the distance travelled during the 4th day

We notice that the difference between the distance travelled on the (n+1)-th day and the distance travelled on the n-th day doubles every day. In fact:

$d_2-d_1=0.3\\d_3-d_2=2\cdot 0.3 = 0.6\\d_4-d_3=2\cdot 0.6 = 1.2$