What is the soloution of x^2-1/x^2+5x+4<_0
1 answer:
<span>You have the following inequality given in the problem shown above:
(x^2-1)/(x^2+5x+4)</span>≤<span>0
1. To solve it, you must factor it and then you must make the study of the signs.
2. Once you do the proccedure mentioned above, you obtain the following solutions:
-4<x<-1
-1<x</span>≤<span>1
3. You can graph the inaquality given in the problem, as you can see in the figure attached.</span>
(x^2-1)/(x^2+5x+4)</span>≤<span>0
1. To solve it, you must factor it and then you must make the study of the signs.
2. Once you do the proccedure mentioned above, you obtain the following solutions:
-4<x<-1
-1<x</span>≤<span>1
3. You can graph the inaquality given in the problem, as you can see in the figure attached.</span>
You might be interested in
Answer:
The limit leads to a determinate form.
Step-by-step explanation:
The following are indeterminate forms.
Given the limit of a function , to show if the given limit is determinate or indeterminate form, we will need to substitute the value of -
into the function as shown,
where a is any constant, therefore
Since we are able to get a finite value i.e 0, this shows that the limit does exist and leads to a determinate form
This figure is what we would call a composite figure. A composite figure is a shape that is made up of multiple other shapes. when looking at the pond and the border that surrounds it, we can see that this heart shape is made up of two semi circles and a square. To find the area of the border, we will use the area formulas for semi circles and for squares:
Semi Circle Area Formula:
÷ 2
Square Area Formula: length x width
Pi = 3.14
Radius = Half of the diameter, in this case 6
After finding the area of each shape, we can subtract the 2 ft width to find the area of the flower bed border.
Semi Circle Area Formula:
Square Area Formula: length x width
Pi = 3.14
Radius = Half of the diameter, in this case 6
After finding the area of each shape, we can subtract the 2 ft width to find the area of the flower bed border.