Write the absolute value inequality in the form |x-b| c that has the solution set x<-5 or x>7.

1 answer:

4 0

Answer:
|x - 1| > 6
Step-by-step explanation:
b is the centre:
(-5 + 7)/2
2/2 = 1
b = 1
c is the distance from the centre
c = 7 - 1 = 6
Since it's an or case, > with the modulus
|x - 1| > 6

You might be interested in

Find the cost per visit to the nearest cent body fit $238 per year 95

AfilCa [17]

The answer of this question is $2.51

4 0

3 years ago

Write the equation of a hyperbola with vertices (0, -4) and (0, 4) and foci (0, -5) and (0, 5).

andre [41]

Check the picture below. So, more or less looks like so.

notice, the center is clearly at the origin, and notice how long the "a" component is, also, bear in mind that, is opening towards the y-axis, that means the fraction with the "y" variable is the positive one.

Also notice, the "c" distance from the center to either foci, is just 5 units.

$\bf \textit{hyperbolas, vertical traverse axis }\\\\
\cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1
\qquad 
\begin{cases}
center\ ({{ h}},{{ k}})\\
vertices\ ({{ h}}, {{ k}}\pm a)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{{{ a }}^2+{{ b }}^2}
\end{cases}\\\\
-------------------------------\\\\$

$\bf \begin{cases}
h=0\\
k=0\\
a=4\\
c=5
\end{cases}\implies \cfrac{(y-{{ 0}})^2}{{{ 4}}^2}-\cfrac{(x-{{ 0}})^2}{{{ b}}^2}=1\implies \cfrac{y^2}{16}-\cfrac{x^2}{b^2}=1
\\\\\\
c=\sqrt{a^2+b^2}\implies c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\\\\\\
\sqrt{5^2-4^2}=b\implies \boxed{3=b}
\\\\\\
\cfrac{y^2}{16}-\cfrac{x^2}{3^2}=1\implies \boxed{\cfrac{y^2}{16}-\cfrac{x^2}{9}=1}$

7 0

3 years ago

Compare the process of solving |x – 1| + 1 < 15 to that of solving |x – 1| + 1 > 15. Check all of the following you includ

zepelin [54]

The required Comparison of the inequalities are

The |x – 1| + 1 > 15 represents the value of x lies between 13<x<15.

The range of values encompassing the region's junction is (-13, 15).

If x is more than or equal to 15, then x-11+1>15 indicates the value of x is greater than or equal to 13. None of the regions in the intersection are empty.

<h3>What is inequality?</h3>

When comparing two numbers, an inequality indicates whether one is less than, larger than, or not equal to the other.

We take into account the various variables of the inequality

|x – 1| + 1 > 15

Therefore

|-x-1|+1-1<15-1

|-x-1|-1 <14

13<x<15

The required region lies between the inequality -13 <x< 15.

Simplify the inequality Ix-11+1 > 15 we get,

|x-1|+1 > 15

|x+1| +1-1 >15-1

|x-1| > 14

x> 15

x<-13

If x has a value between -13 and x + 15, then the expression "|x-1|+1+115" is true. The range "(-13, 15)" contains the intersection of the region.

If "|x-1|+1>15" then either "x >15" or "x-13" applies to the value of x. This region's intersection is unoccupied.