What is the solution of x+8/5x-1>0?
2 answers:
Answer:
x<-8 or x>1/5
Step-by-step explanation:
First we replace > symbol by = sign
To solve for x we multiply 5x-1 on both sides
x+8 =0
x=-8
5x-1=0 solve for x
x= 1/5
WE got two values x=-8 and x= 1/5
Now we make a number line and check with each interval
x<-8, -8<x<1/5, x>1/5
Pick x= -9 and check with the given inequality
1/46 >0 is true, so x<-8 satisfies our inequality
Pick x= 0 and check with the given inequality
-8 >0 is false , -8<x<1/5 does not satifies our inequality
Pick x= 2 and check with the given inequality
10/9>0 is true , x>1/5 satisfies our inequality
You might be interested in
----------------------------------------------------------------------------------------------------
Check the picture attached.
the secants of the circle, BA and DC, intersect at point P, as shown.
i)
let m(P)=a
ii)
let m(CDA)=b, then the measure of ar CA is 2b, as "the measure of an inscribed angle is half the measure of the arc it intercepts"
iii)
m(DAB)= m(P)+m(CDA)= a+b, as DAB is an exterior angle of the triangle PAD.
iv)
then the measure of arc DB is 2m(DAB)=2(a+b)=2a+2b
v)
This is always true,
----------------------------------------------------------------------------------------------------
In our problem, there is a slight difference, we have 1 secant and one tangent, but it is just a special case and nothing changes.
As an exercise, we can repeat the steps of the general proof, by first joining U and T.
----------------------------------------------------------------------------------------------------
by the theorem we just proved,
m(R)=(m(TU)-m(SU))/2
30°=(100°-m(SU))/2
60°=100°-m(SU)
m(SU)=40°
Check the picture attached.
the secants of the circle, BA and DC, intersect at point P, as shown.
i)
let m(P)=a
ii)
let m(CDA)=b, then the measure of ar CA is 2b, as "the measure of an inscribed angle is half the measure of the arc it intercepts"
iii)
m(DAB)= m(P)+m(CDA)= a+b, as DAB is an exterior angle of the triangle PAD.
iv)
then the measure of arc DB is 2m(DAB)=2(a+b)=2a+2b
v)
This is always true,
----------------------------------------------------------------------------------------------------
In our problem, there is a slight difference, we have 1 secant and one tangent, but it is just a special case and nothing changes.
As an exercise, we can repeat the steps of the general proof, by first joining U and T.
----------------------------------------------------------------------------------------------------
by the theorem we just proved,
m(R)=(m(TU)-m(SU))/2
30°=(100°-m(SU))/2
60°=100°-m(SU)
m(SU)=40°