Answer:
2 vertical asymptotes occurring at x = 5 and x = -1
Step-by-step explanation:
given
recall that asymptotic occur at the locations that will make the equation undefined. In this case, the asymptote will occur at x-locations which will cause the denominator to become zero (and hence undefined)
Equating the denominator to zero,
(x-5)(x+1) = 0
(x-5) =0
x = 5 (first asymptote)
or (x+1) = 0
x = -1 (2nd asymptote)
I believe you will need to find what -0.125 of an inch is, subtract that from 16, then divide that into 11.
I think its c
Step-by-step explanation:
1)x+30=40
x=40-30
Answer :x=10
2)30-20+2x=10
10+2x=10
2x=0
Answer: x=0
3)3x-10+13=-2x+28
3x+3=-2x+28
3x+2x+3=28
3x+2x=28-3
5x=28-3
5x=25
x=5
4)13x-23-45=-7x+12
13x-68=-7x+12
13x+7x-68=12
13x+7x=12+68
20x=12+68
20x=80
x=4
5)2(x+4)=10x+24
2x+8=10x+24
2x-10x+8=24
2x-10x=24-8
-8x=24-8
-8x=16
x=-2
6)3(x-5)=1-(2x-4)
3x-15=1-(2x-4)
3x-15=1-2x+4
3x-15=5-2x
3x+2x-15=5
3x+2x=5+15
5x=5+15
5x=20
x=4
7)5x-10=15
5x=15+10
5x=25
x=5
8)x+4=x+6
4=6
no solution
PLEASE MARK ME AS BRAINLIEST
Answer:
65 and 155
Step-by-step explanation:
I just did the answer and I got it right so trust me
