1)3(k+5)=-2(3k-6)
3k+15=-6k+12
3k=-6k+12-15
3k=-6k-3
3k+6k=-3
9k=-3
k=-3/9=-1/3
so, k=-1/3
2)1/2r+4=3/4r-3/2
1/2r=3/4r-3/2-4
1/2r=3/4r-11/2
1/2r-3/4r=-11/2
-r/4=-11/2
4*(-r/4)=4*(-11/2)
-r=-22
r=22
3)2a-9=2a+5
2a=2a+5+9
2a=2a+14
2a-2a=14
0=14
no solution
4)8m+2+4m=2(6m+1)
12m+2=12m+2
12m=12m+2-2
12m=12m
12m-12m=0
0=0
true for all m
Answer:
the remaining mass of unobtanium -41 after a 5 seconds, 10 seconds, 15 seconds and 20 seconds is 24 gr, 12 gr , 5.33 gr and 3 gr respectively
Step-by-step explanation:
the equation that governs the remaining mass m of unobtanium -41 after a time t is
m=m₀*2^(-t/T) , where t is in seconds ,m₀ represents initial mass and T=half-life
Therefore
a) for t=5 s
m=48 gr*2^(-t/5 s) = 48 gr*2^(-5 s/5 s) = 48 gr/2 = 24 gr
b) for t=10 s
m=48 gr*2^(-t/5 s) = 48 gr*2^(-10 s/5 s) = 48 gr/4 = 12 gr
b) for t=15 s
m=48 gr*2^(-t/5 s) = 48 gr*2^(-15 s/5 s) = 48 gr/9 = 5.33 gr
b) for t=20 s
m=48 gr*2^(-t/5 s) = 48 gr*2^(-20 s/5 s) = 48 gr/16 = 3 gr
thus the remaining mass of unobtanium -41 after a 5 seconds, 10 seconds, 15 seconds and 20 seconds is 24 gr, 12 gr , 5.33 gr and 3 gr respectively
Answer:
The value of x is less than 16.
Step-by-step explanation:
x<16= x is less than 16
x>16= x is greater than 16
x≤16= x is at most 16
x≥16=x is at least 16
Hope this helps!
Answer:
Distance is 8.485
Midpoint is (3,4)
Step-by-step explanation:
Use the points (-1,7) and (5,1)
Insert them into the equation and solve for D
Insert points into second equation to find midpoint
