1 and 50
2 and 25
2 and 50
5 and 50
10 and 50
25 and 50
From 1st condition:
3^(x+1)=81
Take log on both sides,
➡ log [3^(x+1)]=log81
Use the property: log(x^a)=a logx
➡ (x+1) log 3=log(3⁴)
➡ (x+1) log 3=4×log 3
➡ (x+1)=4 (log 3 /log 3)
➡ x+1=4
➡ x=3➡↪➡↪ ANS
From 2nd condition, (Same process are repeated)
81^(x-y)=3
➡ 3^4(x-y)=3
➡ log [3^4(3-y)]=log 3
➡ 4(3-y) log 3=log 3
➡ (3-y)=(log 3)/(4log 3)
➡ 3-y=(1/4)
➡ y=11/4 ➡↪➡↪ ANS
F=ir^t
139=134r^10
139/134=r^10
r=(139/134)^(1/10) then:
f=134(139/134)^(t/10) so in 2014, t=24 so
f=134(139/134)^(2.4)
f≈146 million (to nearest million)
Some will say that you have to use the exponential function, but it really gives you the same answer...even for continuous compounding :)...
A=Pe^(kt)
139=134e^(10k)
139/134=e^(10k)
ln(139/134)=10k
k=ln(139/134)/10 so
A=134e^(t*ln(139/134)/10) when t=24
A=134e^(2.4*ln(139/134))
A≈146 million (to nearest million)
The only real reason or advantage to using A=Pe^(kt) is when you start getting into differential equations...
Answer:
resell, return, or store the goods
Step-by-step explanation:
Answer:
a = 5
Step-by-step explanation:
a/4 = 15/12
We can use cross products to solve
a* 12 = 4*15
12a = 60
Divide each side by 12
12a/12 = 60/12
a =5
