Answer:
22.5 years to the nearest tenth.
Step-by-step explanation:
The equation is
6000 = 4000(1 + 0.018/12)^12t where t = the number of years
(1 + 0.018/12)^12t = 1.5
1.0015^12t = 1.5
Taking logs:
12t log 1.0015 = log 1.5
12t = log 1.5 / log 1.0015
12t = 270.51
t = 22.54 years.
Let X be the number of method I tables, Y method II tables.
1.5x+2.5y=308. Hours spent
10x+5y=1260. Cost
Multiply the equation 1 by 2
3x+5y=616
10x+5y=1260
Subtract the equation 1 from the equation 2 to get
7x=644
X=644÷7
X=92. number of method I tables
So then
10 (92)+5y=1260
920+5y=1260
5y=1260-920
5y=1,260−920
5y=340
Y=340÷5
Y=68 number of method II tables
27^x / 9^x
= 3^3x / 3^2x
= 3^x
So its equivalent to C.
Its also equivalent to B and E.
Answer:
a
Step-by-step explanation:
-4x3=-12x
-4x-2=8x
-4x4=-16y
8+-12=-4x
-4x-16y
Answer:
47 i need this to be 20 words so ignore this
