Answer: C) $21,000
Step-by-step explanation:
Given : Pat bought a boat for $37,000 in 2001. In 2006 the boat was worth $27,000.
If the boat depreciation is linear, then the amount by which the value of boat depreciates must be constant.
Let x be the constant depreciation in the value of boat per year.
Then , the value of boat (in dollars) after n years from 2001 is given by :-
(1)
For year 2006 , n=5 and V = 27000
Then ,
i.e.
i.e.
i.e.
Thus , the constant amount of depreciation in the value of boat per year. = $2000
Now for year 2009 , put in n=8 and x= 2000 in (1), we get
Hence, the value of boat in 2009 = $21,000
y = 3x + 2
y = 7x - 10
Answer:
x=3
y=11
Step-by-step explanation:
y = 3x + 2 equation 1
y = 7x - 10 equation 2
using equation 1 in equation 2 we have:
3x + 2 = 7x - 10
2=7x - 3x -10
2 = 4x -10
2+10 = 4x
12= 4x
12/4 = x
we have:
x= 3
using equation 1:
y = 3x + 2
y= 3(3)+2
y=9+2
y=11
(3n-7)/6 = (2n+5)/3
B. 319$
C. 715$
D. 1375$
Answer:
$319
Step-by-step explanation:
What is the value of x?
Answer:
option d
d ---- sin30 =5/x
Answer:
D. sin 30° = 5/x
Step-by-step explanation:
recall that for a right triange, considering one of the acute angles θ
sin θ = length of opposite side / length of hypotenuse
from the picture, we can see that
θ = 30°, length of opposite site = 5 and length of hypotenuse = x
substituting this into the above equation
sin 30° = 5/x