Answer:
-20,476
Step-by-step explanation:
Answer:
Let x = the charge in 1st city before taxes
Let y = the charge in 2nd city before taxes
Set up equation before taxes.
y = x - 1500 eq1
Set up equation for total tax paid.
0.065x + 0.06y = 378.75 eq2
Substitute eq1 into eq2.
0.065x + 0.06(x - 1500) = 378.75
0.065x + 0.06x - 90 = 378.75
0.125x - 90 = 378.75
0.125x = 468.75
x = 3750
Substitute this value of x into eq1.
y = 3750 - 1500
y = 2250
The hotel charge in city one is $3750 and the hotel charge in city two is $2250
Answer:
Option D.The decrease in the value of the car, which is 8%
Step-by-step explanation:
we have a exponential function of the form
where
y is the value of the car
x is the time in years
a is the initial value
b is the base
r is the rate of decrease
b=1+r
In this problem we have
a=$24,000 initial value of the car
b=0.92
so
0.92=1+r
r=0.92-1=-0.08=-8%-----> is negative because is a rate of decrease
You can predict that it's odd because the end of the 2 numbers is 1 and 1 x 1 is 1 and one is an odd number. (answer is odd)
X=2h, y=3k
Substitute these values into equations.
y+2x = 4 ------> 3k+2*2h=4 -----> 3k +4h =4
2/y - 3/2x = 1-----> 2/3k -3/(2*2h) = 1 ------> 2/3k - 3/4h =1
We have a system of equations now.
3k +4h =4 ------> 3k = 4-4h ( Substitute 3k in the 2nd equation.)
2/3k - 3/4h =1
2/(4-4h) -3/4h = 1
2/(2(2-2h)) - 3/4h = 1
1/(2-2h) -3/4h - 1=0
4h/4h(2-2h) -3(2-2h)/4h(2-2h) - 4h(2-2h)/4h(2-2h) =0
(4h- 3(2-2h) - 4h(2-2h))/4h(2-2h) = 0
Numerator should be = 0
4h- 3(2-2h) - 4h(2-2h)=0
Denominator cannot be = 0
4h(2-2h)≠0
Solve equation for numerator=0
4h- 3(2-2h) - 4h(2-2h)=0
4h - 6+6h-8h+8h² =0
8h² +2h -6=0
4h² + h-3 =0
(4h-3)(h+1)=0
4h-3=0, h+1=0
h=3/4 or h=-1
Check which
4h(2-2h)≠0
1) h= 3/4 , 4*3/4(2-2*3/4)=3*(2-6)= -12 ≠0, so we can use h= 3/4
2)h=-1, 4(-1)(2-2*(-1)) =-4*4=-16 ≠0, so we can use h= -1, also.
h=3/4, then 3k = 4-4*3/4 =4 - 3=1 , 3k =1, k=1/3
h=-1, then 3k = 4-4*(-1) =8 , 3k=8, k=8/3
So,
if h=3/4, then k=1/3,
and if h=-1, then k=8/3 .
