We start with
$1,187.92
With the interest and the late fee charge
$1,187.92 (1 + 0.1225/12) + 30 = $1,230.05
With the interest minus the payment of $125
$1,230.05 (1 + 0.1225/12) - $125 = $1,117.60
The new principal after the latest payment is $1,117.60.
Answer:
First, we write the augmented matrix.
⎡
⎢
⎣
1
−
1
1
2
3
−
1
3
−
2
−
9
|
8
−
2
9
⎤
⎥
⎦
Next, we perform row operations to obtain row-echelon form.
−
2
R
1
+
R
2
=
R
2
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
3
−
2
−
9
|
8
−
18
9
⎤
⎥
⎦
−
3
R
1
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
0
1
−
12
|
8
−
18
−
15
⎤
⎥
⎦
The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R
2
and \displaystyle {R}_{3}R
3
.
Interchange
R
2
and
R
3
→
⎡
⎢
⎣
1
−
1
1
8
0
1
−
12
−
15
0
5
−
3
−
18
⎤
⎥
⎦
Then
−
5
R
2
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
57
|
8
−
15
57
⎤
⎥
⎦
−
1
57
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
1
|
8
−
15
1
⎤
⎥
⎦
The last matrix represents the equivalent system.
x
−
y
+
z
=
8
y
−
12
z
=
−
15
z
=
1
Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).First, we write the augmented matrix.
⎡
⎢
⎣
1
−
1
1
2
3
−
1
3
−
2
−
9
|
8
−
2
9
⎤
⎥
⎦
Next, we perform row operations to obtain row-echelon form.
−
2
R
1
+
R
2
=
R
2
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
3
−
2
−
9
|
8
−
18
9
⎤
⎥
⎦
−
3
R
1
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
0
1
−
12
|
8
−
18
−
15
⎤
⎥
⎦
The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R
2
and \displaystyle {R}_{3}R
3
.
Interchange
R
2
and
R
3
→
⎡
⎢
⎣
1
−
1
1
8
0
1
−
12
−
15
0
5
−
3
−
18
⎤
⎥
⎦
Then
−
5
R
2
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
57
|
8
−
15
57
⎤
⎥
⎦
−
1
57
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
1
|
8
−
15
1
⎤
⎥
⎦
The last matrix represents the equivalent system.
x
−
y
+
z
=
8
y
−
12
z
=
−
15
z=1
Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).
Total distance 5 km; at 5km / 0.65 h =
Second part distance: x; at 6 km/h, during t2
First part distance: 5 - x; at 8.75 km/h, during t1
V = d/t ⇒ t = d/V
t2 = x/6
t1=[5-x]/8.75
t2 + t1 = 0.65
x/6 + [5-x]/8.75 = 0.65
x/6 + 5/8.75 - x/8.75 = 13/20
x/6 - x/8.75 = 13/20 - 5/8.75
x/6 - 4x/35 =13/20 - 20/35
35x - 24x = (35*6)(35*13 - 20*20)/(20*35)
11 x = 16.5
x = 16.5/11 = 1.5 km
Answer:
say what
Step-by-step explanation:
If your functions are
f(x) = 16·x
g(x) = 16·(1/2x)
The output values of g(x) are one-half the output values of f(x) for the same value of x.
If your functions are
f(x) = 16^x
g(x) = 16^(1/2x)
The output vaules of g(x) are the square root of the output values of f(x) for the same value of x.
_____
You seem to have left out the relevant operator symbol, so we don't really know your intent. Copy/paste operations will do that. Editing is sometimes needed.
