Answer:
Dominic can ride a mile every 4 minutes so if total miles is represented by m, then the solution for how many minutes he's riding for a specific amount of time would be 4m (4 x m). For example, if you want to find out how many minutes it would take to ride 1 mile, substitute m for 1. It would be 4 x 1 = 4 so it takes a total of 4 minutes. If you want to find out how many minutes it took to ride 6 miles, substitute m for 6. It would be 4 x 6 = 24 so it would take a total of 24 minutes to ride 6 miles. If you want to find out how many minutes it took to ride 12 miles, substitute m for 12. It would be 4 x 12 = 48 so it would take a total of 48 minutes to ride 12 miles. I think you gave a lack of information so your question is incomplete but I hope this is applicable and helps anyways!
The cross products of the proportion are equal.
According to the question, Sonny's equation can be written as follows.
16/x = 48/15
Now, the variable x is replaced by the constant 5. So the equation can now be written as follows.
16/5 = 48/15
When two fractions are equal to one another as shown above, the constants are cross multiplied with one another in order to form an equation.
16*15= 48*5
The values of both the products are equal which is 240.
Therefore, because the cross products of the proportion are equal.
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Answer:
Number of week day hours purchased is 5
Step-by-step explanation:
Total number of Parking hours purchased = 26 parking hours
Parking cost on weekdays = $2 per hour
Parking costs on weekends = $10 per hour
Total amount spent on parking = $220.
To Find:
Number of week days purchased = ?
Solution:
Let
The number of week days purchased be x
The number of weekends purchased be y
We know that the total hours purchased is 26
So,
x+y = 26
y = 26-x------------------------------------------------------(1)
Now the total cost is 220
(Total number of weekdays X cost per weekday ) +(Total number of weekends + cost per weekend) =220
Substituting the values
=>
= 220
=>
=>2x + 260 -10x =220
=>260 -8x = 220
=>260 -220 =8x
=>40 = 8x
=>x=
x= 5-------------------------------------------(2)
Now substituting (2) in (1) we get
y= 26-5
y= 21
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).
