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Vikki [24]
2 years ago
15

A sandwich store charges $20 to have 3 turkey subs delivered and $26 to have 4 delivered. Is the relationship between number of

turkey subs delivered and the amount charged proportional?
no it's not but how do I know it's not? a simple answer will do.​
Mathematics
1 answer:
BabaBlast [244]2 years ago
8 0

Answer:

Step-by-step explanation:

The cost per sandwich when 3 are delivered is $20/3 or $6.67 each.

The cost per sandwich when 4 are delivered is $26/4 or $6.50 each.

The average cost per sandwich changes with the number delivered, so it is not proportional.  The cost decreases per sandwich as the volume of sandwiches increases.  One can't predict the cost of 6 delivered sandwiches based on the information provided.  If it were proportional, the cost could be calculated on the basis of the slope of the line.  If all sandwiches cost $6.50 each, regardless of the volume ordered, the slope of a line relating number ordered and total cost would have a constant slope of 6.50.  That isn't the case in this problem.  The slope keeps changing.

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The expression \dfrac12bh
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Answer:

2 one.

Step-by-step explanation:

3 0
3 years ago
What id the mean of the set? 10, 15, 14, 8, 18, 11, 12, 12, 10, 10, 17, 16
Elan Coil [88]
The Mean is of the data is 12.75
5 0
3 years ago
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How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?
BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:  

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}

Where n = 8 and r = 0,1....8

When r = 0

\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1

When r = 1

\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8

When r = 2

\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8

When r = 3

\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56

When r = 4

\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70

When r = 5

\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56

When r = 6

\left[\begin{array}{c}8&&6\end{array}\right] = \frac{8!}{(8-6)!6!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} = 28

When r = 7

\left[\begin{array}{c}8&&7\end{array}\right] = \frac{8!}{(8-7)!7!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8

When r = 8

\left[\begin{array}{c}8&&8\end{array}\right] = \frac{8!}{(8-8)!8!} = \frac{8!}{8!0!} = 1

The full sequence is: 1,8,28,56,70,56,28,8,1

And the number of terms is 9

3 0
3 years ago
Select the correct answer.
Sati [7]
The answer would be A because if you combine all the like terms you will get A as a result.

-2x^2+8x-9+4x+7x^2+2
-2x^2+7x^2 combine both of these and it will get you 5x^2
And then you would do 8x+4x which would be 12x and lastly you would -9+2 which would be -7 so the answer is A.

Hope this helps!!
3 0
2 years ago
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Hi can you pls help me pls yes do all
ruslelena [56]
Base x Height / 2 = Area of triangle.
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