You have a penny and a dime. You flip them simultaneously. How many possible outcomes are there?
2 answers:
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If the company wants to keep its production costs under $175,000, then 5.6 ≤ x ≤ 24.13 constraint is reasonable for the model given that the function C(x) = −0.74x² + 22x + 75 ,the production cost C, in thousands of dollars for a tech company to manufacture a calculator, where x is the number of calculators produced, in thousands. This can be obtained by using the given graph of the function.
<h3>Which constraint is reasonable for the model:</h3>A constraint is a condition of an optimization problem that should be satisfied the condition.
From the we have the function,
⇒ C(x) = −0.74x² + 22x + 75
the production cost C, in thousands of dollars for a tech company to manufacture a calculator, x is the number of calculators produced, in thousands.
In the graph the dotted line is the line where C(x) is $175,000. Above this line every the value is greater than $175,000.
The points where this line, that is C(x) = y = 175, intersect the graph of the given function C(x) = −0.74x² + 22x + 75 is (5.6, 175) and (24.13, 175).
- This means that above the point (5.6, 175) the graph has the value greater than 175000 and below the point the graph has the value below 175000.
- Similarly, below the point (24.13, 175) the graph has the value greater than $175,000 and above the point the graph has the value below $175,000.
Therefore, x ≥ 5.6 and x ≤ 24.13
⇒ 5.6 ≤ x ≤ 24.13
Hence if the company wants to keep its production costs under $175,000, then 5.6 ≤ x ≤ 24.13 constraint is reasonable for the model given that the function C(x) = −0.74x² + 22x + 75 ,the production cost C, in thousands of dollars for a tech company to manufacture a calculator, where x is the number of calculators produced, in thousands.
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When a number is divisible, that means that why you divide that number by another, you will get a whole number. For example, 4 is divisible by 2 because 4 ÷ 2 = 2. The number 4 is not divisible by 3 because 4 ÷ 3 = 1
(which is not a whole number).
We'll start with 3 -
To find numbers that are divisible by 3, you can use the 3 Divisibility Rule. First, take the number you are given and add together all of the digits. Then, look at that sum and see if it can be evenly divided by 3. If it is, then that number is divisible by 3. For the numbers that you are given in the problem:
23 = 2 + 3 = 5 (not divisible by 3)
28 = 2 + 8 = 10 (not divisible by 3)
33 = 3 + 3 = 9 (9÷3 = 3, so 33 is divisible by 3)
42 = 4 + 2 = 6 (6÷3 = 2, so 42 is divisible by 3)
51 = 5 + 1 = 6 (6÷3 = 2, so 42 is divisible by 3)
59 = 5 + 9 = 14 (not divisible by 3)
86 = 8 + 6 = 14 (not divisible by 3)
96 = 9 + 6 = 15 (15÷3 = 5, so 96 is divisible by 3
Now let's look at numbers divisible by 7. There are rules for divisibility of 7, but they are more for 3-digit numbers. Since your numbers are small enough, we can use basic multiplication facts to check divisibility.
23 (7x3 = 21 and 7x4 =28, 23 is between these and is not divisible by 7)
28 (7x4 =28, so 28 is divisible by 7)
33 (7x4 = 28 and 7x5 = 35, 33 is between these and is not divisible by 7)
42 (7x6 = 42, so 42 is divisible by 7)
51 (7x7 = 49 and 7x8 = 56, 51 is between these and is not divisible by 7)
59 (7x8 = 56 and 7x9 = 63, 59 is between these and is not divisible by 7)
86 (7x12 = 84 and 7x13 = 91, 86 is between these and is not divisible by 7)
96 (7x13 = 91 and 7x14 = 98, 96 is between these and is not divisible by 7)
Now let's look at what we just figured out from the divisibility rules above.
Subway stops divisible by 3: 33, 42, 51, 96
Subway stops divisible by 7: 28, 42
The only number in common in those two lists is 42, so that is the stop that Rajiv should get off of the subway.
We'll start with 3 -
To find numbers that are divisible by 3, you can use the 3 Divisibility Rule. First, take the number you are given and add together all of the digits. Then, look at that sum and see if it can be evenly divided by 3. If it is, then that number is divisible by 3. For the numbers that you are given in the problem:
23 = 2 + 3 = 5 (not divisible by 3)
28 = 2 + 8 = 10 (not divisible by 3)
33 = 3 + 3 = 9 (9÷3 = 3, so 33 is divisible by 3)
42 = 4 + 2 = 6 (6÷3 = 2, so 42 is divisible by 3)
51 = 5 + 1 = 6 (6÷3 = 2, so 42 is divisible by 3)
59 = 5 + 9 = 14 (not divisible by 3)
86 = 8 + 6 = 14 (not divisible by 3)
96 = 9 + 6 = 15 (15÷3 = 5, so 96 is divisible by 3
Now let's look at numbers divisible by 7. There are rules for divisibility of 7, but they are more for 3-digit numbers. Since your numbers are small enough, we can use basic multiplication facts to check divisibility.
23 (7x3 = 21 and 7x4 =28, 23 is between these and is not divisible by 7)
28 (7x4 =28, so 28 is divisible by 7)
33 (7x4 = 28 and 7x5 = 35, 33 is between these and is not divisible by 7)
42 (7x6 = 42, so 42 is divisible by 7)
51 (7x7 = 49 and 7x8 = 56, 51 is between these and is not divisible by 7)
59 (7x8 = 56 and 7x9 = 63, 59 is between these and is not divisible by 7)
86 (7x12 = 84 and 7x13 = 91, 86 is between these and is not divisible by 7)
96 (7x13 = 91 and 7x14 = 98, 96 is between these and is not divisible by 7)
Now let's look at what we just figured out from the divisibility rules above.
Subway stops divisible by 3: 33, 42, 51, 96
Subway stops divisible by 7: 28, 42
The only number in common in those two lists is 42, so that is the stop that Rajiv should get off of the subway.