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I am Lyosha [343]
3 years ago
7

A hardware store orders at most $2,500 worth of concrete and sand. The store needs to make a profit of at least $2,750 on the or

der. The possible combinations of concrete and sand for this order are given by this system of inequalities, where c = pounds of concrete and s = pounds of sand: 6c + 2s ≤ 2,500 4.50c + 3s ≥ 2,750 Which graph's shaded region represents the possible combinations of concrete and sand for this order?

Mathematics
1 answer:
yaroslaw [1]3 years ago
5 0

Answer: The below graph's shaded region represents the possible combinations of concrete and sand for this order

Step-by-step explanation:

Here, the system of inequality that represents the given situation is,

6c + 2s ≤ 2,500 --------(1)

4.50c + 3s ≥ 2,750 -------(2)

Where c = pounds of concrete,

s = pounds of sand

Now, The related equation of the above inequalities are,

6c + 2s = 2500

4.50c + 3s = 2750

By solving the above equations,

We get, c =222.\bar{2}, s = 583.\bar{3}

Thus, the intersection point of the inequalities = (222.\bar{2}, 583.\bar{3})

Now, At (0,0) ,

inequality (1), 6×0 + 2×0≤ 2500 ⇒ 0 ≤ 2500 ( true)

Hence, the shaded region of inequality contains the origin.

Similarly, At (0,0), inequality (2) is false,

⇒ The shaded region of inequality (2) does not contain the origin.

Thus, By the above explanation we can plot the shaded region of the system of the given inequalities.

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Answer:

For 1,000 bricks, the amount charged will be the same

Step-by-step explanation:

Let the number of bricks for which they charge the same amount be $x

Thus;

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5 0
3 years ago
At a farm, there are two huge pits for storing hay. 90 tons of hay is stored in the first pit, 75 tons in the second pit. Then,
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Answer:

63 tons

Step-by-step explanation:

The problem statement asks for the tons of hay removed from the first pit. It is convenient to let a variable (x) represent that amount. This is said to be 3 times the amount removed from the second pit, so that amount must be x/3.

The amount remaining in the first pit is 90-x.

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Since the first pit remaining amount is half the second pit remaining amount, we can write the equation ...

... 90 -x = (1/2)(75 -x/3)

... 180 -2x = 75 -x/3 . . . . multiply by 2

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... 63 = x . . . . . . . . . . . . . divide by 5

63 tons of hay were taken from the first pit.

_____

<em>Check</em>

After removing 63 tons from the first pit, there are 27 tons remaining. After removing 63/3 = 21 tons from the second pit, there are 54 tons remaining. 27 is half of 54, so the answer checks OK.

6 0
3 years ago
50 points for this helpppp plzz
maxonik [38]

Answer:

ok

Step-by-step explanation:

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