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motikmotik
3 years ago
14

An airport is located at the point (-4, 3). The noise of planes landing and taking off can be heard up to 3 miles away. Write th

e inequality of a circle that represents the situation
Mathematics
1 answer:
Anestetic [448]3 years ago
8 0

Answer:

The correct answer is: ( x + 4) ^ {2} + (y - 3) ^ {2} \leq  9

Step-by-step explanation:

The airport is located at the point (-4 , 3). Thus we can consider the center of the circle to be this particular point.

Since the noise can be heard till 3 miles away, this implies we can consider the radius of the circle to be 3.

We all know the general equation of circle with center at (\alpha , \beta) with radius r is given by:

(x - \alpha ) ^{2} + (y - \beta )^{2} = r^{2}

Here the value of \alpha is (-4) ; value of \beta is 3 ; and value of r is 3.

Now since the noise of landing and taking off of the planes would be within the circle, hence we use less than equal to (\leq) sign instead of equal to sign.

Thus the general equation of noise of the planes can be given by the inequality

( x - (- 4)) ^ {2} + (y - 3) ^ {2} \leq  3^{2}\\= ( x + 4) ^ {2} + (y - 3) ^ {2} \leq  3^{2}

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2 years ago
Nemecek Brothers make a single product on two separate production lines, A and B. Its labor force is equivalent to 1000 hours pe
frez [133]

Answer:

(a) The inequality for the number of items, x, produced by the labor, is given as follows;

250 ≤ x ≤ 600

(b) The inequality for the cost, C is $1,000 ≤ C ≤ $3,000

Step-by-step explanation:

The total time available for production = 1000 hours per week

The time it takes to produce an item on line A = 1 hour

The time it takes to produce an item on line B = 4 hour

Therefore, with both lines working simultaneously, the time it takes to produce 5 items = 4 hours

The number of items produced per the weekly labor = 1000/4 × 5 = 1,250 items

The minimum number of items that can be produced is when only line B is working which produces 1 item per 4 hours, with the weekly number of items = 1000/4 × 1 = 250 items

Therefore, the number of items, x, produced per week with the available labor is given as follows;

250 ≤ x ≤ 1250

Which is revised to 250 ≤ x ≤ 600 as shown in the following answer

(b) The cost of producing a single item on line A = $5

The cost of producing a single item on line B = $4

The total available amount for operating cost = $3,000

Therefore, given that we can have either one item each from lines A and B with a total possible item

When the minimum number of possible items is produced by line B, we have;

Cost = 250 × 4 = $1,000

When the maximum number of items possible, 1,250, is produced, whereby we have 250 items produced from line B and 1,000 items produced from line A, the total cost becomes;

Total cost = 250 × 4 + 1000 × 5 = 6,000

Whereby available weekly outlay = $3000, the maximum that can be produced from line A alone is therefore;

$3,000/$5 = 600 items = The maximum number of items that can be produced

The inequality for the cost, C, becomes;

$1,000 ≤ C ≤ $3,000

The time to produce the maximum 600 items on line A alone is given as follows;

1 hour/item × 600 items = 600 hours

The inequality for the number of items, x, produced by the labor, is therefore, given as follows;

250 ≤ x ≤ 600

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