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3 years ago

I NEED REAL HELP IF NOT REAL GET BANNED

Mathematics

1 answer:

kari74 [83]3 years ago

6 0

Answer:

Every minute, the hour hand moves 1/2 of a degree, and the minute hand moves 6 degrees. So in 48 minutes, the hour hand moves 24 degrees and the minute hand moves 288 degrees. 288-180-24=84. So, 84 degrees.

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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

8 0

3 years ago

37. Verify Green's theorem in the plane for f (3x2- 8y2) dx + (4y - 6xy) dy, where C is the boundary of the

Nastasia [14]

I'll only look at (37) here, since

• (38) was addressed in 24438105

• (39) was addressed in 24434477

• (40) and (41) were both addressed in 24434541

In both parts, we're considering the line integral

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy$

and I assume C has a positive orientation in both cases

(a) It looks like the region has the curves y = x and y = x ² as its boundary***, so that the interior of C is the set D given by

$D = \left\{(x,y) \mid 0\le x\le1 \text{ and }x^2\le y\le x\right\}$

• Compute the line integral directly by splitting up C into two component curves,

C₁ : x = t and y = t ² with 0 ≤ t ≤ 1

C₂ : x = 1 - t and y = 1 - t with 0 ≤ t ≤ 1

Then

$\displaystyle \int_C = \int_{C_1} + \int_{C_2} \\\\ = \int_0^1 \left((3t^2-8t^4)+(4t^2-6t^3)(2t))\right)\,\mathrm dt \\+ \int_0^1 \left((-5(1-t)^2)(-1)+(4(1-t)-6(1-t)^2)(-1)\right)\,\mathrm dt \\\\ = \int_0^1 (7-18t+14t^2+8t^3-20t^4)\,\mathrm dt = \boxed{\frac23}$

*** Obviously this interpretation is incorrect if the solution is supposed to be 3/2, so make the appropriate adjustment when you work this out for yourself.

• Compute the same integral using Green's theorem:

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy = \iint_D \frac{\partial(4y-6xy)}{\partial x} - \frac{\partial(3x^2-8y^2)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \int_0^1\int_{x^2}^x 10y\,\mathrm dy\,\mathrm dx = \boxed{\frac23}$

(b) C is the boundary of the region

$D = \left\{(x,y) \mid 0\le x\le 1\text{ and }0\le y\le1-x\right\}$

• Compute the line integral directly, splitting up C into 3 components,

C₁ : x = t and y = 0 with 0 ≤ t ≤ 1

C₂ : x = 1 - t and y = t with 0 ≤ t ≤ 1

C₃ : x = 0 and y = 1 - t with 0 ≤ t ≤ 1

Then

$\displaystyle \int_C = \int_{C_1} + \int_{C_2} + \int_{C_3} \\\\ = \int_0^1 3t^2\,\mathrm dt + \int_0^1 (11t^2+4t-3)\,\mathrm dt + \int_0^1(4t-4)\,\mathrm dt \\\\ = \int_0^1 (14t^2+8t-7)\,\mathrm dt = \boxed{\frac53}$