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hoa [83]
3 years ago
15

Write an equation of the line passing through ( 2, 5) and parallel to y=8x+5

Mathematics
1 answer:
solmaris [256]3 years ago
3 0

Answer:

IM DUM

Step-by-step explanation:

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Over a 24-hour period, the tide in a harbor can be modeled by one period of a sinusoidal function. The tide measures 4.35 ft at
netineya [11]

Answer:

  f(x) = 4.35 +3.95·sin(πx/12)

Step-by-step explanation:

For problems of this sort, a sine function is used that is of the form ...

  f(x) = A + Bsin(2πx/P)

where A is the average or middle value of the oscillation, B is the one-sided amplitude, P is the period in the same units as x.

It is rare that a tide function has a period (P) of 24 hours, but we'll use that value since the problem statement requires it. The value of A is the middle value of the oscillation, 4.35 ft in this problem. The value of B is the amplitude, given as 8.3 ft -4.35 ft = 3.95 ft. Putting these values into the form gives ...

  f(x) = 4.35 + 3.95·sin(2πx/24)

The argument of the sine function can be simplified to πx/12, as in the Answer, above.

8 0
3 years ago
6 pints of glaze it takes 7/8 of a long for a bowl now many bowls can you glaze
lubasha [3.4K]

Answer:

ok so umm i dont get what ur looking for?

Step-by-step explanation:

3 0
3 years ago
The figure shown below is a rhombus. Give the angle value (numerical, not the letters.
Rzqust [24]

Answer:

90 deg

Step-by-step explanation:

Recall that for a rhombus, the diagonals always intersect at right angles.

Hence angle PXM is 90 deg

6 0
3 years ago
Read 2 more answers
Last month, a coral reef grew 5000 millimeters taller. How much did it grow in meters? Be sure to include the correct unit in yo
nirvana33 [79]
The correct answer is 5 meters.
4 0
3 years ago
A piece of wire 11 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria
forsale [732]

Answer:

11 meters

Step-by-step explanation:

First, we can say that the square has a side length of x. The perimeter of the square is 4x, and that is how much wire goes into the square. To maximize the area, we should use all the wire possible, so the remaining wire goes into the triangle, or (11-4x).

The area of the square is x², and the area of an equilateral triangle with side length a is (√3/4)a². Next, 11-4x is equal to the perimeter of the triangle, and since it is equilateral, each side has (11-4x)/3 length. Plugging that in for a, we get the area of the equilateral triangle is

(√3/4)((11-4x)/3)²

= (√3/4)(11/3 - 4x/3)²

= (√3/4)(121/9  - 88x/9 + 16x²/9)

= (16√3/36)x² - (88√3/36)x + (121√3/36)

The total area is then

(16√3/36)x² - (88√3/36)x + (121√3/36) + x²

= (16√3/36 + 1)x² -  (88√3/36)x + (121√3/36)

Because the coefficient for x² is positive, the parabola would open up and the derivative of the parabola would be the local minimum. Therefore, to find the maximum area, we need to go to the absolute minimum/maximum points of x (x=0 or x=2.75)

When x=0, each side of the triangle is 11/3 meters long and its area is

(√3/4)a² ≈ 5.82

When x=2.75, each side of the square is 2.75 meters long and its area is

2.75² = 7.5625

Therefore, a maximum is reached when x=2.75, or the wire used for the square is 2.75 * 4 = 11 meters

3 0
3 years ago
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