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sp2606 [1]
3 years ago
9

In New York City at the spring equinox there are 12 hours 8 minutes of daylight. The

Mathematics
1 answer:
slamgirl [31]3 years ago
5 0

Answer:

  independent: day number; dependent: hours of daylight

  d(t) = 12.133 +2.883sin(2π(t-80)/365.25)

  1.79 fewer hours on Feb 10

Step-by-step explanation:

a) The independent variable is the day number of the year (t), and the dependent variable is daylight hours (d).

__

b) The average value of the sinusoidal function for daylight hours is given as 12 hours, 8 minutes, about 12.133 hours. The amplitude of the function is given as 2 hours 53 minutes, about 2.883 hours. Without too much error, we can assume the year length is 365.25 days, so that is the period of the function,

March 21 is day 80 of the year, so that will be the horizontal offset of the function. Putting these values into the form ...

  d(t) = (average value) +(amplitude)sin(2π/(period)·(t -offset days))

  d(t) = 12.133 +2.883sin(2π(t-80)/365.25)

__

c) d(41) = 10.34, so February 10 will have ...

  12.13 -10.34 = 1.79

hours less daylight.

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4 years ago
PLEASE HELP!!! Maria is saving money so she can go to a professional football game. Maria has $25. She saves $5 a week. Graph a
Kruka [31]
Equation would be: Y = 25 + 5x
Where, x = number of weeks
So, when x = 1, Y = 25 + 5(1) = 25+5 = 30
When x = 2, Y = 25 + 5(2) = 25 + 10 = 35
x = 3, Y = 25 + 5(3) = 25 + 15 = 40

So, Mark the coordinates: (0, 25), (1, 30), (2, 35), (3, 40), (4, 45), (5, 50)...
And draw a line...Graph is done!

Hope this helps!
6 0
3 years ago
A piece of wire of length 6363 is​ cut, and the resulting two pieces are formed to make a circle and a square. Where should the
Lerok [7]

Answer:

a.

35.2792 cm from one end (The square)

And 27.7208 cm from the other end (The circle)

b. See (b) explanation below

Step-by-step explanation:

Given

Length of Wire ,= 63cm

Let L be the length of one side of the square

Circumference of a circle = 2πr

Perimeter of a square = 4L

a. To minimise

4L + 2πr = 63 ----- make r the subject of formula

2πr = 63 - 4L

r = (63 - 4L)/2π

r = (31.5 - 2L)/π

Let X = Area of the Square. + Area of the circle

X = L² + πr²

Substitute (31.5 - 2L)/π for r

So,

X² = L² + π((31.5 - 2L)/π)²

X² = L² + π(31.5 - 2L)²/π²

X² = L² + (31.5 - 2L)²/π

X² = L² + (992.25 - 126L + 4L²)/π

X² = L² + 992.25/π - 126L/π +4L²/π ------ Collect Like Terms

X² = 992.25/π - 126L/π + 4L²/π + L²

X² = 992.25/π - 126L/π (4/π + 1)L² ---- Arrange in descending order of power

X² = (4/π + 1)L² - 126L/π + 992.25/π

The coefficient of L² is positive so this represents a parabola that opens upward, so its vertex will be at a minimum

To find the x-cordinate of the vertex, we use the vertex formula

i.e

L = -b/2a

L = - (-126/π) / (2 * (4/π + 1)

L = (126/π) / ( 2 * (4 + π)/π)

L = (126/π) /( (8 + 2π)/π)

L = 126/π * π/(8 + 2π)

L = (126)/(8 + 2π)

L = 63/(4 + π)

So, for the minimum area, the side of a square will be 63/(4 + π)

= 8.8198 cm ---- Approximated

We will need to cut the wire at 4 times the side of the square. (i.e. the four sides of the square)

I.e.

4 * (63/(4 + π)) cm

Or

35.2792 cm from one end.

Subtract this result from 63, we'll get the other end.

i.e. 63 - 35.2792

= 27.7208 cm from the other end

b. To maximize

Now for the maximum area.

The problem is only defined for 0 ≤ L ≤ 63/4 which gives

0 ≤ L ≤ 15.75

When L=0, the square shrinks to 0 and the whole 63 cm wire is made into a circle.

Similarly, when L =15.75 cm, the whole 63 cm wire is made into a square, the circle shrinks to 0.

Since the parabola opens upward, the maximum value is at one endpoint of the interval, either when

L=0 or when L = 15.75.

It is well known that if a piece of wire is bent into a circle or a square, the circle will have more area, so we will assume that the maximum area would be when we "cut" the wire 0, or no, centimeters from the

end, and bend the whole wire into a circle. That is we don't cut the wire at

all.

7 0
3 years ago
Three angles of a quadrilateral are 75,120, 90,then the fourth angle..​
Roman55 [17]

Answer:

75

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Geometry</u>

  • Sum of Angles in a Quadrilateral: 360

Step-by-step explanation:

<u>Step 1: Set up</u>

75 + 120 + 90 + ∠4 = 360

<u>Step 2: Solve</u>

  1. Add:                                                                                                                   285 + ∠4 = 360
  2. [Subtraction Property of Equality] Subtract 285 on both sides:                   ∠4 = 75
5 0
3 years ago
How many variables are displayed in a scatterplot?
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8 0
3 years ago
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