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SVEN [57.7K]
2 years ago
6

The graph below shows the average Valentine’s Day spending between 2003 and 2012

Mathematics
1 answer:
Lapatulllka [165]2 years ago
7 0
The average rate of change of a graph between two intervals is given by the difference in value of the values on the graph of the two interval divided by the difference between the two intervals.

Part A.

From the graph the average Valentine's day spending in 2005 is 98 while the average Valentine's day spending in 2007 is 120.

The average rate of change in spending between 2005 and 2007 is given by

\frac{120-98}{2007-2005} = \frac{22}{2} =\$11/year



Part B

From the graph the average Valentine's day spending in 2004 is 100 while the average Valentine's day spending in 2010 is 103.

The average rate of change in spending between 2004 and 2010 is given by

\frac{103-100}{2010-2004} = \frac{3}{6} =\$0.5/year



Part C:

From the graph the average Valentine's day spending in 2009 is 102 while the average Valentine's day spending in 2010 is 103.

The average rate of change in spending between 2009 and 2010 is given by

\frac{103-102}{2010-2009} = \$1/year

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Step-by-step explanation:

Consider the provided information.

Angle A C B is 90 degrees and angle A B C is 35 degrees.

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Use the trigonometric function: \cos\theta=\frac{Adjacent}{Hypotenuse}

\cos35^0=\frac{5}{c}

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The total area of the polygon is 176 square feet. Find the value of x.
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Answer:

x = 6 ft

Step-by-step explanation:

Total area of the given polygon = Area of triangle 1 + Area of rectangle 2 + Area of triangle 3

Area of triangle 1 = Area of triangle 3 = \frac{1}{2}(\text{Base})(\text{Height})

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Total area of the given polygon = 4x + 128 + 4x

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The segments shown below could form a triangle.
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By the Triangle Inequality Theorem, the sum of two sides should be greater than the length of the third side, while the difference of these two sides should be less than the length of this third side. Normally you would take the absolute value of the difference of these two side as you wouldn't know which is greater than the other!

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