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yuradex [85]
4 years ago
14

The table compares the daily average temperature in a park (in degrees Celsius) and the number of people who visited the park th

at day. Can the number of people who visited the park be represented as a function of the daily average temperature?

Mathematics
2 answers:
lions [1.4K]4 years ago
8 0

Answer:

yes

Step-by-step explanation:

Simora [160]4 years ago
7 0

Answer:

YES, the number of people who visited the park can be represented as a function of the daily average temperature.

Step-by-step explanation:

From the table of values given which defines the number of visitors to the park as a function of daily temperature, the temperature is the domain values (independent variable), while the number of visitors is the range values (dependent variable).

For any relation to be considered a function, as a principle, any given domain value must have exactly one range value. In other words, there must not be two different range values assign to one domain value.

This means, for the relation between temperature and number of visitors to be considered a function, it must not have a domain value that is mapped to more than one range value.

Examining the table given, if we map each domain value to the range value, we will find out that there are only 1 exact range value for a domain value.

<em>Therefore, we can say: YES, the number of people who visited the park can be represented as a function of the daily average temperature.</em>

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Match each set of points with the quadratic function whose graph passes through those points.
AnnZ [28]

Answer:

f(x) = x² - 2x - 15 passes through (-3 , 0) , (0 , -15) , (5 , 0)

f(x) = -x² - 2x + 15 passes through (-2 , 15) , (-1 , 16) , (0 , 15)

f(x) = -x² + 2x - 15 passes through (0 , -15) , (1 , -14) , (2 , -15)

Step-by-step explanation:

* Lets explain how to solve this question

- To find the points whose graph passes through them substitute the

  x-coordinate in the function if the answer is the same with the

  y-coordinate of the point then the graph passes through this point

  lets do that

- Check the first set of points with the first function

# Pint (0 , -15)

∵ f(x) = x² - 2x - 15

∴ f(0) = (0)² - 2(0) - 15 = -15 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (0 , -15)

# Pint (1 , -14)

∵ f(x) = x² - 2x - 15

∴ f(0) = (1)² - 2(1) - 15 = -16 ⇒ not same value of y-coordinate

∴ The graph of the function does not pass through point (1 , -14)

∴ The graph does not pass through this set of points

- Check the second set of points with the first function

# Pint (-2 , 15)

∵ f(x) = x² - 2x - 15

∴ f(0) = (-2)² - 2(-2) - 15 = 4 + 4 - 15 -7 ⇒ not same value of y-coordinate

∴ The graph of the function does not pass through point (-2 , 15)

∴ The graph does not pass through this set of points

- Check the third set of points with the first function

# Pint (-3 , 0)

∵ f(x) = x² + 2x - 15

∴ f(0) = (-3)² - 2(-3) - 15 = 9 + 6  -15 = 0 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (-3 , 0)

# Pint (0 , -15)

∵ f(x) = x² - 2x - 15

∴ f(0) = (0)² - 2(0) - 15 = -15 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (0 , -15)

# Pint (5 , 0)

∵ f(x) = x² + 2x - 15

∴ f(0) = (5)² - 2(5) - 15 = 25 - 10  -15 = 0 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (5 , 0)

∴ The graph passes through this set of points

* f(x) = x² - 2x - 15 passes through (-3 , 0) , (0 , -15) , (5 , 0)

- Check the first set of points with the second function

# Pint (0 , -15)

∵ f(x) = -x² - 2x + 15

∴ f(0) = -(0)² - 2(0) + 15 = 15 ⇒ not same value of y-coordinate

∴ The graph of the function does not passes through point (0 , -15)

∴ The graph does not pass through this set of points

- Check the second set of points with the second function

# Pint (-2 , 15)

∵ f(x) = -x² - 2x + 15

∴ f(0) = -(-2)² - 2(-2) + 15 = -4 + 4 + 15 = 15 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (-2 , 15)

# Pint (-1 , 16)

∵ f(x) = -x² - 2x + 15

∴ f(0) = -(-1)² - 2(-1) + 15 = -1 + 2 + 15 = 16 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (-1 , 16)

# Pint (0 , 15)

∵ f(x) = -x² - 2x + 15

∴ f(0) = -(0)² - 2(0) + 15 = 15 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (0 , 15)

∴ The graph passes through this set of points

* f(x) = -x² - 2x + 15 passes through (-2 , 15) , (-1 , 16) , (0 , 15)

- Now we have the first set of points and the third function

∴ The graph passes through this set of points

∴ f(x) = -x² + 2x - 15 passes through (0 , -15) , (1 , -14) , (2 , -15)

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4 years ago
M 30 degrees <br> 9x-10<br> 10x+10
ivann1987 [24]

Answer:

m∠W = 80°

Step-by-step explanation:

To find m∠W, remember that:

One exterior angle of a triangle is equal to the sum of the other two interior angles.

Now apply this information.

Notice that point U has an exterior angle. It will be equivalent to the sum of m∠V and m∠W, or in algebraic from: U = m∠V + m∠W

Substitute the information we know.

U = m∠V + m∠W

10x + 10 = 30 + 9x - 10          Simplify the right side's like terms.

10x + 10 = 20 + 9x          Isolate "x"

10x - 9x + 10 = 20 + 9x - 9x          Subtract 9x from both sides

10x - 9x + 10 = 20          9x cancels out on the right side

x + 10 = 20          Simplified left side, 10x - 9x = x

x + 10 - 10 = 20 - 10          Subtract 10 from both sides

x = 20 - 10          "x" is isolated. Simplify right side.

x = 10          Solved for "x"

To find m∠W, substitute the value of "x" into the equation of its angle.

m∠W = 9x - 10

m∠W = 9(10) - 10          Multiply 9 and 10, then subtract 10 from the product.

m∠W = 80

Write with the units, degrees symbol °

Therefore the measure of ∠W is 80°.

3 0
4 years ago
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45-16x is the answer
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Where r is the radius of the cylinder and h is the height of the cylinder.
AnnyKZ [126]
<h3>Given:-</h3>
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\\  \\

<h3>To find:</h3>
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\\  \\

we know:-

\bigstar \boxed{ \rm Surface \: Area \: of \: Cylinder=2\pi rh+2\pi {r}^{2} }

\\  \\

So:-

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\\

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\\

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\\

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\\

⇢SA of cylinder = 2 × π × 112

\\

⇢SA of cylinder = 2 × 112 × π

\\

⇢SA of cylinder = 224π inches ²

\\

.°. <u>Surface</u><u> </u><u>Area</u><u> </u><u>of</u><u> </u><u>cylinder</u><u> </u><u>is</u><u> </u><u>224π inches</u><u>²</u><u>.</u>

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