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PolarNik [594]
4 years ago
9

The credit remaining on a phone card (in dollars) is a linear function of the total calling time made with the card (in minutes)

. The remaining credit after
44
minutes of calls is
$17.96
, and the remaining credit after
77
minutes of calls is
$12.68
. What is the remaining credit after
100
minutes of calls?
Mathematics
1 answer:
Svetach [21]4 years ago
3 0
What's the slope of the linear equation?
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F(x) = X + 8/x^2+10x+16​
zaharov [31]

The answer is

Xmin= -10

Xmax= 10

Ymin= -10

Ymax= 10

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3 years ago
What would the answers be for 2. And 3.?
Ne4ueva [31]

Answer:

H and B

Step-by-step explanation:

For number 3 its B because pie = irrational

5 0
3 years ago
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What is the slope of the line represented by the equation f(t) =2t-6
VARVARA [1.3K]

Answer:

The slope is 2.

Step-by-step explanation:

The given line has equation: f(t)=2t-6.

This is a linear equation in t.

The slope is the coefficient of the independent variable t.

The coefficient of t in  f(t)=2t-6 is 2.

Therefore the slope is 2.

Alternatively,  f(t)=2t-6 is of the form  f(t)=mt+c, where m=2 is the slope.

8 0
4 years ago
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Help in Number 9 please??
andrey2020 [161]
The vertex is (0, 0)
equation is y = x^2
5 0
3 years ago
The equation of function h is h... PLEASE HELP MATH
Flura [38]

Answer:

Part A: the value of h(4) - m(16) is -4

Part B: The y-intercepts are 4 units apart

Part C: m(x) can not exceed h(x) for any value of x

Step-by-step explanation:

Let us use the table to find the function m(x)

There is a constant difference between each two consecutive values of x and also in y, then the table represents a linear function

The form of the linear function is m(x) = a x + b, where

  • a is the slope of the function
  • b is the y-intercept

The slope = Δm(x)/Δx

∵ At x = 8, m(x) = 2

∵ At x = 10, m(x) = 3

∴ The slope = \frac{3-2}{10-8}=\frac{1}{2}

∴ a = \frac{1}{2}

- Substitute it in the form of the function

∴ m(x) = \frac{1}{2} x + b

- To find b substitute x and m(x) in the function by (8 , 2)

∵ 2 = \frac{1}{2} (8) + b

∴ 2 = 4 + b

- Subtract 4 from both sides

∴ -2 = b

∴ m(x) = \frac{1}{2} x - 2

Now let us answer the questions

Part A:

∵ h(x) = \frac{1}{2} (x - 2)²

∴ h(4) = \frac{1}{2} (4 - 2)²

∴ h(4) = \frac{1}{2} (2)²

∴ h(4) =  \frac{1}{2}(4)

∴ h(4) = 2

∵ m(x) = \frac{1}{2} x - 2

∴ m(16) =  \frac{1}{2} (16) - 2

∴ m(16) = 8 - 2

∴ m(16) = 6

- Find now h(4) - m(16)

∵ h(4) - m(16) = 2 - 6

∴ h(4) - m(16) = -4

Part B:

The y-intercept is the value of h(x) at x = 0

∵ h(x) = \frac{1}{2} (x - 2)²

∵ x = 0

∴ h(0) = \frac{1}{2} (0 - 2)²

∴ h(0) =  \frac{1}{2} (-2)² =  

∴ h(0) = 2

∴ The y-intercept of h(x) is 2

∵ m(x) = \frac{1}{2} x - 2

∵ x = 0

∴ m(0) = \frac{1}{2} (0) - 2 = 0 - 2

∴ m(0) = -2

∴ The y-intercept of m(x) is -2

- Find the distance between y = 2 and y = -2

∴ The difference between the y-intercepts of the graphs = 2 - (-2)

∴ The difference between the y-intercepts of the graphs = 4

∴ The y-intercepts are 4 units apart

Part C:

The minimum/maximum point of a quadratic function f(x) = a(x - h) + k is point (h , k)

Compare this form with the form of h(x)

∵ h = 2 and k = 0

∴ The minimum point of the graph of h(x) is (2 , 0)

∵ k is the minimum value of f(x)

∴ 0 is the minimum value of h(x)

∴ The domain of h(x) is all real numbers

∴ The range of h(x) is h(x) ≥ 2

∵ m(8) = 2

∵ m(14) = 5

∵ h(8) = \frac{1}{2} (8 - 2)² = 18

∵ h(14) = \frac{1}{2} (14 - 2)² = 72

∴ h(x) is always > m(x)

∴ m(x) can not exceed h(x) for any value of x

<em>Look to the attached graph for more understand</em>

The blue graph represents h(x)

The green graph represents m(x)

The blue graph is above the green graph for all values of x, then there is no value of x make m(x) exceeds h(x)

7 0
3 years ago
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