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3 years ago

What is the set fee, or initial value, of this function

Mathematics

1 answer:

Ber [7]3 years ago

6 0

Answer:
The set fee would be $15

Explanation:
The set fee is the starting value. This means that it is the value of the y at x = 0 (y-intercept).

To get the set fee, we would first need to get the equation of the line.

Equation of the linear line has the following general formula:
y = mx + c
where m is the slope and c is the y-intercept

1- getting the slope:
we are given two points which are:
(20,25) and (50,40)
the slope = $\frac{y2-y1}{x2-x1}= \frac{40-25}{50-20} = 0.5$

The equation now is:
y = 0.5x + c

2- getting the value of the y-intercept:
To get the value of the c, we will use any of the given points, substitute in the equation and solve for c.
I will choose the point (20,25)
y = 0.5x + c
25 = 0.5(20) + c
25 = 10 + c
c = 15

The equation of the line representing the scenario is:
y = 0.5x + 15

Now, we know that the value of the c is the y-intercept which is the initial value of the function at x=0.
In our situation, this represents the set fee.

Hope this helps :)

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The coordinates of the vertices of a polygon are (-2, 1). (-3, 3), (-1, 5), (2, 4), and (2, 1). What is the perimeter of the pol

kobusy [5.1K]

Answer:

$15.2\ units$

Step-by-step explanation:

step 1

Plot the vertices of the polygon to better understand the problem

we have

$A(-2, 1). B(-3, 3), C(-1, 5), D(2, 4),E(2, 1)$

using a graphing tool

The polygon is a pentagon (the number of sides is 5)

see the attached figure

The perimeter is equal to

$P=AB+BC+CD+DE+AE$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 2

Find the distance AB

$A(-2, 1). B(-3, 3)$

substitute in the formula

$d=\sqrt{(3-1)^{2}+(-3+2)^{2}}$

$d=\sqrt{(2)^{2}+(-1)^{2}}$

$d_A_B=\sqrt{5}=2.24\ units$

step 3

Find the distance BC

$B(-3, 3), C(-1, 5)$

substitute in the formula

$d=\sqrt{(5-3)^{2}+(-1+3)^{2}}$

$d=\sqrt{(2)^{2}+(2)^{2}}$

$d_B_C=\sqrt{8}=2.83\ units$

step 4

Find the distance CD

$C(-1, 5), D(2, 4)$

substitute in the formula

$d=\sqrt{(4-5)^{2}+(2+1)^{2}}$

$d=\sqrt{(-1)^{2}+(3)^{2}}$

$d_C_D=\sqrt{10}=3.16\ units$

step 5

Find the distance DE

$D(2, 4),E(2, 1)$

substitute in the formula

$d=\sqrt{(1-4)^{2}+(2-2)^{2}}$

$d=\sqrt{(-3)^{2}+(0)^{2}}$

$d_D_E=\sqrt{9}\ units$

$d_D_E=3\ units$

step 6

Find the distance AE

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substitute in the formula

$d=\sqrt{(1-1)^{2}+(2+2)^{2}}$

$d=\sqrt{(0)^{2}+(4)^{2}}$

$d_A_E=\sqrt{16}\ units$

$d_A_E=4\ units$

step 7

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$P=AB+BC+CD+DE+AE$

substitute the values

$P=2.24+2.83+3.16+3+4=15.23\ units$

Round to the nearest tenth of a unit

$P=15.2\ units$

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3 years ago

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Answer:

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Answer:

The graph in the attached figure

Step-by-step explanation:

The complete question is

Which graph best represents the solution to the following system?

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Adds 2y both sides

$5x \leq 10+2y$

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$5x-10 \leq 2y$

Divide by 2 both sides

$2.5x-5 \leq y$

Rewrite

$y \geq 2.5x-5$

The solution of the inequality A is the shaded area above the solid line

The equation of the solid line is $y=2.5x-5$

The slope of the solid line is positive $m=2.5$

The y-intercept of the solid line is (0,-5)

The x-intercept of the solid line is (2,0)

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The solution of the inequality B is the shaded area below the dashed line

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$x=5$ or $8.5$

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