What is the relationship among proportional relationships lines, rates of change, and slope?

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Answer 1

Answer:

First of all, let's define some concepts. In mathematics, proportional relationships happen when two values always change by the same multiple. That is, when one doubles, the other doubles as well. So, you can always reduce a proportional relationship to the same equation as follows:

$y=kx \n where \ k \ is \ the \ proportionality \ constant$

This constant is also the slope of the straight line.

On the other hand, the average rate of change between any two points $(x_(1),f(x_(1)) \ and \ (x_(2),f(x_(2))$ is the slope of the line through the two points. The line through the two points is called the secant line and its slope is denoted as $m_(sec)$ , so:

$ARC=(f(x_(2))-f(x_(1)))/(x_(2)-x_(1)) \n \n \therefore ARC=(Change \ in \ y)/(Change \ in \ x)=m_(sec)$

Finally, The slope of a nonvertical line is the number of units the line rises (or falls)vertically for each unit of horizontal change from left to right, so:

$Slope=(change \ in \ y)/(change \ in \ x)=(rise)/(run)$

So, we can say that the relationship of these three concepts is that they all talk about the slope of a curve.

Answer 2

Answer:
It introduces the relationship between two variables and is called correlation. Proportionality or variation is state of relationship or correlation between two variables It has two types: direct variation or proportion which states both variables are positively correlation. It is when both the variables increase or decrease together. On the contrary, indirect variation or proportion indicates negative relationship or correlation. Elaborately, the opposite of what happens to direct variation. One increases with the other variables, you got it, decreases. This correlations are important to consider because you can determine and identify how two variables relates with one another. Notice x = y (direct), y=1/x (indirect)

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Solve each equation for x over the set of real numbers.

3x+5/(x-1)(x^4+7)

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$3x+5/(x-1)(x^4+7)$

$3x+5(x^4+7)/x-1$

True or false. You can replace any letter in function notation with any number.

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false is the answer hope i helped

Which of the following could not be true for a function?Domain is {2}, Range is {2}
Domain is {2, 3}, Range is {2}
Domain is {2}, Range is {2, 3}
Domain is {2, 3}, Range is {2, 3}

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C. hope this helps :::)

Answer:

I agree with C

Step-by-step explanation:

GEOMETRY ASSISTANCE NEEDED?

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Secant and tangent theorem: Square of measure of tangent is equal to the product of length of whole secant and length of exterior secant segment, i.e.

$\text{ tangent}^2=\text{whole secant}\cdot \text{exterior secant}$ .

The whole secant with exterior secant 5 in is 5+y in. Using this theorem you can state that

$9^2=(5+y)\cdot 5$ .

Solve this equation:

$81=25+5y,\n 5y=81-25,\n 5y=56,\n y=11.2$ in.

Answer: y=11.2 in.

4 Algebra problems, can I please get some help? (With steps to solve each one!)

Answers

a) a³b²   c a
------- x -----   ÷ -----
   c²d² ab c²d³

First perform multiplication:

(a³b²c / abc²d²) ÷ a / c²d³

In dividing fractions. Get the reciprocal of the 2nd fraction and multiply it to the 1st fraction.

a/c²d³ is the 2nd fraction. Its reciprocal is c²d³/a

So,
a³b²c c²d³ a³b²c³d³
------------ x   --------- = -------------- = abcd *cancel out like terms
abc²d² a a²bc²d²

b) (6x / 4x -16) ÷ (4x / x² -16)
(6x / 4x-16) × (x²-16 / 4x)
6x(x² - 16) / 4x(4x-16)
6x³ - 96 / 16x² - 64x

c) 3x² - 6x x + 3x² *use distributive property of multiplication
--------------- × -------------
  3x + 1 x² -4x + 4

3x² (x+3x²) - 6x (x + 3x²)   ⇒   3x³ + 9x⁴ - 6x² - 18x³
3x (x² - 4x + 4) + 1(x² -4x +4) ⇒ 3x³ -12x² + 12x + x² -4x + 4

  9x⁴ + 3x³ - 18x³ - 6x²   ⇒   9x⁴ - 15x³ - 6x²
3x³ - 12x² + x² + 12x - 4x + 4   ⇒ 3x³ - 11x² + 8x + 4

d) 2x² - 10x + 12 2 + x
---------------------- × -----------
x² - 4   3 - x

2(2x² - 10x + 12) + x (2x² - 10x + 12) ⇒ 4x² - 20x + 24 + 2x³ -10x² + 12x
3(x² - 4) - x(x² -4) 3x² - 12 - x³ + 4x

2x³ + 4x² - 10x² - 20x + 12x + 24 ⇒ 2x³ - 6x² - 8x + 24
-x³ + 3x² + 4x - 12 -x³ + 3x² + 4x - 12

The midpoint of cd is -9,-10, endpoint c has coordinates of -1,9 what is the endpoint of d?

Answers

$A(x_A;\ y_A);\ B(x_B;\ y_B)\n\nThe\ midpoint:M_(AB)\left((x_A+x_B)/(2);\ (y_A+y_B)/(2)\right)\n\n\nM_(CD)(-9;-10);\ C(-1;\ 9);\ D(x;\ y)\n\n(-1+x)/(2)=-9\ \ \ \ |multiply\ both\ sides\ by\ 2\n-1+x=-18\ \ \ \ |add\ 1\ to\ both\ sides\nx=-17\n\n(9+y)/(2)=-10\ \ \ \ \ |multiply\ both\ sides\ by\ 2\n9+y=-20\ \ \ \ |subtract\ 9\ from\ both\ sides\ny=-29\n\nAnswer{\boxed{C(-17;-29)}$

From c to the midpoint the coordinates have changed from (-1;9) to (-9;-10). So the difference is (-9-(-1); -10-(9)) that equals (-8;-19). And you should add (-8;-19) to the coordinates of the middle point (-9;-10). So we get (-9+(-8); -10+(-19)) that is (-17;-29). This is the endpoint of d. Answer: d (-17;-29).

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