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goblinko [34]
3 years ago
6

A relation is plotted as a linear function on the coordinate plane starting at point C (0,−1)(0,−1) and ending at point D (2,−11

)(2,−11) .
What is the rate of change for the linear function and what is its initial value?

the rate of change is ______ and the intivial value is______
Mathematics
1 answer:
Alika [10]3 years ago
4 0
To find the rate of change, or slope, pick two points. The slope, is change in y coordinates over change in x coordinates.
slope =  \frac{y2-y1}{x2-x1}.

In this case, the slope is 
\frac{-11+1}{2-0} =  \frac{-10}{2} = -5.
Remember that subtracting a negative number equals adding the number without the negative sign.
a-(-b) = a+b

Now for the y-intercept. The y-intercept is where the graph intersects the y-axis, where x = 0.
You have the coordinate (0, -1), where x = 0. So the y-intercept is -1.

Putting these values into the slope-intercept form y = mx+b, the equation is
y = -5x - 1
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2 years ago
A rectangle has width that is 6 meters less than the length. The area of the rectangle is 280 square meters. Find the dimensions
salantis [7]

Dimensions are length 20 meter and width 14 meter

<em><u>Solution:</u></em>

Let "a" be the length of rectangle

Let "b" be the width of rectangle

Given that,

<em><u>A rectangle has width that is 6 meters less than the length</u></em>

Width = length - 6

b = a - 6

The area of the rectangle is 280 square meters

<em><u>The area of the rectangle is given by formula:</u></em>

Area = length \times width

<em><u>Substituting the values we get,</u></em>

Area = a \times (a-6)\\\\280 = a^2-6a\\\\a^2-6a -280=0

<em><u>Solve the above equation by quadratic formula</u></em>

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\\quad x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=1,\:b=-6,\:c=-280:\quad a_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:1\left(-280\right)}}{2\cdot \:1}

a =\frac{6 \pm \sqrt{36+1120}}{2}\\\\a = \frac{6 \pm \sqrt{1156}}{2}\\\\a = \frac{6 \pm 34}{2}\\\\Thus\ we\ have\ two\ solutions\\\\a = \frac{6+34}{2} \text{ or } a = \frac{6-34}{2}\\\\a = 20 \text{ or } a = -14

Since, length cannot be negative, ignore a = -14

<em><u>Thus solution of length is a = 20</u></em>

Therefore,

width = length - 6

width = 20 - 6 = 14

Thus dimensions are length 20 meter and width 14 meter

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Answer:

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Step-by-step explanation:

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