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3 years ago

Y₂-y₁=m(x₂-x₁) solve for x

1 answer:

6 0

$y_2-y_1=m(x_2-x_1)\ \ \ \ |divide\ both\ sides\ by\ m\neq0\\\\x_2-x_1=\dfrac{y_2-y_1}{m}\ \ \ \ \ |add\ x_1\ to\ both\ sides\\\\\boxed{x_2=\dfrac{y_2-y_1}{m}}\\\\x_2-x_1=\dfrac{y_2-y_1}{m}\ \ \ \ |subtract\ x_2\ from\ both\ sides\\\\-x_1=\dfrac{y_2-y_1}{m}-x_2\ \ \ \ \ |change\ signs\\\\\boxed{x_1=x_2-\dfrac{y_2-y_1}{m}}$

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Step-by-step explanation:

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David just accepted a job at a new company where he will make an annual salary of $69000. David was told that for each year he s

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Answer:

After 4 years working for the company, he would make $79,000 of salary.

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[te]S(t) = 69000 + 2500t[/tex]

Step-by-step explanation:

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After 4 years working for the company, he would make $79,000 of salary.

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4 years ago

1. Find the area of the region bounded above by y = e^x, bounded below by y = x, and bounded on the sides

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The area of the region bounded above by y= eˣ bounded by y = x, and bounded on the sides; x =0; and x = 1 is given as e¹ - 1.5.

<h3>What is the significance of "Area under the curve"?</h3>

This is the condition in which one process increases a quantity at a certain rate and another process decreases the same quantity at the same rate, and the "area" (actually the integral of the difference between those two rates integrated over a given period of time) is the accumulated effect of those two processes.

<h3>What is the justification for the above answer?</h3>

Area = $\int\limits^1_0 {(e^x-x)} \, dx$

= $\int\limits^1_0 {e^x-x} \, dx - \int\limits^1_0 {(x)} \, dx$

= e¹-(1/2-0); or

Area = e -1.5 Squared Unit

The related Graph is attached accordingly.

Learn more about area bounded by curve:
brainly.com/question/27866606
#SPJ1

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2 years ago