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3 years ago

Which correctly describes how the graph of the inequality 6y − 3x > 9 is shaded? -Above the solid line

Mathematics

2 answers:

baherus [9]3 years ago

4 0

The statement which correctly describes the shaded region for the inequality is $\fbox{\begin\\\ Above the dashed line\\\end{minispace}}$

Further explanation:

In the question it is given that the inequality is $6y-3x>9$ .

The equation corresponding to the inequality $6y-3x>9$ is $6y-3x=9$ .

The equation $6y-3x=9$ represents a line and the inequality $6y-3x>9$ represents the region which lies either above or below the line $6y-3x=9$ .

Transform the equation $6y-3x=9$ in its slope intercept form as $y=mx+c$ , where $m$ represents the slope of the line and $c$ represents the $y$ -intercept.

$y$ -intercept is the point at which the line intersects the $y$ -axis.

In order to convert the equation $6y-3x=9$ in its slope intercept form add $3x$ to equation $6y-3x=9$ .

$6y-3x+3x=9+3x$

$6y=9+3x$

Now, divide the above equation by $6$ .

$\fbox{\begin\\\math{y=\dfrac{x}{2}+\dfrac{1}{2}}\\\end{minispace}}$

Compare the above final equation with the general form of the slope intercept form $\fbox{\begin\\\math{y=mx+c}\\\end{minispace}}$ .

It is observed that the value of $m$ is $\dfrac{1}{2}$ and the value of $c$ is $\dfrac{3}{2}$ .

This implies that the $y$ -intercept of the line is $\dfrac{3}{2}$ so, it can be said that the line passes through the point $\fbox{\begin\\\ \left(0,\dfrac{3}{2}\right)\\\end{minispace}}$ .

To draw a line we require at least two points through which the line passes so, in order to obtain the other point substitute $0$ for $y$ in $6y=9+3x$ .

$0=9+3x$

$3x=-9$

$\fbox{\begin\\\math{x=-3}\\\end{minispace}}$

This implies that the line passes through the point $\fbox{\begin\\\ (-3,0)\\\end{minispace}}$ .

Now plot the points $(-3,0)$ and $\left(0,\dfrac{3}{2}\right)$ in the Cartesian plane and join the points to obtain the graph of the line $6y-3x=9$ .

Figure $1$ shows the graph of the equation $6y-3x=9$ .

Now to obtain the region of the inequality $6y-3x>9$ consider any point which lies below the line $6y-3x=9$ .

Consider $(0,0)$ to check if it satisfies the inequality $6y-3x>9$ .

Substitute $x=0$ and $y=0$ in $6y-3x>9$ .

$(6\times0)-(3\times0)>9$

$0>9$

The above result obtain is not true as $0$ is not greater than $9$ so, the point $(0,0)$ does not satisfies the inequality $6y-3x>9$ .

Now consider $(-2,2)$ to check if it satisfies the inequality $6y-3x>9$ .

Substitute $x=-2$ and $y=2$ in the inequality $6y-3x>9$ .

$(6\times2)-(3\times(-2))>9$

$12+6>9$

$18>9$

The result obtain is true as $18$ is greater than $9$ so, the point $(-2,2)$ satisfies the inequality $6y-3x>9$ .

The point $(-2,2)$ lies above the line so, the region for the inequality $6y-3x>9$ is the region above the line $6y-3x=9$ .

The region the for the inequality $6y-3x>9$ does not include the points on the line $6y-3x=9$ because in the given inequality the inequality sign used is $>$ .

Figure $2$ shows the region for the inequality $\fbox{\begin\\\math{6y-3x>9}\\\end{minispace}}$ .

Therefore, the statement which correctly describes the shaded region for the inequality is $\fbox{\begin\\\ Above the dashed line\\\end{minispace}}$

Learn more:

A problem to determine the range of a function brainly.com/question/3852778
A problem to determine the vertex of a curve brainly.com/question/1286775
A problem to convert degree into radians brainly.com/question/3161884

Answer details:

Grade: High school

Subject: Mathematics

Chapter: Linear inequality

Keywords: Linear, equality, inequality, linear inequality, region, shaded region, common region, above the dashed line, graph, graph of inequality, slope, intercepts, y-intercept, 6y-3x=9, 6y-3x>9, slope intercept form.

Nesterboy [21]3 years ago

3 0

Answer:

Above the dashed line

Step-by-step explanation:

We are given that an inequality equation

$6y-3x > 9$

We have to find the statement which correctly describes the graph of inequality .

First we change inequality equation into equality

$6y-3x=9$

$2y-x=3$

Substitute x=0 then we get

$y=\frac{3}{2}$

Substitute y=0 then we get

$x=-3$

Therefore, the line passing through the point (-3,0) and (0,3/2).

Substitute x=0 and y=0 in inequality equation then we get

$0-0=0\ngtr 9$

Therefore, the given equation is false.

When equation is false then the shaded region above the dashed line.

Answer:Above the dashed line

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2 years ago

Hello again! This is another Calculus question to be explained.

podryga [215]

Answer:

See explanation.

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Functions

Function Notation

Exponential Property [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Property [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the following and are trying to find the second derivative at x = 2:

$\displaystyle f(2) = 2$

$\displaystyle \frac{dy}{dx} = 6\sqrt{x^2 + 3y^2}$

We can differentiate the 1st derivative to obtain the 2nd derivative. Let's start by rewriting the 1st derivative:

$\displaystyle \frac{dy}{dx} = 6(x^2 + 3y^2)^\big{\frac{1}{2}}$

When we differentiate this, we must follow the Chain Rule: $\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \Big[ 6(x^2 + 3y^2)^\big{\frac{1}{2}} \Big] \cdot \frac{d}{dx} \Big[ (x^2 + 3y^2) \Big]$

Use the Basic Power Rule:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} (2x + 6yy')$

We know that y' is the notation for the 1st derivative. Substitute in the 1st derivative equation:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 6y(6\sqrt{x^2 + 3y^2}) \big]$

Simplifying it, we have:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]$

We can rewrite the 2nd derivative using exponential rules:

$\displaystyle \frac{d^2y}{dx^2} = \frac{3\big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]}{\sqrt{x^2 + 3y^2}}$

To evaluate the 2nd derivative at x = 2, simply substitute in x = 2 and the value f(2) = 2 into it:

$\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = \frac{3\big[ 2(2) + 36(2)\sqrt{2^2 + 3(2)^2} \big]}{\sqrt{2^2 + 3(2)^2}}$

When we evaluate this using order of operations, we should obtain our answer:

$\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = 219$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

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2 years ago