Illustrate the compound inequality $6x - 4y < 10$ and $y > 2$
Since the compound inequality is joined by $and$, then we need to find the intersection of the graphs.
To begin, we graph each of the two inequalities $6x - 4y < 10 \text{ and } y > 2$ seperately as shown below
Then, we use heavy shading to identify the intersection of the graphs.
To verify this, we choose a test point on the intersection of the region. Let's say point $(1,3)$. So, we have
$
\begin{equation}
\begin{aligned}
6x - 4y &< 10 && \text{and} & y &> 2 \\
\\
6(1) - 4(3) &< 10 && \text{and} & 3 &> 2\\
\\
6 - 12 &< 10 \\
\\
-6 &< 10
\end{aligned}
\end{equation}
$
We can see that the ordered pairs we choose inside the intersection of the graph switches both inequalities.
Friday, August 4, 2017
Intermediate Algebra, Chapter 3, 3.4, Section 3.4, Problem 28
Subscribe to:
Post Comments (Atom)
Summarize the major research findings of "Toward an experimental ecology of human development."
Based on findings of prior research, the author, Bronfenbrenner proposes that methods for natural observation research have been applied in ...
-
Find the indefinite integral $\displaystyle \int \sec^4 \left( \frac{x}{2} \right) dx$. Illustrate by graphing both the integrand and its an...
-
Determine $\displaystyle \frac{dy}{dx}$ of $y^5 + x^2y^3 = 1 + x^4 y$ by Implicit Differentiation. $\displaystyle \frac{d}{dx}(y^5) + ...
-
Determine the area of the region bounded by the hyperbola $9x^2 - 4y^2 = 36$ and the line $ x= 3$ By using vertical strips, Si...
-
Find the integral $\displaystyle \int^1_0 \frac{1}{\sqrt{16 t^2 + 1}} dt$ If we let $u = 4t$, then $du = 4dt$, so $\displaystyle dt = \frac{...
-
Determine the integral $\displaystyle \int \frac{\sin^3 (\sqrt{x})}{\sqrt{x}} dx$ Let $u = \sqrt{x}$, then $\displaystyle du = \frac{1}{2 \s...
-
Gertrude's comment "The lady protests too much, methinks" in act 3, scene 2, of Shakespeare's Hamlet exposes her own guilt...
-
Given y=cos(2x), y=0 x=0,x=pi/4 so the solid of revolution about x-axis is given as V = pi * int _a ^b [R(x)^2 -r(x)^2] dx here R(x) =cos(2x...
No comments:
Post a Comment