Question
Show the solution set of inequations 4x – 5y ≤ 20 graphically
✓
Answer
| Given inequality | 4x – 5y ≤ 20 |
| Corresponding equality | 4x – 5y = 20 |
| Intersection of line with X-axis | A(5, 0) |
| Intersection of line with Y-axis | B(0, – 4) |
| Origin test | 4(0) – 5(0) ≤ 20 i.e., 0 ≤ 20 which is true |
| Region | Origin side of the line |
The shaded portion represents the graphical solution.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Let $R_0$ denote the set of all non-zero real numbers and let $A=R_0 \times R_0$.
If ' $k$ ' is a binary operation on adefined by, $(a, b)^*(c, d)=(a c, b d)$ for all $(a, b),(c, d) \in A$
Find the identity element in A
→If ' $k$ ' is a binary operation on adefined by, $(a, b)^*(c, d)=(a c, b d)$ for all $(a, b),(c, d) \in A$
Find the identity element in A
Find the vector equation of the line which passes through the point with position vector $4 \bar{i}-\bar{j}+2 \hat{k}$ and is in the direction of $-2 \bar{i}+\bar{j}+\hat{k}$.
→Solve the differential equation $\sec ^2 x \cdot \tan y d x+\sec ^2 y \cdot \tan x \cdot d y=0$
→Evaluate: $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin ^2 x d x$
→For what value of x, the following matrix is singular?
$\begin{vmatrix}5-\text{x}&\text{x}+1\\2&4\end{vmatrix}$
→$\begin{vmatrix}5-\text{x}&\text{x}+1\\2&4\end{vmatrix}$
Write necessary condition for a point x = c to be an extreme point of the function f(x).
Let $\text{A}=\begin{bmatrix}2&4\\3&2\end{bmatrix},\text{B}=\begin{bmatrix}1&3\\-2&5\end{bmatrix}$ and $\text{C}=\begin{bmatrix}-2&5\\3&4\end{bmatrix}.$ Find each of the following:
$3\text{A}-2\text{B}+3\text{C}$
→$3\text{A}-2\text{B}+3\text{C}$
Find the angle between planes $\bar{r} \cdot(\hat{i}+\hat{j}-2 \hat{k})=8$ and $\bar{r} \cdot(-2 \hat{i}+\hat{j}+\hat{k})=3$
→If $A=\left[\begin{array}{cc}2 & -3 \\ 3 & 5\end{array}\right]$ then find $A^{-1}$ by adjoint method.
→Find the angle between the lines $\vec{\text{r}}=\big(2\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}\big)+\lambda\big(3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}\big)$ and $\vec{\text{r}}=7\hat{\text{i}}-6\hat{\text{k}}+\mu\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big).$
→