LINEAR INEQUALITIES:-
SOLVE FOR X:-
Answers
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|x - 1| = (x - 1) when x ≥ 1
= - (x - ) when x < 1
|x - 2| = (x - 2) when x ≥ 2
= - (x - 2) when x < 2
|x - 3| = (x -3) when x ≥ 3
= - (x - 3) when x < 3
______________________
for x < 1
|x - 1| + |x - 2| + |x - 3| ≥ 6
= - (x - 1) - (x - 2) - (x - 3) ≥ 6
= - 3x + 6 ≥ 6
add ( - 6) on both sides
= > - 3x ≥ 0
multiply by ( - 1) both sides (equality will be reversed) = > 3x ≤ 0 divide both sides by 3
= > x ≤ 0
- - - - - - - - - - - - - - - - - -
for 1 ≤ x < 2
|x - 1| + |x - 2| + |x - 3| ≥ 6
= (x - 1) - (x - 2) - (x - 3) ≥ 6
= - x + 4 ≥ 6
subtract 4 on both sides
= > - x ≥ 2
multiply by ( - 1) both sides (equality will change)
= x ≤ - 2
- - - - - - - - - - - - - - - - - - - - -
for 2 ≤ x < 3
|x - 1| + |x - 2| + |x - 3| ≥ 6
= (x - 1) + (x - 2) - (x - 3) ≥ 6
= x ≥ 6
- - - - - - - - - - - - - - - - - - - - - -
for x ≥ 3
|x - 1| + |x - 2| + |x - 3| > 6
= (x - 1) + (x - 2) + (x - 3) > 6
= 3x - 6 ≥ 6
add 6 on both sides
= > 3x ≥ 12 divide by 3 on both sides
= > x ≥ 4
- - - - - - - - - - - - - - - - - - - - - -
on combining the ranges we get
x ≤ 0 or x ≥ 4
Step-by-step explanation:
\begin{gathered}\dashrightarrow\sf\:\:(Diagonal)^2=(Length)^2+(Breadth)^2\\\\\\\dashrightarrow\sf\:\:(BD)^2=(BC)^2+(CD)^2\\\\\\\dashrightarrow\sf\:\:(BD)^2=(24\:cm)^2+(7\:cm)^2\\\\\\\dashrightarrow\sf\:\:(BD)^2=576\:cm^2+49\:cm^2\\\\\\\dashrightarrow\sf\:\:(BD)^2=625\:cm^2\\\\\\\dashrightarrow\sf\:\:BD=\sqrt{625\:cm^2}\\\\\\\dashrightarrow\sf\:\:BD=\sqrt{25\:cm \times 25\:cm}\\\\\\\dashrightarrow\:\:\underline{\boxed{\sf BD=25\:cm}}\qquad\bigg\lgroup\bf Diagonal\bigg\rgroup\end{gathered}
⇢(Diagonal)
2
=(Length)
2
+(Breadth)
2
⇢(BD)
2
=(BC)
2
+(CD)
2
⇢(BD)
2
=(24cm)
2
+(7cm)
2
⇢(BD)
2
=576cm
2
+49cm
2
⇢(BD)
2
=625cm
2
⇢BD=
625cm
2
⇢BD=
25cm×25cm
⇢
BD=25cm
⎩
⎪
⎪
⎪
⎧
Diagonal
⎭
⎪
⎪
⎪
⎫
⠀
\therefore\:\underline{\textsf{Hence, Length of Diagonal is C) \textbf{25 cm}}}.∴
Hence, Length of Diagonal is C) 25 cm
.