Calculator search results

Formula
Solve the inequality
Answer
circle-check-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
Graph
$0.5 \left ( x - 2 \right ) \leq 0.7 x - 1.2$
$0.5 \left ( x - 2 \right ) \leq 0.7 x - 1.2$
Solution of inequality
$x \geq 1$
$0.5 \left( x-2 \right) \leq 0.7x-1.2$
$x \geq 1$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ 0.5 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \leq 0.7 x - 1.2$
$ $ Multiply each term in parentheses by $ 0.5$
$\color{#FF6800}{ 0.5 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 0.5 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \leq 0.7 x - 1.2$
$\color{#FF6800}{ 0.5 } \color{#FF6800}{ x } + 0.5 \times \left ( - 2 \right ) \leq 0.7 x - 1.2$
$ $ Calculate the multiplication expression $ $
$\color{#FF6800}{ \dfrac { x } { 2 } } + 0.5 \times \left ( - 2 \right ) \leq 0.7 x - 1.2$
$\dfrac { x } { 2 } + \color{#FF6800}{ 0.5 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \leq 0.7 x - 1.2$
$ $ Multiply $ 0.5 $ and $ - 2$
$\dfrac { x } { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \leq 0.7 x - 1.2$
$\dfrac { x } { 2 } - 1 \leq \color{#FF6800}{ 0.7 } \color{#FF6800}{ x } - 1.2$
$ $ Calculate the multiplication expression $ $
$\dfrac { x } { 2 } - 1 \leq \color{#FF6800}{ \dfrac { 7 x } { 10 } } - 1.2$
$\dfrac { x } { 2 } - 1 \leq \dfrac { 7 x } { 10 } \color{#FF6800}{ - } \color{#FF6800}{ 1.2 }$
$ $ Convert decimals to fractions $ $
$\dfrac { x } { 2 } - 1 \leq \dfrac { 7 x } { 10 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 6 } { 5 } }$
$\color{#FF6800}{ \dfrac { x } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \leq \color{#FF6800}{ \dfrac { 7 x } { 10 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 6 } { 5 } }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 10 } \leq \color{#FF6800}{ 7 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 12 }$
$5 x - 10 \leq \color{#FF6800}{ 7 } \color{#FF6800}{ x } - 12$
$ $ Move the variable to the left-hand side and change the symbol $ $
$5 x - 10 \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \leq - 12$
$5 x \color{#FF6800}{ - } \color{#FF6800}{ 10 } - 7 x \leq - 12$
$ $ Move the constant to the right side and change the sign $ $
$5 x - 7 x \leq - 12 \color{#FF6800}{ + } \color{#FF6800}{ 10 }$
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } \leq - 12 + 10$
$ $ Organize the expression $ $
$\color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \leq - 12 + 10$
$- 2 x \leq \color{#FF6800}{ - } \color{#FF6800}{ 12 } \color{#FF6800}{ + } \color{#FF6800}{ 10 }$
$ $ Add $ - 12 $ and $ 10$
$- 2 x \leq \color{#FF6800}{ - } \color{#FF6800}{ 2 }$
$\color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \leq \color{#FF6800}{ - } \color{#FF6800}{ 2 }$
$ $ Change the symbol of the inequality of both sides, and reverse the symbol of the inequality to the opposite direction $ $
$2 x \geq 2$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } \geq \color{#FF6800}{ 2 }$
$ $ Divide both sides by the same number $ $
$\color{#FF6800}{ x } \geq \color{#FF6800}{ 1 }$
$ $ 그래프 보기 $ $
Inequality
Solution search results
search-thumbnail-If the sum of two consecutive numbers is $45$ and one number is $X$ .This statement in the form of equation $1s:$ $\left(1$ Point) $\right)$ $○5x+1$ $1eft\left(x+1$ $r1gnt\right)=45s$ $○sx+1ef\left(x+2$ $r1gnt\right)=145s$ $sx+1x=45s$
7th-9th grade
Algebra
Check solution
search-thumbnail-$s|ef\left(-1n$ $\left($ }\right)^{50}\ $\right)$ \ | | is\ equal\ to\ $S$ $s1S$ $S-1S$ $s2S$ $s50s$
7th-9th grade
Other
Check solution
search-thumbnail-The is the statement of the Mean Value Theorem from your $te\times tb00k$ $Tneoren$ $m4.5$ $Ne8n$ Value Theorem Let f be continuous over $leclose$ interval $\left(a.b\right)aod$ $i$ $eo$ $xd$ $csnlc$ $\left(ab\right)$ $mmd$ exists at least one point cE $\left(a.b\right)$ $si1dhtba$ $f^{'}\left(c\right)=\dfrac {f\left(b\right)-f\left(a\right)} {b-a}$ What is the geometric interpretation of the conclusion of the theorem? O The tangent line to the graph of \(f\left(x\right)\) at $l\left(c1\right)$ is parallel to the secant line connecting \(\left(a,f\left(a\right)\right)\) and \(\left(b,f\left(b\right)\right)\). O The tangent line to the graph of $\left(f\left(lef\left(x\left(right\right)$ at \(c\) is the secant line connecting \(\left(a,f\left(a\right)\right)\) $ana$ \(\left(b,f\left(b\right)\right)\). O The tangent line to the graph $of$ $\left(f\left(leF\left(x\left(right\right)\left(\right)at1\left(c1\right)is$ perpendicular to the secant line connecting \(\left(a,f\left(a\right)\right)\) and A(\left(b,f\left(b\right)\right)\). O The tangent line to the graph of $\left(fleH\left(x\right)nght\right)\right)\right)$ $at$ $\left(c\right)\right)ishoizonta|$ $Tnere$ is more than one tangent line to the graph $0no$ $\left(flcf\left(x\right)night\right)\left($ $at1\left(c1\right)$
Calculus
Check solution
search-thumbnail-What is the inverse of $g\left(x\right)=\sqrt{x-1} $ $Og\left(x\right)=\sqrt{x+1} $ $+1$ $g^{'}\left(x\right)=5-$ $y_{9}$ $g\sqrt{ef\left(x\left(nght\right)} =\left(f_{tac\left(x+1\right)0sq\pi \left(x\right)}$ $Og^{'}\left(x\right)=x^{2}+1$ $○g\left(x\right)=x^{2}-1$ « Previous Ne
Other
Check solution
search-thumbnail-What is the inverse of $g\left(x\right)=\sqrt{x-1} $ $Og\left(x\right)=\sqrt{x+1} $ $+1$ $g^{'}\left(x\right)=5-$ $y_{9}$ $g\sqrt{ef\left(x\left(nght\right)} =\left(f_{tac\left(x+1\right)0sq\pi \left(x\right)}$ $Og^{'}\left(x\right)=x^{2}+1$ $○g\left(x\right)=x^{2}-1$ « Previous Ne
Other
Check solution
Have you found the solution you wanted?
Try again
Try more features at QANDA!
Search by problem image
Ask 1:1 question to TOP class teachers
AI recommend problems and video lecture
apple logogoogle play logo