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Für x lösen
x\in \left(1,5\right)
x
∈
(
1
,
5
)
Diagramm
Quiz
Algebra
5 ähnliche Probleme wie:
3 < 2 x + 1 < 11
3
<
2
x
+
1
<
1
1
Ähnliche Aufgaben aus Websuche
What is the solution of the inequality -3 < 2x+1 < 5
What is the solution of the inequality
−
3
<
2
x
+
1
<
5
https://math.stackexchange.com/questions/2197882/what-is-the-solution-of-the-inequality-3-2x1-5
Subtracting 1 from each side gives -4<2x<4 Dividing by 2 from each side gives -2<x<2 which is equal to |x|<2
Subtracting
1
from each side gives
−
4
<
2
x
<
4
Dividing by
2
from each side gives
−
2
<
x
<
2
which is equal to
∣
x
∣
<
2
How do you solve the inequality \displaystyle-{5}{<}{2}{x}+{1}{<}{15} ?
How do you solve the inequality
−
5
<
2
x
+
1
<
1
5
?
https://socratic.org/questions/how-do-you-solve-the-inequality-5-2x-1-15
\displaystyle{]}-{3},{7}{[} Explanation: \displaystyle−{5}{<}{2}{x}+{1}{<}{15} \displaystyle−{6}{<}{2}{x}{<}{14} \displaystyle−{3}{<}{x}{<}{7}
]
−
3
,
7
[
Explanation:
−
5
<
2
x
+
1
<
1
5
−
6
<
2
x
<
1
4
−
3
<
x
<
7
How do I solve the polynomial inequality \displaystyle{1}{<}{2}{x}+{3}{<}{11} ?
How do I solve the polynomial inequality
1
<
2
x
+
3
<
1
1
?
https://socratic.org/questions/how-do-i-solve-the-polynomial-inequality-1-2x-3-11
Larry Leitch-Casey Aug 24, 2014 Hey there! To solve polynomial inequalities like these (where there is a binomial in the middle), always remember to do inverse operations on BOTH sides! ...
Larry Leitch-Casey Aug 24, 2014 Hey there! To solve polynomial inequalities like these (where there is a binomial in the middle), always remember to do inverse operations on BOTH sides! ...
How do you solve \displaystyle{13}{<}{2}{x}-{1}{<}{17} ?
How do you solve
1
3
<
2
x
−
1
<
1
7
?
https://socratic.org/questions/how-do-you-solve-13-2x-1-17
\displaystyle{7}{<}{x}{<}{9} Explanation: Solve this as two separate inequalities first, and then combine them at the end. \displaystyle{13}{<}{2}{x}-{1}{<}{17} can be split into: \displaystyle{\left({13}{<}{2}{x}-{1}\right)}\ \text{ and }\ {\left({2}{x}-{1}{<}{17}\right)} ...
7
<
x
<
9
Explanation: Solve this as two separate inequalities first, and then combine them at the end.
1
3
<
2
x
−
1
<
1
7
can be split into:
(
1
3
<
2
x
−
1
)
and
(
2
x
−
1
<
1
7
)
...
How do you solve and graph \displaystyle-{5}{<}{2}{x}+{1}{<}{4} ?
How do you solve and graph
−
5
<
2
x
+
1
<
4
?
https://socratic.org/questions/how-do-you-solve-and-graph-5-2x-1-4
\displaystyle-{3}{<}{x}{<}\frac{{3}}{{2}} Shade the area between a dashed line for \displaystyle{x}=-{3} and a dashed line for \displaystyle{x}=\frac{{3}}{{2}} Explanation: We want ...
−
3
<
x
<
2
3
Shade the area between a dashed line for
x
=
−
3
and a dashed line for
x
=
2
3
Explanation: We want ...
How do you solve and write the following in interval notation: \displaystyle{2}{<}{2}{x}+{4}{<}{10} ?
How do you solve and write the following in interval notation:
2
<
2
x
+
4
<
1
0
?
https://socratic.org/questions/how-do-you-solve-and-write-the-following-in-interval-notation-2-2x-4-10
\displaystyle{\left(-{1},{3}\right)} Explanation: As \displaystyle{2}{<}{2}{x}+{4}{<}{10} we have \displaystyle{2}{<}{2}{x}+{4} or \displaystyle{2}-{4}{<}{2}{x} i.e. \displaystyle-{1}{<}{x} ...
(
−
1
,
3
)
Explanation: As
2
<
2
x
+
4
<
1
0
we have
2
<
2
x
+
4
or
2
−
4
<
2
x
i.e.
−
1
<
x
...
Weitere Elemente
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In die Zwischenablage kopiert
Beispiele
Quadratische Gleichung
{ x } ^ { 2 } - 4 x - 5 = 0
x
2
−
4
x
−
5
=
0
Trigonometrie
4 \sin \theta \cos \theta = 2 \sin \theta
4
sin
θ
cos
θ
=
2
sin
θ
Lineare Gleichung
y = 3x + 4
y
=
3
x
+
4
Arithmetisch
699 * 533
6
9
9
∗
5
3
3
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
[
2
5
3
4
]
[
2
−
1
0
1
3
5
]
Simultane Gleichung
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
{
8
x
+
2
y
=
4
6
7
x
+
3
y
=
4
7
Differenzierung
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
d
x
d
(
x
−
5
)
(
3
x
2
−
2
)
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
∫
0
1
x
e
−
x
2
d
x
Grenzwerte
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}
x
→
−
3
lim
x
2
+
2
x
−
3
x
2
−
9
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