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Daftar
10 \left| x-3 \right| > 40
1
0
∣
x
−
3
∣
>
4
0
Atasi untuk x
x\in \left(-\infty,-1\right)\cup \left(7,\infty\right)
x
∈
(
−
∞
,
−
1
)
∪
(
7
,
∞
)
Grafik
Pertidaksamaan Grafik
Kedua Sisi Grafik dalam 2D
Kuis
5 soal serupa dengan:
10 \left| x-3 \right| > 40
1
0
∣
x
−
3
∣
>
4
0
Soal yang Mirip dari Pencarian Web
How do you solve \displaystyle{\left|{2}{x}-{3}\right|}={x}+{5} and find any extraneous solutions?
How do you solve
∣
2
x
−
3
∣
=
x
+
5
and find any extraneous solutions?
https://socratic.org/questions/how-do-you-solve-2x-3-x-5-and-find-any-extraneous-solutions
\displaystyle{x}={8} or \displaystyle{x}=-\frac{{2}}{{3}} (neither solution is extraneous) Explanation: Consider the two possibilities for \displaystyle{2}{x}-{3} : \displaystyle{\left.\begin{array}{ccc} \text{if }\ {2}{x}-{3}\ge{0}&{\left(\text{xxx}\right)}&\text{if }\ {2}{x}-{3}{<}{0}\\{\left|{{2}{x}-{3}}\right|}={x}+{5}&&{\left|{{2}{x}-{3}}\right|}={x}+{5}\\\rightarrow{2}{x}-{3}={x}+{5}&&\rightarrow{3}-{2}{x}={x}+{5}\\\rightarrow{x}={8}&&\rightarrow-{2}={3}{x}\\&&\rightarrow{x}=-\frac{{2}}{{3}}\end{array}\right.} ...
x
=
8
or
x
=
−
3
2
(neither solution is extraneous) Explanation: Consider the two possibilities for
2
x
−
3
:
if
2
x
−
3
≥
0
∣
2
x
−
3
∣
=
x
+
5
→
2
x
−
3
=
x
+
5
→
x
=
8
(
xxx
)
if
2
x
−
3
<
0
∣
2
x
−
3
∣
=
x
+
5
→
3
−
2
x
=
x
+
5
→
−
2
=
3
x
→
x
=
−
3
2
...
How do you solve \displaystyle{\left|{{2}{x}-{3}}\right|}\le{4} ?
How do you solve
∣
2
x
−
3
∣
≤
4
?
https://socratic.org/questions/how-do-you-solve-abs-2x-3-4
KillerBunny Feb 4, 2015 First of all, you have to determine the absolute value. Since \displaystyle{\left|{a}\right|}={a} if \displaystyle{a}{>}{0} and \displaystyle-{a} if \displaystyle{a}{<}{0} ...
KillerBunny Feb 4, 2015 First of all, you have to determine the absolute value. Since
∣
a
∣
=
a
if
a
>
0
and
−
a
if
a
<
0
...
How do you solve \displaystyle{\left|{x}-{3}\right|}+{1}={3} ?
How do you solve
∣
x
−
3
∣
+
1
=
3
?
https://socratic.org/questions/how-do-you-solve-x-3-1-3
\displaystyle{x}={5} and \displaystyle{x}={1} Explanation: First, isolate the absolute value term on the left side of the equation: \displaystyle{\left|{{x}-{3}}\right|}+{1}-{1}={3}-{1} ...
x
=
5
and
x
=
1
Explanation: First, isolate the absolute value term on the left side of the equation:
∣
x
−
3
∣
+
1
−
1
=
3
−
1
...
How do you solve \displaystyle{\left|{{x}-{3}}\right|}-{6}={2} ?
How do you solve
∣
x
−
3
∣
−
6
=
2
?
https://socratic.org/questions/how-do-you-solve-abs-x-3-6-2
See a solution process below: Explanation: First, add \displaystyle{\left({6}\right)} to each side of the equation to isolate the absolute value function while keeping the equation balanced: \displaystyle{\left|{{x}-{3}}\right|}-{6}+{\left({6}\right)}={2}+{\left({6}\right)} ...
See a solution process below: Explanation: First, add
(
6
)
to each side of the equation to isolate the absolute value function while keeping the equation balanced:
∣
x
−
3
∣
−
6
+
(
6
)
=
2
+
(
6
)
...
How do you graph and solve \displaystyle{\left|{x}-{4}\right|}-{5}{<}{1} ?
How do you graph and solve
∣
x
−
4
∣
−
5
<
1
?
https://socratic.org/questions/how-do-you-graph-and-solve-x-4-5-1
Mark the ends of the closed interval \displaystyle{\left[-{2},{10}\right]} . The segment in-between (sans the ends) is the graph for the inequality that is equivalent to \displaystyle{x}{>}-{2}{\quad\text{and}\quad}{x}{<}{10} ...
Mark the ends of the closed interval
[
−
2
,
1
0
]
. The segment in-between (sans the ends) is the graph for the inequality that is equivalent to
x
>
−
2
and
x
<
1
0
...
How do you solve and graph \displaystyle-{0.1}\le{3.4}{x}-{1.8}{<}{6.7} ?
How do you solve and graph
−
0
.
1
≤
3
.
4
x
−
1
.
8
<
6
.
7
?
https://socratic.org/questions/how-do-you-solve-and-graph-0-1-3-4x-1-8-6-7
Douglas K. Aug 8, 2017 Given: \displaystyle-{0.1}\le{3.4}{x}-{1.8}{<}{6.7} Add 1.8 to everything: \displaystyle{1.7}\le{3.4}{x}{<}{8.5} Divide everything by 3.4: \displaystyle\frac{{1.7}}{{3.4}}\le{x}{<}\frac{{8.5}}{{3.4}} ...
Douglas K. Aug 8, 2017 Given:
−
0
.
1
≤
3
.
4
x
−
1
.
8
<
6
.
7
Add 1.8 to everything:
1
.
7
≤
3
.
4
x
<
8
.
5
Divide everything by 3.4:
3
.
4
1
.
7
≤
x
<
3
.
4
8
.
5
...
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{ x } ^ { 2 } - 4 x - 5 = 0
x
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−
4
x
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5
=
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Trigonometri
4 \sin \theta \cos \theta = 2 \sin \theta
4
sin
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cos
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y
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Matriks
\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
]
Persamaan simultan
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
{
8
x
+
2
y
=
4
6
7
x
+
3
y
=
4
7
Diferensial
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
d
x
d
(
x
−
5
)
(
3
x
2
−
2
)
Integral
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
∫
0
1
x
e
−
x
2
d
x
Limit
\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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