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-\frac{1}{3}\approx -0.333333333
−
3
1
≈
−
0
.
3
3
3
3
3
3
3
3
3
Lösungsschritte anzeigen
Lösungsschritte
2/3-1
2
/
3
−
1
Wandelt 1 in einen Bruch \frac{3}{3} um.
Wandelt
1
in einen Bruch
3
3
um.
\frac{2}{3}-\frac{3}{3}
3
2
−
3
3
Da \frac{2}{3} und \frac{3}{3} denselben Nenner haben, subtrahieren Sie diese, indem Sie ihre Zähler subtrahieren.
Da
3
2
und
3
3
denselben Nenner haben, subtrahieren Sie diese, indem Sie ihre Zähler subtrahieren.
\frac{2-3}{3}
3
2
−
3
Subtrahieren Sie 3 von 2, um -1 zu erhalten.
Subtrahieren Sie
3
von
2
, um
−
1
zu erhalten.
-\frac{1}{3}
−
3
1
Faktorisieren
\frac{-1}{3} \approx -0.333333333
3
−
1
≈
−
0
.
3
3
3
3
3
3
3
3
3
Quiz
Arithmetic
5 ähnliche Probleme wie:
2/3-1
2
/
3
−
1
Ähnliche Aufgaben aus Websuche
How do you simplify \displaystyle-\frac{{2}}{{{3}-{i}}} ?
How do you simplify
−
3
−
i
2
?
https://socratic.org/questions/how-do-you-simplify-2-3-i
Multiply the numerator and denominator by the conjugate of the denominator to find \displaystyle-\frac{{2}}{{{3}-{i}}}=-\frac{{3}}{{5}}-\frac{{1}}{{5}}{i} Explanation: The conjugate of a ...
Multiply the numerator and denominator by the conjugate of the denominator to find
−
3
−
i
2
=
−
5
3
−
5
1
i
Explanation: The conjugate of a ...
How to show that x^2 \equiv 23 \equiv 5 \pmod 9 is not solvable?
How to show that
x
2
≡
2
3
≡
5
(
m
o
d
9
)
is not solvable?
https://math.stackexchange.com/questions/413334/how-to-show-that-x2-equiv-23-equiv-5-pmod-9-is-not-solvable
Hint: if it had a solution mod \,9\, then that would yield a solution mod \,3,\, which do not exist, i.e. \ 9\mid x^2-23\,\Rightarrow\,3\mid x^2-2,\, but \,{\rm mod}\ 3\!: x^2\in \{0,\pm1\}^2 \equiv \{0,1\} \not\ni 2.
Hint: if it had a solution mod
9
then that would yield a solution mod
3
,
which do not exist, i.e.
9
∣
x
2
−
2
3
⇒
3
∣
x
2
−
2
,
but
m
o
d
3
:
x
2
∈
{
0
,
±
1
}
2
≡
{
0
,
1
}
∋
2
.
Combinatorial proof of \binom{3n}{n} \frac{2}{3n-1} as the answer to a coin-flipping problem
Combinatorial proof of
(
n
3
n
)
3
n
−
1
2
as the answer to a coin-flipping problem
https://math.stackexchange.com/questions/60991/combinatorial-proof-of-binom3nn-frac23n-1-as-the-answer-to-a-coin-fl
Here's the full combinatorial argument. Raney's lemma was about half of it, I'd say. The argument shows that S(n,2) = \binom{3n-1}{n} \frac{3}{3n-1} (which is equivalent to the two formulations I ...
Here's the full combinatorial argument. Raney's lemma was about half of it, I'd say. The argument shows that
S
(
n
,
2
)
=
(
n
3
n
−
1
)
3
n
−
1
3
(which is equivalent to the two formulations I ...
probability that B hits the target and C does not [closed]
probability that
B
hits the target and
C
does not [closed]
https://math.stackexchange.com/questions/2122107/probability-that-b-hits-the-target-and-c-does-not
Let E_1, E_2 and E_3 denote the events: A hits B , B hits A and C hits A respectively and E denote the event: A is hit. As E_2 and E_3 are independent, we have P (E) = P (E_2 \cup E_3) = 1- P (E_2' \cap E_3') = 1-(\frac{1}{2})(\frac {2}{3}) = \frac {2}{3} ...
Let
E
1
,
E
2
and
E
3
denote the events:
A
hits
B
,
B
hits
A
and
C
hits
A
respectively and
E
denote the event:
A
is hit. As
E
2
and
E
3
are independent, we have
P
(
E
)
=
P
(
E
2
∪
E
3
)
=
1
−
P
(
E
2
′
∩
E
3
′
)
=
1
−
(
2
1
)
(
3
2
)
=
3
2
...
CART for regression: Avoid impossible combinations
CART for regression: Avoid impossible combinations
https://stats.stackexchange.com/questions/67470/cart-for-regression-avoid-impossible-combinations
Force rpart to split down to a single observation per node: rpart(a~b, d, control=rpart.control(minsplit=1, minbucket=1, cp=-1)) (The value cp=-1 means to stop splitting if the improvement in cp ...
Force rpart to split down to a single observation per node: rpart(a~b, d, control=rpart.control(minsplit=1, minbucket=1, cp=-1)) (The value cp=-1 means to stop splitting if the improvement in cp ...
Unique critical point does not imply global maximum/global minimum
Unique critical point does not imply global maximum/global minimum
https://math.stackexchange.com/questions/121326/unique-critical-point-does-not-imply-global-maximum-global-minimum
Your intuitive picture is probably as follows: Imaging filling water into the hollow at the origin. As the water rises, since there are points lower than the origin, you expect the water to start ...
Your intuitive picture is probably as follows: Imaging filling water into the hollow at the origin. As the water rises, since there are points lower than the origin, you expect the water to start ...
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In die Zwischenablage kopiert
\frac{2}{3}-\frac{3}{3}
Wandelt 1 in einen Bruch \frac{3}{3} um.
\frac{2-3}{3}
Da \frac{2}{3} und \frac{3}{3} denselben Nenner haben, subtrahieren Sie diese, indem Sie ihre Zähler subtrahieren.
-\frac{1}{3}
Subtrahieren Sie 3 von 2, um -1 zu erhalten.
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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