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Minimo comune multiplo
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Algebra
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Risolvere una variabile
Fattore
Espandi
Calcolo delle frazioni
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Elenco
Calcola
\frac{25}{6}\approx 4.166666667
6
2
5
≈
4
.
1
6
6
6
6
6
6
6
7
Visualizza i passaggi della soluzione
Procedura della soluzione
\frac{4-3}{6}+2^2
6
4
−
3
+
2
2
Sottrai 3 da 4 per ottenere 1.
Sottrai
3
da
4
per ottenere
1
.
\frac{1}{6}+2^{2}
6
1
+
2
2
Calcola 2 alla potenza di 2 e ottieni 4.
Calcola
2
alla potenza di
2
e ottieni
4
.
\frac{1}{6}+4
6
1
+
4
Converti 4 nella frazione \frac{24}{6}.
Converti
4
nella frazione
6
2
4
.
\frac{1}{6}+\frac{24}{6}
6
1
+
6
2
4
Poiché \frac{1}{6} e \frac{24}{6} hanno lo stesso denominatore, calcolane l'addizione sommando i numeratori.
Poiché
6
1
e
6
2
4
hanno lo stesso denominatore, calcolane l'addizione sommando i numeratori.
\frac{1+24}{6}
6
1
+
2
4
E 1 e 24 per ottenere 25.
E
1
e
2
4
per ottenere
2
5
.
\frac{25}{6}
6
2
5
Scomponi in fattori
\frac{5 ^ {2}}{2 \cdot 3} = 4\frac{1}{6} \approx 4.166666667
2
⋅
3
5
2
=
4
6
1
≈
4
.
1
6
6
6
6
6
6
6
7
Quiz
Arithmetic
5 problemi simili a:
\frac{4-3}{6}+2^2
6
4
−
3
+
2
2
Problemi simili da ricerca Web
6^-3/6^-5
6
−
3
/
6
−
5
https://www.tiger-algebra.com/drill/6~-3/6~-5/
6(-3)/6(-5) Final result : 36 Reformatting the input : Changes made to your input should not affect the solution: (1): "^-5" was replaced by "^(-5)". 1 more similar replacement(s) Step by step ...
6(-3)/6(-5) Final result : 36 Reformatting the input : Changes made to your input should not affect the solution: (1): "^-5" was replaced by "^(-5)". 1 more similar replacement(s) Step by step ...
How do you use the ratio test to test the convergence of the series \displaystyle∑\frac{{{2}{n}^{{2}}}}{{{n}!}} from n=1 to infinity?
How do you use the ratio test to test the convergence of the series
∑
n
!
2
n
2
from n=1 to infinity?
https://socratic.org/questions/how-do-you-use-the-ratio-test-to-test-the-convergence-of-the-series-2n-2-n-from-
The infinite series converges (see below). Explanation: Let \displaystyle{a}_{{{n}}}=\frac{{{2}{n}^{{2}}}}{{{n}!}} . Then \displaystyle\frac{{\left|{a}_{{{n}+{1}}}\right|}}{{\left|{a}_{{{n}}}\right|}}=\frac{{{2}{\left({n}+{1}\right)}^{{2}}}}{{{\left({n}+{1}\right)}!}}\cdot\frac{{{n}!}}{{{2}{n}^{{2}}}}=\frac{{{\left({n}+{1}\right)}^{{2}}}}{{{n}^{{2}}{\left({n}+{1}\right)}}}=\frac{{{n}+{1}}}{{n}^{{2}}} ...
The infinite series converges (see below). Explanation: Let
a
n
=
n
!
2
n
2
. Then
∣
a
n
∣
∣
a
n
+
1
∣
=
(
n
+
1
)
!
2
(
n
+
1
)
2
⋅
2
n
2
n
!
=
n
2
(
n
+
1
)
(
n
+
1
)
2
=
n
2
n
+
1
...
How to find laurent series expansion of (2z+3)/(z+2)^2
How to find laurent series expansion of
(
2
z
+
3
)
/
(
z
+
2
)
2
https://math.stackexchange.com/questions/2199623/how-to-find-laurent-series-expansion-of-2z3-z22
Rewrite as follows to get the principal part of the Laurent series at z=-2 in standard form: \frac{2z+\color{red}{3}}{\left(z+2\right)^2} =\frac{2z+\color{red}{4-1}}{\left(z+2\right)^2} = \frac{2\left(z+2\right)}{\left(z+2\right)^2}-\frac{1}{\left(z+2\right)^2} = \frac{\color{blue}{2}}{z+2}-\frac{1}{\left(z+2\right)^2} ...
Rewrite as follows to get the principal part of the Laurent series at
z
=
−
2
in standard form:
(
z
+
2
)
2
2
z
+
3
=
(
z
+
2
)
2
2
z
+
4
−
1
=
(
z
+
2
)
2
2
(
z
+
2
)
−
(
z
+
2
)
2
1
=
z
+
2
2
−
(
z
+
2
)
2
1
...
(34362)^(1/3)
(
3
4
3
6
2
)
(
1
/
3
)
https://www.tiger-algebra.com/drill/(34362)~(1/3)/
∛(34,362) - What Is the cubed root of 34,362 ?This Is the same as (34,362)⅓ Step by Step solution We attempt To simplify the cubed root Of 34,362 Step 1: Factor 34,362 34,362 can be ...
∛(34,362) - What Is the cubed root of 34,362 ?This Is the same as (34,362)⅓ Step by Step solution We attempt To simplify the cubed root Of 34,362 Step 1: Factor 34,362 34,362 can be ...
How do you change \displaystyle{4}^{{-{{3}}}}=\frac{{1}}{{64}} into log form?
How do you change
4
−
3
=
6
4
1
into log form?
https://socratic.org/questions/how-do-you-change-4-3-1-64-into-log-form
\displaystyle{{\log}_{{4}}{\left(\frac{{1}}{{64}}\right)}}=-{3} Explanation: As \displaystyle{a}^{{n}}={b} in logarithmic form is written as \displaystyle{{\log}_{{a}}{b}}={n} Hence, \displaystyle{4}^{{-{3}}}=\frac{{1}}{{64}} ...
lo
g
4
(
6
4
1
)
=
−
3
Explanation: As
a
n
=
b
in logarithmic form is written as
lo
g
a
b
=
n
Hence,
4
−
3
=
6
4
1
...
v^2=25/81
v
2
=
2
5
/
8
1
http://tiger-algebra.com/drill/v~2=25/81/
v2=25/81 Two solutions were found : v = 5/9 = 0.556 v = -5/9 = -0.556 Rearrange: Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation ...
v2=25/81 Two solutions were found : v = 5/9 = 0.556 v = -5/9 = -0.556 Rearrange: Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation ...
Altri Elementi
Condividi
Copia
Copiato negli Appunti
\frac{1}{6}+2^{2}
Sottrai 3 da 4 per ottenere 1.
\frac{1}{6}+4
Calcola 2 alla potenza di 2 e ottieni 4.
\frac{1}{6}+\frac{24}{6}
Converti 4 nella frazione \frac{24}{6}.
\frac{1+24}{6}
Poiché \frac{1}{6} e \frac{24}{6} hanno lo stesso denominatore, calcolane l'addizione sommando i numeratori.
\frac{25}{6}
E 1 e 24 per ottenere 25.
Problemi analoghi
4 - 3 \times 6 + 2
4
−
3
×
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+
2
(4 - 3) \times 6 + 2
(
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−
3
)
×
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+
2
4 - 3 \times (6 + 2) ^ 2
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−
3
×
(
6
+
2
)
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\frac{4-3}{6}+2^2
6
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2
5-4(7-9(5-1)) \times 3^3 -4
5
−
4
(
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−
9
(
5
−
1
)
)
×
3
3
−
4
12-2(7-4)^2 \div 4
1
2
−
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(
7
−
4
)
2
÷
4
\frac{ \left( 4-3 \right) + { \left( 1+2 \right) }^{ 2 } }{ 6+ \left( 7-5 \right) }
6
+
(
7
−
5
)
(
4
−
3
)
+
(
1
+
2
)
2
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