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Math Problems
Algebra 2
Sum of finite series starts from 1
Solve the equation for
x
x
x
.
\newline
x
−
7
6
=
2
x
−
3
6
\frac{x-7}{6}=\frac{2 x-3}{6}
6
x
−
7
=
6
2
x
−
3
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∑
k
=
1
n
(
k
−
1
)
!
(
k
+
2
)
!
\sum_{k=1}^{n} \frac{(k-1)!}{(k+2)!}
∑
k
=
1
n
(
k
+
2
)!
(
k
−
1
)!
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Simplify:
∑
n
=
1
∞
12
(
2
5
)
n
−
1
\sum_{n=1}^{\infty}12\left(\frac{2}{5}\right)^{n-1}
n
=
1
∑
∞
12
(
5
2
)
n
−
1
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∑
k
=
1
n
∑
i
=
1
k
a
i
=
∑
i
=
1
n
(
n
−
i
+
1
)
a
i
\sum_{k=1}^{n} \sum_{i=1}^{k} a_{i}=\sum_{i=1}^{n}(n-i+1) a_{i}
∑
k
=
1
n
∑
i
=
1
k
a
i
=
∑
i
=
1
n
(
n
−
i
+
1
)
a
i
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2
2
2
)
∑
n
=
3
∞
(
−
1
)
n
(
2
n
3
+
2
n
)
n
\sum_{n=3}^{\infty}(-1)^{n}\left(\frac{2 n}{3+2 n}\right)^{n} \quad
∑
n
=
3
∞
(
−
1
)
n
(
3
+
2
n
2
n
)
n
Does it abcolatelf (warege ??
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5
x
+
7
2
=
3
2
x
−
14
5 x+\frac{7}{2}=\frac{3}{2} x-14
5
x
+
2
7
=
2
3
x
−
14
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2
y
+
5
3
=
26
3
−
y
2 y+\frac{5}{3}=\frac{26}{3}-y
2
y
+
3
5
=
3
26
−
y
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∑
k
=
1
n
1
k
(
n
k
−
1
)
5
k
=
∑
k
=
1
n
1
n
+
1
(
n
+
1
k
)
5
k
\sum_{k=1}^{n} \frac{1}{k}\binom{n}{k-1} 5^{k}=\sum_{k=1}^{n} \frac{1}{n+1}\binom{n+1}{k} 5^{k}
∑
k
=
1
n
k
1
(
k
−
1
n
)
5
k
=
∑
k
=
1
n
n
+
1
1
(
k
n
+
1
)
5
k
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25
25
25
.
∑
n
=
1
∞
8
n
n
2
(
n
+
1
)
!
\sum_{n=1}^{\infty} \frac{8^{n} n^{2}}{(n+1)!}
∑
n
=
1
∞
(
n
+
1
)!
8
n
n
2
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24
24
24
.
∑
n
=
1
∞
(
2
n
)
!
(
n
!
)
2
\sum_{n=1}^{\infty} \frac{(2 n)!}{(n!)^{2}}
∑
n
=
1
∞
(
n
!
)
2
(
2
n
)!
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2
2
2
.
∑
k
=
1
n
(
2
3
)
k
\sum_{k=1}^{n}\left(\frac{2}{3}\right)^{k}
∑
k
=
1
n
(
3
2
)
k
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2
2
2
.
∑
k
=
1
π
(
2
3
)
k
\sum_{k=1}^{\pi}\left(\frac{2}{3}\right)^{k}
∑
k
=
1
π
(
3
2
)
k
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∑
3
n
(
27
2
i
+
6
5
π
)
=
2295
π
+
12
10
π
\sum_{3}^{n}\left(\frac{27}{2} i+\frac{6}{5 \pi}\right)=\frac{2295 \pi+12}{10 \pi}
∑
3
n
(
2
27
i
+
5
π
6
)
=
10
π
2295
π
+
12
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9
9
9
)
∑
n
=
1
∞
(
−
1
)
n
n
(
n
+
1
)
\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n(n+1)}
∑
n
=
1
∞
n
(
n
+
1
)
(
−
1
)
n
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w can
−
4.25
+
(
−
8.75
)
-4.25+(-8.75)
−
4.25
+
(
−
8.75
)
be expressed as the sum of its integer and decimal parts?
\newline
4
+
8
+
0.25
+
0.75
4+8+0.25+0.75
4
+
8
+
0.25
+
0.75
\newline
4
+
(
−
8
)
+
0.25
+
(
−
0.75
)
4+(-8)+0.25+(-0.75)
4
+
(
−
8
)
+
0.25
+
(
−
0.75
)
\newline
−
4
+
(
−
8
)
+
0.25
+
0.75
-4+(-8)+0.25+0.75
−
4
+
(
−
8
)
+
0.25
+
0.75
\newline
−
4
+
(
−
8
)
+
(
−
0.25
)
+
(
−
0.75
)
-4+(-8)+(-0.25)+(-0.75)
−
4
+
(
−
8
)
+
(
−
0.25
)
+
(
−
0.75
)
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lim
x
→
0
(
x
+
2
)
3
−
8
x
\lim _{x \rightarrow 0} \frac{(x+2)^{3}-8}{x}
lim
x
→
0
x
(
x
+
2
)
3
−
8
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7
7
7
.
\newline
∑
n
=
1
∞
12
(
2
5
)
n
−
1
\sum_{n=1}^{\infty} 12\left(\frac{2}{5}\right)^{n-1}
n
=
1
∑
∞
12
(
5
2
)
n
−
1
\newline
converyes
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What is the sum of the series
\newline
5
−
5
2
+
5
5
3
−
25
4
+
⋯
+
(
−
1
)
n
5
n
+
1
+
…
\sqrt{5}-\frac{5}{2}+\frac{5\sqrt{5}}{3}-\frac{25}{4}+\dots+(-1)^{n}\frac{\sqrt{5}}{n+1}+\dots
5
−
2
5
+
3
5
5
−
4
25
+
⋯
+
(
−
1
)
n
n
+
1
5
+
…
?
\newline
(A)
ln
(
1
+
5
)
\ln(1+\sqrt{5})
ln
(
1
+
5
)
\newline
(B)
e
5
e_{\sqrt{5}}
e
5
\newline
(C)
ln
(
5
)
\ln(\sqrt{5})
ln
(
5
)
\newline
(D)
5
\sqrt{5}
5
\newline
(E) The series diverges.
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∑
i
=
1
n
(
i
+
2
)
2
\sum_{i=1}^{n}(i+2)^{2}
∑
i
=
1
n
(
i
+
2
)
2
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Simplify:
1
+
sec
(
−
x
)
sin
(
−
x
)
+
tan
(
−
x
)
=
−
csc
x
\frac{1+\sec(-x)}{\sin(-x)+\tan(-x)}=-\csc x
s
i
n
(
−
x
)
+
t
a
n
(
−
x
)
1
+
s
e
c
(
−
x
)
=
−
csc
x
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Simplify:
k
−
3
k
=
k
+
4
5
k
−
1
k
\frac{k-3}{k}=\frac{k+4}{5k}-\frac{1}{k}
k
k
−
3
=
5
k
k
+
4
−
k
1
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Which of the following would be the correct result from the following expression?
\newline
12
3
(
10
)
−
1
2
(
8
)
+
10
1
(
2
)
+
D
(
16
)
123_{(10)}-12_{(8)}+101_{(2)}+D_{(16)}
12
3
(
10
)
−
1
2
(
8
)
+
10
1
(
2
)
+
D
(
16
)
\newline
(A)
130
130
130
\newline
(B)
133
133
133
\newline
(C)
131
131
131
\newline
(D)
132
132
132
\newline
(E)
136
136
136
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x
+
7
−
8
x
3
=
17
6
−
5
x
2
x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}
x
+
7
−
3
8
x
=
6
17
−
2
5
x
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Identify the part of the circle given the equation.
\newline
10
10
10
.
(
x
−
10
)
2
+
(
y
−
5
)
2
=
28
(x-10)^{2}+(y-5)^{2}=28
(
x
−
10
)
2
+
(
y
−
5
)
2
=
28
\newline
radius :
□
\square
□
\sqrt{ }
□
\square
□
\newline
circumference:
□
\square
□
(round to tenths)
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5
(
s
+
3
)
+
2
S
−
1
=
2
(
2
S
−
1
)
+
28
5(s+3)+2 S-1=2(2 S-1)+28
5
(
s
+
3
)
+
2
S
−
1
=
2
(
2
S
−
1
)
+
28
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15
+
(
60
:
4
)
+
270
15+(60:4)+270
15
+
(
60
:
4
)
+
270
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lim
x
→
+
∞
−
x
+
ln
(
1
+
x
e
x
)
=
+
∞
\lim _{x \rightarrow+\infty}-x+\ln \left(1+x e^{x}\right)=+\infty
lim
x
→
+
∞
−
x
+
ln
(
1
+
x
e
x
)
=
+
∞
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5
5
5
.
sin
A
+
cos
A
sin
A
−
cos
A
+
sin
A
−
cos
A
sin
A
+
cos
A
=
2
2
sin
2
A
−
1
\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}=\frac{2}{2 \sin ^{2} A-1}
s
i
n
A
−
c
o
s
A
s
i
n
A
+
c
o
s
A
+
s
i
n
A
+
c
o
s
A
s
i
n
A
−
c
o
s
A
=
2
s
i
n
2
A
−
1
2
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(
a
+
2
i
)
(
1
+
b
i
)
=
17
−
19
i
(a+2 i)(1+b i)=17-19 i
(
a
+
2
i
)
(
1
+
bi
)
=
17
−
19
i
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Practice
\newline
Badges
\newline
Formatting Help
\newline
Simplify.
\newline
(
−
5
)
⋅
(
a
−
2
b
+
3
c
−
4
d
)
−
(
−
3
)
⋅
(
4
a
−
3
b
+
2
c
−
d
)
(-5) \cdot(a-2 b+3 c-4 d)-(-3) \cdot(4 a-3 b+2 c-d)
(
−
5
)
⋅
(
a
−
2
b
+
3
c
−
4
d
)
−
(
−
3
)
⋅
(
4
a
−
3
b
+
2
c
−
d
)
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29
(
27
+
16
(
26
+
9
(
x
−
7
)
)
)
+
17
(
8
+
(
x
−
9
)
(
36
+
8
+
(
2
x
−
7
)
(
x
−
5
)
)
)
−
38
x
=
(
18
+
(
x
−
19
)
(
37
+
6
+
9
(
x
−
4
)
)
)
29(27+16(26+9(x-7))) + 17(8+ (x-9)(36+8 +(2x-7)(x-5))) - 38x= (18+ (x-19)(37+6+ 9(x-4)))
29
(
27
+
16
(
26
+
9
(
x
−
7
)))
+
17
(
8
+
(
x
−
9
)
(
36
+
8
+
(
2
x
−
7
)
(
x
−
5
)))
−
38
x
=
(
18
+
(
x
−
19
)
(
37
+
6
+
9
(
x
−
4
)))
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−
6
r
−
5
(
−
10
r
+
6
)
≤
3
r
+
9
+
2
r
-6 r-5(-10 r+6) \leq 3 r+9+2 r
−
6
r
−
5
(
−
10
r
+
6
)
≤
3
r
+
9
+
2
r
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7
7
7
.
∑
n
=
1
∞
(
−
1
)
n
+
1
n
2
n
−
1
\sum_{n=1}^{\infty} \frac{(-1)^{n+1} \sqrt{n}}{2 n-1}
∑
n
=
1
∞
2
n
−
1
(
−
1
)
n
+
1
n
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(b)
(
1
−
x
2
)
(
1
−
y
2
)
+
4
x
y
\left(1-x^{2}\right)\left(1-y^{2}\right)+4 x y
(
1
−
x
2
)
(
1
−
y
2
)
+
4
x
y
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17
17
17
.
∑
k
=
3
n
(
k
2
−
3
)
\sum_{k=3}^{n}\left(k^{2}-3\right)
∑
k
=
3
n
(
k
2
−
3
)
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81
−
7
×
−
42
18
×
27
−
6
×
−
15
45
\frac{81}{-7} \times \frac{-42}{18} \times \frac{27}{-6} \times \frac{-15}{45}
−
7
81
×
18
−
42
×
−
6
27
×
45
−
15
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Simplify:
−
4
5
×
3
7
×
15
16
×
(
−
14
9
)
\frac{-4}{5} \times \frac{3}{7} \times \frac{15}{16} \times\left(\frac{-14}{9}\right)
5
−
4
×
7
3
×
16
15
×
(
9
−
14
)
\newline
(a)
1
1
1
\newline
(b)
0
0
0
\newline
(c)
2
2
2
\newline
(d)
1
2
\frac{1}{2}
2
1
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7
7
7
.
6
n
−
5
3
=
12
n
−
10
6
\frac{6 n-5}{3}=\frac{12 n-10}{6}
3
6
n
−
5
=
6
12
n
−
10
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lim
x
⃗
3
−
x
+
3
x
2
−
2
x
−
3
\lim _{\vec{x} 3} \frac{-x+3}{x^{2}-2 x-3}
lim
x
3
x
2
−
2
x
−
3
−
x
+
3
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3
3
3
. If
1
+
2
+
3
+
1+2+3+
1
+
2
+
3
+
\newline
b)
(
2
,
3
,
5
,
1
,
x
(2,3,5,1, x
(
2
,
3
,
5
,
1
,
x
\newline
a)
46656
46656
46656
\newline
+
n
=
36
+n=36
+
n
=
36
then find
13
13
13
c)
(
4
/
9
,
25
,
49
,
121
)
(4 / 9,25,49,121)
(
4/9
,
25
,
49
,
121
)
\newline
b)
1296
1296
1296
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3
x
−
2
x
−
3
=
12
9
−
x
r
\frac{3}{x}-\frac{2}{x-3}=\frac{12}{9-x^{r}}
x
3
−
x
−
3
2
=
9
−
x
r
12
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Question
\newline
Find the sum of
10
x
2
+
7
x
+
6
10 x^{2}+7 x+6
10
x
2
+
7
x
+
6
and
6
x
+
5
6 x+5
6
x
+
5
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9
9
9
Figuren visar grafen till funktionen
y
=
f
(
x
)
y=f(x)
y
=
f
(
x
)
.
\newline
Bestäm med hjälp av grafen
\newline
a)
f
(
2
)
f(2)
f
(
2
)
\newline
b) lösningen till ekvationen
f
(
x
)
=
2
f(x)=2
f
(
x
)
=
2
.
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(b)
3
x
−
[
2
x
+
4
(
3
a
−
x
)
]
3 x-[2 x+4(3 a-x)]
3
x
−
[
2
x
+
4
(
3
a
−
x
)]
.
Get tutor help
Express
9
x
−
5
2
(
2
x
−
1
)
(
x
−
1
)
\frac{9x-5}{2(2x-1)(x-1)}
2
(
2
x
−
1
)
(
x
−
1
)
9
x
−
5
in partial fractions.
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Express
9
x
−
5
2
(
2
x
−
1
)
(
x
−
1
)
\frac{9 x-5}{2(2 x-1)(x-1)}
2
(
2
x
−
1
)
(
x
−
1
)
9
x
−
5
in partial frac
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9
x
+
5
=
9 x+5=
9
x
+
5
=
\newline
Express
5
x
−
13
(
x
−
1
)
(
x
−
3
)
2
\frac{5 x-13}{(x-1)(x-3)^{2}}
(
x
−
1
)
(
x
−
3
)
2
5
x
−
13
in the form
A
(
x
−
\frac{A}{(x-}
(
x
−
A
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5
x
+
6
y
−
4
(
x
−
6
y
)
5x+6y-4(x-6y)
5
x
+
6
y
−
4
(
x
−
6
y
)
Get tutor help
4
4
4
. Tentukan jarak antara
\newline
a. Titik pusat
O
(
0
,
0
,
0
)
O(0,0,0)
O
(
0
,
0
,
0
)
dan garis
x
−
4
3
=
y
−
2
4
=
4
−
z
5
\frac{x-4}{3}=\frac{y-2}{4}=\frac{4-z}{5}
3
x
−
4
=
4
y
−
2
=
5
4
−
z
.
\newline
b. Garis
x
−
2
−
1
=
y
−
3
4
=
z
2
\frac{x-2}{-1}=\frac{y-3}{4}=\frac{z}{2}
−
1
x
−
2
=
4
y
−
3
=
2
z
dan garis
x
+
1
1
=
y
−
2
0
=
z
+
1
2
\frac{x+1}{1}=\frac{y-2}{0}=\frac{z+1}{2}
1
x
+
1
=
0
y
−
2
=
2
z
+
1
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∏
k
=
1
2
k
×
lim
x
→
0
e
16
x
−
1
x
+
∑
k
=
0
2
sin
(
2
π
k
3
)
\prod_{k=1}^{2} k \times \lim _{x \rightarrow 0} \frac{e^{16 x}-1}{x}+\sum_{k=0}^{2} \sin \left(\frac{2 \pi k}{3}\right)
∏
k
=
1
2
k
×
lim
x
→
0
x
e
16
x
−
1
+
∑
k
=
0
2
sin
(
3
2
πk
)
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1
2
3
...
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