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Math Problems
Algebra 2
Sum of finite series starts from 1
Find the sum.
\newline
Problem
19
19
19
\newline
−
110
+
(
−
102
)
+
(
−
94
)
+
⋯
+
66
+
74
-110+(-102)+(-94)+\cdots+66+74
−
110
+
(
−
102
)
+
(
−
94
)
+
⋯
+
66
+
74
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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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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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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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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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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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Simplify each of the following expressions:
\newline
(i)
(
3
+
3
)
(
2
+
2
)
(3+\sqrt{3})(2+\sqrt{2})
(
3
+
3
)
(
2
+
2
)
\newline
(ii)
(
3
+
3
)
(
3
−
3
)
(3+\sqrt{3})(3-\sqrt{3})
(
3
+
3
)
(
3
−
3
)
\newline
(iii)
(
5
+
2
)
2
(\sqrt{5}+\sqrt{2})^{2}
(
5
+
2
)
2
\newline
(iv)
(
5
−
2
)
(
5
+
2
)
(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})
(
5
−
2
)
(
5
+
2
)
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Solve the following equation for
x
x
x
. Express your answer in the simplest form.
\newline
−
(
6
x
−
2
)
+
4
=
3
2
(
−
4
x
+
8
)
−
6
-(6 x-2)+4=\frac{3}{2}(-4 x+8)-6
−
(
6
x
−
2
)
+
4
=
2
3
(
−
4
x
+
8
)
−
6
\newline
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Solve the equation for
m
(
8
m
−
5
)
(
6
m
+
9
)
=
0
m(8 m-5)(6 m+9)=0
m
(
8
m
−
5
)
(
6
m
+
9
)
=
0
\newline
(A)
−
9
-9
−
9
\newline
(B)
5
5
5
\newline
(C)
5
8
\frac{5}{8}
8
5
\newline
(D)
−
3
2
-\frac{3}{2}
−
2
3
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200000
=
P
×
(
1
+
13
100
)
10
200000=P \times\left(1+\frac{13}{100}\right)^{10}
200000
=
P
×
(
1
+
100
13
)
10
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Solve.
\newline
y
=
cos
(
x
−
3
π
2
)
−
1
y=\cos \left(x-\frac{3 \pi}{2}\right)-1
y
=
cos
(
x
−
2
3
π
)
−
1
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−
3
x
(
36
x
2
−
25
)
(
x
2
−
2
)
=
0
-3 x\left(36 x^{2}-25\right)\left(x^{2}-2\right)=0
−
3
x
(
36
x
2
−
25
)
(
x
2
−
2
)
=
0
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Question
3
3
3
\newline
Solve the equation
x
x
−
1
+
2
5
=
3
2
\frac{x}{x-1}+\frac{2}{5}=\frac{3}{2}
x
−
1
x
+
5
2
=
2
3
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following. Express your answers i
\newline
(b)
25
3
×
(
1
125
)
−
3
\quad \sqrt[3]{25} \times\left(\frac{1}{\sqrt{125}}\right)^{-3}
3
25
×
(
125
1
)
−
3
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(
−
36
)
×
(
−
35
76
)
×
(
19
15
)
×
(
3
−
2
)
−
1
(-36) \times\left(\frac{-35}{76}\right) \times\left(\frac{19}{15}\right) \times\left(\frac{3}{-2}\right)^{-1}
(
−
36
)
×
(
76
−
35
)
×
(
15
19
)
×
(
−
2
3
)
−
1
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given that
3
x
−
7
=
−
6
y
3x-7=-6y
3
x
−
7
=
−
6
y
,evaluate
3
x
−
7
4
(
x
+
4
y
)
−
28
3
\frac{3x-7}{4(x+4y)-\frac{28}{3}}
4
(
x
+
4
y
)
−
3
28
3
x
−
7
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d)
15
x
4
=
9
4
−
3
−
x
2
\frac{15 x}{4}=\frac{9}{4}-\frac{3-x}{2}
4
15
x
=
4
9
−
2
3
−
x
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∑
i
=
4
7
i
2
\sum_{i=4}^{7} i^{2}
∑
i
=
4
7
i
2
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x
+
2
−
4
x
−
2
+
x
+
7
−
6
x
−
2
=
1
\sqrt{x+2-4 \sqrt{x-2}}+\sqrt{x+7-6 \sqrt{x-2}}=1
x
+
2
−
4
x
−
2
+
x
+
7
−
6
x
−
2
=
1
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Solve.
\newline
−
6
x
2
+
6
−
2
x
=
x
-6 x^{2}+6-2 x=x
−
6
x
2
+
6
−
2
x
=
x
\newline
Choose
1
1
1
answer:
\newline
(A)
x
=
5
±
57
16
x=\frac{5 \pm \sqrt{57}}{16}
x
=
16
5
±
57
\newline
(B)
x
=
−
4
±
34
3
x=\frac{-4 \pm \sqrt{34}}{3}
x
=
3
−
4
±
34
\newline
(C)
x
=
−
7
±
3
41
−
16
x=\frac{-7 \pm 3 \sqrt{41}}{-16}
x
=
−
16
−
7
±
3
41
\newline
(D)
x
=
1
±
17
−
4
x=\frac{1 \pm \sqrt{17}}{-4}
x
=
−
4
1
±
17
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Which equation has the correct sign on the product?
\newline
A.
(
−
8
)
(
−
8
)
=
−
64
(-8)(-8)=-64
(
−
8
)
(
−
8
)
=
−
64
\newline
B.
(
−
18
)
4
=
−
72
(-18) 4=-72
(
−
18
)
4
=
−
72
\newline
C.
45
(
−
4
)
=
180
45(-4)=180
45
(
−
4
)
=
180
\newline
D.
13
⋅
3
=
−
39
13 \cdot 3=-39
13
⋅
3
=
−
39
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∑
n
=
1
∞
n
+
1
n
!
\sum_{n=1}^{\infty} \frac{n+1}{n !}
n
=
1
∑
∞
n
!
n
+
1
\newline
study if it converges or diverces.
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what is this equal to
\newline
∑
n
=
1
∞
1
n
=
1
+
1
2
+
1
3
+
⋯
\sum_{n=1}^{\infty}\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots
∑
n
=
1
∞
n
1
=
1
+
2
1
+
3
1
+
⋯
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lim
x
→
∞
−
4
x
2
+
3
x
+
6
2
x
2
+
1
x
−
8
=
_
_
_
_
_
\lim _{x \rightarrow \infty} \frac{-4 x^{2}+3 x+6}{2 x^{2}+1 x-8}=\_\_\_\_\_
lim
x
→
∞
2
x
2
+
1
x
−
8
−
4
x
2
+
3
x
+
6
=
_____
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lim
x
→
∞
75
x
+
25
x
3
+
x
+
2
=
_
_
_
_
_
\lim _{x \rightarrow \infty} \frac{75 x+25}{x^{3}+x+2}=\_\_\_\_\_
lim
x
→
∞
x
3
+
x
+
2
75
x
+
25
=
_____
Get tutor help
Which expression has a product of
1
1
3
1 \frac{1}{3}
1
3
1
?
\newline
(A)
1
2
5
×
3
4
1 \frac{2}{5} \times \frac{3}{4}
1
5
2
×
4
3
\newline
(B)
2
2
3
×
1
2
2 \frac{2}{3} \times \frac{1}{2}
2
3
2
×
2
1
\newline
(C)
2
1
3
×
2
3
2 \frac{1}{3} \times \frac{2}{3}
2
3
1
×
3
2
\newline
(D)
3
1
4
×
1
4
3 \frac{1}{4} \times \frac{1}{4}
3
4
1
×
4
1
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Factor
18
p
−
36
18 p-36
18
p
−
36
to identify the equivalent expressions.
\newline
Choose
2
2
2
answers:
\newline
(A)
3
(
9
p
−
12
)
3(9p-12)
3
(
9
p
−
12
)
\newline
(B)
9
(
2
p
−
4
p
)
9(2p-4p)
9
(
2
p
−
4
p
)
\newline
(C)
2
(
9
p
−
18
)
2(9p-18)
2
(
9
p
−
18
)
\newline
(D)
18
(
p
−
2
)
18(p-2)
18
(
p
−
2
)
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2
k
−
5
32
k
2
+
8
k
⋅
5
k
+
2
7
\frac{2 k-5}{32 k^{2}+8 k} \cdot \frac{5 k+2}{7}
32
k
2
+
8
k
2
k
−
5
⋅
7
5
k
+
2
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Which expression is equivalent to the expression below?
\newline
2
(
8
n
+
8
p
)
−
8
n
2(8 n+8 p)-8 n
2
(
8
n
+
8
p
)
−
8
n
\newline
2
(
8
n
+
8
p
−
8
n
)
2(8 n+8 p-8 n)
2
(
8
n
+
8
p
−
8
n
)
\newline
24
n
+
8
p
24 n+8 p
24
n
+
8
p
\newline
8
n
+
16
p
8 n+16 p
8
n
+
16
p
\newline
18
n
+
10
p
18 n+10 p
18
n
+
10
p
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Calculate in two ways:
(
34
560
+
14
210
)
:
10
=
(
34
560
+
14
210
)
:
10
=
(34\,560+ 14\,210): 10 = (34\,560+ 14\,210): 10 =
(
34
560
+
14
210
)
:
10
=
(
34
560
+
14
210
)
:
10
=
Get tutor help
Solve the given equation
tan
x
sec
x
−
1
+
sin
x
1
−
cos
x
=
2
cosec
x
\frac{\tan x}{\sec x-1}+\frac{\sin x}{1-\cos x}=2 \operatorname{cosec} x
s
e
c
x
−
1
t
a
n
x
+
1
−
c
o
s
x
s
i
n
x
=
2
cosec
x
Get tutor help
Evaluate
∑
n
=
1
∞
(
−
1
)
n
cos
(
1
n
2
)
\sum_{n=1}^{\infty}(-1)^{n} \cos \left(\frac{1}{n^{2}}\right)
∑
n
=
1
∞
(
−
1
)
n
cos
(
n
2
1
)
Get tutor help
3
a
b
−
1
=
2
a
b
−
3
3 a b-1=2 a b-3
3
ab
−
1
=
2
ab
−
3
\newline
Given the equation, what is the value of
a
b
a b
ab
?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
4
-4
−
4
\newline
(B)
−
3
-3
−
3
\newline
(C)
−
2
-2
−
2
\newline
(D)
−
1
-1
−
1
Get tutor help
Solve the equation
log
2
(
4
x
−
3
)
+
log
2
(
3
x
+
5
)
=
3
\log_{2}(4x-3)+\log_{2}(3x+5)=3
lo
g
2
(
4
x
−
3
)
+
lo
g
2
(
3
x
+
5
)
=
3
Get tutor help
Evaluate
−
∑
i
=
1
100
(
−
1
)
i
(
5
!
×
5
4
!
)
1
2
-\sum_{i=1}^{100}(-1)^{i}\left(\frac{5!\times5}{4!}\right)^{\frac{1}{2}}
−
i
=
1
∑
100
(
−
1
)
i
(
4
!
5
!
×
5
)
2
1
Get tutor help
Evaluate
∑
m
≥
1
∞
(
1
−
1
m
)
m
2
=
\sum_{m \geq 1}^{\infty}\left(1-\frac{1}{m}\right)^{m^{2}}=
m
≥
1
∑
∞
(
1
−
m
1
)
m
2
=
Get tutor help
Evaluate
∑
m
=
1
∞
(
1
−
1
m
)
m
2
\sum_{m=1}^{\infty}\left(1-\frac{1}{m}\right)^{m^{2}}
∑
m
=
1
∞
(
1
−
m
1
)
m
2
Get tutor help
Evaluate
∑
m
≥
1
∞
(
2
−
1
m
)
m
2
\sum_{m \geq 1}^{\infty}\left(2-\frac{1}{m}\right)^{m^{2}}
∑
m
≥
1
∞
(
2
−
m
1
)
m
2
Get tutor help
Evaluate
∑
n
=
1
∞
2
+
n
1
−
2
n
\sum_{n=1}^{\infty}\frac{2+n}{1-2n}
n
=
1
∑
∞
1
−
2
n
2
+
n
Get tutor help
Evaluate
∑
k
=
0
9
1000
x
k
=
5
×
60
×
1000
\sum_{k=0}^{9}1000x^{k}=5\times60\times1000
∑
k
=
0
9
1000
x
k
=
5
×
60
×
1000
Get tutor help
What is the sum of the solutions to the equation
(
t
+
3
)
(
t
−
357
)
=
0
(t+3)(t-357)=0
(
t
+
3
)
(
t
−
357
)
=
0
?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
360
-360
−
360
\newline
(B)
−
354
-354
−
354
\newline
(C)
354
354
354
\newline
(D)
360
360
360
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3
(
m
+
2
)
9
=
5
(
m
−
4
)
10
\frac{3(m+2)}{9}=\frac{5(m-4)}{10}
9
3
(
m
+
2
)
=
10
5
(
m
−
4
)
\newline
In the equation above, what is the value of
\newline
m
m
m
?
\newline
A.
6
6
6
\newline
B.
11
11
11
\newline
C.
16
16
16
\newline
D.
28
28
28
Get tutor help
Simplify the given expression
8
(
x
−
2
)
2
+
y
2
−
x
+
y
−
3
x
y
(
1
−
12
x
)
=
4
x
(
2
x
−
9
x
y
)
+
(
y
+
9
)
2
−
3
x
(
y
−
7
)
8(x-2)^{2}+y^{2}-x+y-3xy(1-12x)=4x(2x-9xy)+(y+9)^{2}-3x(y-7)
8
(
x
−
2
)
2
+
y
2
−
x
+
y
−
3
x
y
(
1
−
12
x
)
=
4
x
(
2
x
−
9
x
y
)
+
(
y
+
9
)
2
−
3
x
(
y
−
7
)
Get tutor help
Select the answer which is equivalent to the given expression using your calculator.
\newline
−
5
10
+
6
\frac{-5}{10+\sqrt{6}}
10
+
6
−
5
\newline
−
50
−
5
6
94
\frac{-50-5 \sqrt{6}}{94}
94
−
50
−
5
6
\newline
−
350
−
35
6
94
\frac{-350-35 \sqrt{6}}{94}
94
−
350
−
35
6
\newline
−
50
+
5
6
94
\frac{-50+5 \sqrt{6}}{94}
94
−
50
+
5
6
\newline
−
350
+
35
6
94
\frac{-350+35 \sqrt{6}}{94}
94
−
350
+
35
6
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Select the answer which is equivalent to the given expression using your calculator.
\newline
−
6
17
+
15
\frac{-6}{17+\sqrt{15}}
17
+
15
−
6
\newline
−
51
−
3
15
137
\frac{-51-3 \sqrt{15}}{137}
137
−
51
−
3
15
\newline
−
102
+
6
15
137
\frac{-102+6 \sqrt{15}}{137}
137
−
102
+
6
15
\newline
−
102
−
6
15
137
\frac{-102-6 \sqrt{15}}{137}
137
−
102
−
6
15
\newline
−
51
+
3
15
137
\frac{-51+3 \sqrt{15}}{137}
137
−
51
+
3
15
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Select the answer which is equivalent to the given expression using your calculator.
\newline
−
6
−
16
−
12
\frac{-6}{-16-\sqrt{12}}
−
16
−
12
−
6
\newline
48
+
3
12
122
\frac{48+3 \sqrt{12}}{122}
122
48
+
3
12
\newline
144
+
9
12
122
\frac{144+9 \sqrt{12}}{122}
122
144
+
9
12
\newline
144
−
9
12
122
\frac{144-9 \sqrt{12}}{122}
122
144
−
9
12
\newline
48
−
3
12
122
\frac{48-3 \sqrt{12}}{122}
122
48
−
3
12
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Evaluate the summation below.
\newline
2
∑
k
=
4
9
(
6
−
2
k
)
2 \sum_{k=4}^{9}(6-2 k)
2
k
=
4
∑
9
(
6
−
2
k
)
\newline
Answer:
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Evaluate the summation below.
\newline
4
∑
p
=
0
3
(
9
p
−
2
p
2
)
4 \sum_{p=0}^{3}\left(9 p-2 p^{2}\right)
4
p
=
0
∑
3
(
9
p
−
2
p
2
)
\newline
Answer:
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Evaluate the summation below.
\newline
7
∑
k
=
0
4
(
4
k
−
2
k
2
)
7 \sum_{k=0}^{4}\left(4 k-2 k^{2}\right)
7
k
=
0
∑
4
(
4
k
−
2
k
2
)
\newline
Answer:
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1
2
3
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