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Math Problems
Algebra 2
Sum of finite series starts from 1
Factor completely:
\newline
(
x
−
6
)
(
x
−
9
)
2
−
(
5
x
+
3
)
(
x
−
9
)
(x-6)(x-9)^{2}-(5 x+3)(x-9)
(
x
−
6
)
(
x
−
9
)
2
−
(
5
x
+
3
)
(
x
−
9
)
\newline
Answer:
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Factor completely:
\newline
(
5
x
+
7
)
(
5
x
−
2
)
+
(
5
x
−
9
)
(
5
x
−
2
)
2
(5 x+7)(5 x-2)+(5 x-9)(5 x-2)^{2}
(
5
x
+
7
)
(
5
x
−
2
)
+
(
5
x
−
9
)
(
5
x
−
2
)
2
\newline
Answer:
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Factor completely:
\newline
(
2
x
−
5
)
2
(
x
−
5
)
−
(
2
x
−
5
)
(
5
x
+
7
)
(2 x-5)^{2}(x-5)-(2 x-5)(5 x+7)
(
2
x
−
5
)
2
(
x
−
5
)
−
(
2
x
−
5
)
(
5
x
+
7
)
\newline
Answer:
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∑
k
=
1
∞
(
−
1
)
k
3
k
−
1
2
k
+
1
\sum_{k=1}^{\infty}(-1)^{k} \frac{3 k-1}{2 k+1}
∑
k
=
1
∞
(
−
1
)
k
2
k
+
1
3
k
−
1
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∑
k
=
1
∞
(
1
−
cos
1
k
3
)
\sum_{k=1}^{\infty}\left(1-\cos \frac{1}{k^{3}}\right)
∑
k
=
1
∞
(
1
−
cos
k
3
1
)
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∑
k
=
1
∞
(
−
1
)
k
2
k
−
1
5
k
=
\sum_{k=1}^{\infty}(-1)^{k} \frac{2^{k-1}}{5^{k}}=
∑
k
=
1
∞
(
−
1
)
k
5
k
2
k
−
1
=
Get tutor help
∑
k
=
1
∞
4
(
1
3
)
k
−
1
\sum_{k=1}^{\infty} 4\left(\frac{1}{3}\right)^{k-1}
∑
k
=
1
∞
4
(
3
1
)
k
−
1
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Evaluate.
\newline
∑
n
=
0
∞
n
!
(
cos
1
n
!
−
1
)
\sum_{n=0}^{\infty} n !\left(\cos \frac{1}{\sqrt{n !}}-1\right)
∑
n
=
0
∞
n
!
(
cos
n
!
1
−
1
)
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∑
m
=
0
∞
n
!
(
cos
1
n
!
−
1
)
\sum_{m=0}^{\infty} n !\left(\cos \frac{1}{\sqrt{n !}}-1\right)
∑
m
=
0
∞
n
!
(
cos
n
!
1
−
1
)
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∑
n
=
1
∞
n
+
1
n
n
\sum_{n=1}^{\infty} \frac{n+1}{n \sqrt{n}}
∑
n
=
1
∞
n
n
n
+
1
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Which of the following is an equivalent form of
\newline
(
2
x
+
3
)
−
(
7
x
−
6
)
(2x+3)-(7x-6)
(
2
x
+
3
)
−
(
7
x
−
6
)
?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
5
x
−
3
-5x-3
−
5
x
−
3
\newline
(B)
−
5
x
+
9
-5x+9
−
5
x
+
9
\newline
(C)
4
x
4x
4
x
\newline
(D)
9
x
−
3
9x-3
9
x
−
3
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Solve for
x
x
x
and write your answer in simplest form.
\newline
3
x
+
5
2
(
x
−
3
)
+
3
2
=
8
x
−
8
3 x+\frac{5}{2}(x-3)+\frac{3}{2}=8 x-8
3
x
+
2
5
(
x
−
3
)
+
2
3
=
8
x
−
8
\newline
Answer:
x
=
x=
x
=
Get tutor help
The modulus of
\newline
1
−
i
3
+
i
+
4
i
5
\frac{1-i}{3+i}+\frac{4i}{5}
3
+
i
1
−
i
+
5
4
i
is
\newline
(A)
5
\sqrt{5}
5
unit
\newline
(B)
11
5
\frac{\sqrt{11}}{5}
5
11
unit
\newline
(C)
5
5
\frac{\sqrt{5}}{5}
5
5
unit
\newline
(D)
12
5
\frac{\sqrt{12}}{5}
5
12
unit
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3
(
T
−
R
)
=
10
3(T-R)=10
3
(
T
−
R
)
=
10
, which of the following correctly expresses
T
T
T
in terms of
R
R
R
?
\newline
Choose
1
1
1
answer:
\newline
(A)
T
=
(
R
+
3
)
10
T=\frac{(R+3)}{10}
T
=
10
(
R
+
3
)
\newline
(B)
T
=
(
R
+
10
)
3
T=\frac{(R+10)}{3}
T
=
3
(
R
+
10
)
\newline
(C)
T
=
R
+
3
10
T=R+\frac{3}{10}
T
=
R
+
10
3
\newline
(D)
T
=
R
+
10
3
T=R+\frac{10}{3}
T
=
R
+
3
10
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3
(
T
−
R
)
=
10
3(T-R)=10
3
(
T
−
R
)
=
10
, which of the following correctly expresses
T
T
T
in terms of
R
R
R
?
\newline
Choose
1
1
1
answer:
\newline
(A)
T
=
R
+
3
10
T=\frac{R+3}{10}
T
=
10
R
+
3
\newline
(B)
T
=
R
+
10
3
T=\frac{R+10}{3}
T
=
3
R
+
10
\newline
(C)
T
=
R
+
3
10
T=R+\frac{3}{10}
T
=
R
+
10
3
\newline
(D)
T
=
R
+
10
3
T=R+\frac{10}{3}
T
=
R
+
3
10
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lim
n
→
∞
∑
i
=
1
n
16
i
n
2
\lim_{n \to \infty}\sum_{i=1}^{n}\frac{16i}{n^{2}}
n
→
∞
lim
i
=
1
∑
n
n
2
16
i
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lim
n
→
∞
∑
i
=
1
n
16
i
n
2
\lim_{n \to \infty}\sum_{i=1}^{n}\frac{16i}{n^{2}}
n
→
∞
lim
i
=
1
∑
n
n
2
16
i
Get tutor help
Rewrite in simplest terms:
−
7
(
−
4
t
−
10
u
)
+
9
u
−
9
(
−
u
−
6
t
)
-7(-4 t-10 u)+9 u-9(-u-6 t)
−
7
(
−
4
t
−
10
u
)
+
9
u
−
9
(
−
u
−
6
t
)
\newline
Answer:
Get tutor help
Rewrite in simplest terms:
5
(
−
8
b
−
5
b
+
8
)
−
4
b
5(-8 b-5 b+8)-4 b
5
(
−
8
b
−
5
b
+
8
)
−
4
b
\newline
Answer:
Get tutor help
Rewrite in simplest terms:
(
−
4
x
+
7
y
)
+
(
4
x
+
5
y
)
(-4 x+7 y)+(4 x+5 y)
(
−
4
x
+
7
y
)
+
(
4
x
+
5
y
)
\newline
Answer:
Get tutor help
Rewrite in simplest terms:
(
9
x
+
4
)
−
(
9
x
−
8
)
(9 x+4)-(9 x-8)
(
9
x
+
4
)
−
(
9
x
−
8
)
\newline
Answer:
Get tutor help
3
n
+
2
(
−
2
n
−
1
)
3n+2(-2n-1)
3
n
+
2
(
−
2
n
−
1
)
\newline
Which of the following is equivalent to the given expression?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
n
+
2
-n+2
−
n
+
2
\newline
(B)
n
+
2
n+2
n
+
2
\newline
(C)
−
n
−
2
-n-2
−
n
−
2
\newline
(D)
n
−
2
n-2
n
−
2
Get tutor help
Find
lim
x
→
−
5
(
(
x
+
1
)
2
8
−
x
)
\lim_{x \to -5}\left(\frac{(x+1)^{2}}{8-x}\right)
lim
x
→
−
5
(
8
−
x
(
x
+
1
)
2
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
16
13
\frac{16}{13}
13
16
\newline
(B)
2
2
2
\newline
(C)
12
12
12
\newline
(D) The limit doesn't exist
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What is the sum of the solutions to the equation
\newline
(
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
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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Find
lim
x
→
−
5
(
(
x
+
1
)
2
8
−
x
)
\lim_{x \to -5}\left(\frac{(x+1)^{2}}{8-x}\right)
lim
x
→
−
5
(
8
−
x
(
x
+
1
)
2
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
16
13
\frac{16}{13}
13
16
\newline
(B)
2
2
2
\newline
(C)
12
12
12
\newline
(D) The limit doesn't exist
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3
3
3
n+
2
2
2
(
−
2
-2
−
2
n
−
1
-1
−
1
)
\newline
Which of the following is equivalent to the given expression?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
n
+
2
-n+2
−
n
+
2
\newline
(B)
n
+
2
n+2
n
+
2
\newline
(C)
−
n
−
2
-n-2
−
n
−
2
\newline
(D)
n
−
2
n-2
n
−
2
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∑
n
=
1
∞
ln
(
n
)
−
ln
(
n
+
1
)
\sum_{n=1}^{\infty}\ln(n)-\ln(n+1)
∑
n
=
1
∞
ln
(
n
)
−
ln
(
n
+
1
)
Get tutor help
∑
k
=
1
∞
(
sin
100
)
k
\sum_{k=1}^{\infty}(\sin 100)^{k}
∑
k
=
1
∞
(
sin
100
)
k
Get tutor help
∑
n
=
1
∞
ln
(
n
2
+
1
2
n
2
+
1
)
\sum_{n=1}^{\infty}\ln\left(\frac{n^{2}+1}{2n^{2}+1}\right)
∑
n
=
1
∞
ln
(
2
n
2
+
1
n
2
+
1
)
Get tutor help
∑
n
=
1
∞
3
n
+
1
4
−
n
\sum_{n=1}^{\infty}3^{n+1}4^{-n}
∑
n
=
1
∞
3
n
+
1
4
−
n
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∑
n
=
1
∞
1
4
+
e
−
n
\sum_{n=1}^{\infty}\frac{1}{4+e^{-n}}
n
=
1
∑
∞
4
+
e
−
n
1
Get tutor help
Simplify to create an equivalent expression.
\newline
−
3
(
2
+
4
k
)
+
7
(
2
k
−
1
)
-3(2+4k)+7(2k-1)
−
3
(
2
+
4
k
)
+
7
(
2
k
−
1
)
\newline
Choose
1
1
1
answer:
\newline
(A)
2
k
−
13
2k-13
2
k
−
13
\newline
(B)
8
k
−
13
8k-13
8
k
−
13
\newline
(C)
2
k
+
13
2k+13
2
k
+
13
\newline
(D)
2
k
−
7
2k-7
2
k
−
7
Get tutor help
Let
f
(
x
)
=
x
2
ln
(
x
)
f(x)=\frac{x^{2}}{\ln (x)}
f
(
x
)
=
l
n
(
x
)
x
2
.
\newline
Find
f
′
(
x
)
f^{\prime}(x)
f
′
(
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
2
x
2
2 x^{2}
2
x
2
\newline
(B)
2
x
ln
(
x
)
−
x
(
ln
(
x
)
)
2
\frac{2 x \ln (x)-x}{(\ln (x))^{2}}
(
l
n
(
x
)
)
2
2
x
l
n
(
x
)
−
x
\newline
(C)
2
x
ln
(
x
)
+
x
2 x \ln (x)+x
2
x
ln
(
x
)
+
x
\newline
(D)
2
x
−
1
x
2 x-\frac{1}{x}
2
x
−
x
1
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What is the area of the region between the graphs of
f
(
x
)
=
−
x
2
+
2
x
+
12
f(x)=-x^{2}+2 x+12
f
(
x
)
=
−
x
2
+
2
x
+
12
and
g
(
x
)
=
x
2
−
12
g(x)=x^{2}-12
g
(
x
)
=
x
2
−
12
from
x
=
−
3
x=-3
x
=
−
3
to
x
=
4
x=4
x
=
4
?
\newline
Choose
1
1
1
answer:
\newline
(A)
52
3
13
−
1
−
32
3
\frac{52}{3} \sqrt{13}-1-32 \sqrt{3}
3
52
13
−
1
−
32
3
\newline
(B)
83
3
\frac{83}{3}
3
83
\newline
(C)
7
7
7
\newline
(D)
343
3
\frac{343}{3}
3
343
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Find the sum of the finite series.
\newline
∑
i
=
1
50
(
2
i
−
1
)
(
2
i
+
1
)
\sum_{i=1}^{50} (2i-1)(2i+1)
∑
i
=
1
50
(
2
i
−
1
)
(
2
i
+
1
)
\newline
______
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