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Math Problems
Algebra 2
Sum of finite series starts from 1
3
3
3
.
3
x
2
−
20
=
172
3 x^{2}-20=172
3
x
2
−
20
=
172
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(
1
1
1
)
2
+
9
×
4
3
−
(
15
−
6
3
)
\quad 2+9 \times \frac{4}{3}-\left(15-\frac{6}{3}\right)
2
+
9
×
3
4
−
(
15
−
3
6
)
\newline
(i
1
1
1
)
−
(
2
+
5
)
−
(
9
−
20
)
-(2+5)-(9-20)
−
(
2
+
5
)
−
(
9
−
20
)
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(c)
21
4
x
+
6
=
−
7
\frac{21}{4 x+6}=-7
4
x
+
6
21
=
−
7
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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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1
1
1
. Find each of the following products
\newline
(a)
3
×
(
−
1
)
3 \times(-1)
3
×
(
−
1
)
\newline
(b)
(
−
1
)
×
225
(-1) \times 225
(
−
1
)
×
225
\newline
(c)
(
−
21
)
×
(
−
30
)
(-21) \times(-30)
(
−
21
)
×
(
−
30
)
\newline
(d)
(
−
316
)
×
(
−
1
)
(-316) \times(-1)
(
−
316
)
×
(
−
1
)
\newline
(c)
(
−
15
)
×
0
×
(
−
18
)
(-15) \times 0 \times(-18)
(
−
15
)
×
0
×
(
−
18
)
\newline
(f)
(
−
12
)
×
(
−
11
)
×
(
10
)
(-12) \times(-11) \times(10)
(
−
12
)
×
(
−
11
)
×
(
10
)
\newline
(g)
9
×
(
−
3
)
×
(
−
6
)
9 \times(-3) \times(-6)
9
×
(
−
3
)
×
(
−
6
)
\newline
(
11
11
11
)
(
−
18
)
×
(
−
5
)
×
(
−
4
)
(-18) \times(-5) \times(-4)
(
−
18
)
×
(
−
5
)
×
(
−
4
)
\newline
(i)
(
−
1
)
×
(
−
2
)
×
(
−
3
)
×
4
(-1) \times(-2) \times(-3) \times 4
(
−
1
)
×
(
−
2
)
×
(
−
3
)
×
4
\newline
(j)
(
−
3
)
×
(
−
6
)
×
(
−
2
)
×
(
−
1
)
(-3) \times(-6) \times(-2) \times(-1)
(
−
3
)
×
(
−
6
)
×
(
−
2
)
×
(
−
1
)
\newline
2
2
2
. Verify the following:
\newline
(a)
(
−
1
)
×
225
(-1) \times 225
(
−
1
)
×
225
0
0
0
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(b)
5
−
3
a
−
4
8
+
a
2
5-\frac{3 a-4}{8}+\frac{a}{2}
5
−
8
3
a
−
4
+
2
a
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29
29
29
. Himpunan penyelesaian dari pertidaksamaan
x
−
4
3
≥
2
x
−
8
2
\frac{x-4}{3} \geq \frac{2 x-8}{2}
3
x
−
4
≥
2
2
x
−
8
x
x
x
elemen bilangan asli adalah
\qquad
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315
−
(
19
+
4
)
×
(
18
−
2
)
−
10
315-(19+4) \times(18-2)-10
315
−
(
19
+
4
)
×
(
18
−
2
)
−
10
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−
2
3
×
3
5
+
5
2
−
3
5
×
1
6
-\frac{2}{3} \times \frac{3}{5}+\frac{5}{2}-\frac{3}{5} \times \frac{1}{6}
−
3
2
×
5
3
+
2
5
−
5
3
×
6
1
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m
n
=
∑
k
=
3
∞
2
n
5
−
5
n
3
+
4
n
\frac{m}{n}=\sum_{k=3}^{\infty} \frac{2}{n^{5}-5 n^{3}+4 n}
n
m
=
∑
k
=
3
∞
n
5
−
5
n
3
+
4
n
2
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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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Simplify to create an equivalent expression.
\newline
−
5
(
1
−
5
k
)
−
4
(
2
k
+
5
)
-5(1-5k)-4(2k+5)
−
5
(
1
−
5
k
)
−
4
(
2
k
+
5
)
\newline
Choose
1
1
1
answer:
\newline
(A)
−
17
k
−
25
-17 k-25
−
17
k
−
25
\newline
(B)
−
17
k
+
25
-17 k+25
−
17
k
+
25
\newline
(C)
17
k
+
25
17 k+25
17
k
+
25
\newline
(D)
17
k
−
25
17 k-25
17
k
−
25
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Diberikan fungsi
z
=
x
3
y
z=x^3y
z
=
x
3
y
dimana
x
=
2
t
x=2t
x
=
2
t
dan
y
=
t
2
y=t^2
y
=
t
2
. Nilai
d
z
d
t
\frac{dz}{dt}
d
t
d
z
adalah....
\newline
Select one:
\newline
a.
40
t
40t
40
t
\newline
b.
4
t
2
4t^2
4
t
2
\newline
c.
t
2
t^2
t
2
\newline
d.
40
t
4
40t^4
40
t
4
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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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Next
\newline
Post Test: Multiplication and Civt torited
\newline
2
2
2
\newline
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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6
6
6
.
4
x
2
+
9
x
+
5
=
4 x^{2}+9 x+5=
4
x
2
+
9
x
+
5
=
(
4
x
+
5
)
(
x
+
1
)
(4 x+5)(x+1)
(
4
x
+
5
)
(
x
+
1
)
\newline
\begin{tabular}{|c|c|}
\newline
\hline
4
x
2
4 x^{2}
4
x
2
&
9
x
9 x
9
x
\\
\newline
\hline
5
5
5
& \\
\newline
\hline
\newline
\end{tabular}
\newline
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Let
x
x
x
and
y
y
y
are real numbers. If
3
11
x
−
2
y
×
12
5
x
+
1
=
4
5
2
x
+
y
3^{11 x-2 y} \times 125^{x+1}=45^{2 x+y}
3
11
x
−
2
y
×
12
5
x
+
1
=
4
5
2
x
+
y
, find the value of
x
+
y
x+y
x
+
y
.設
x
x
x
和
y
y
y
皆為實數。若
3
11
x
−
2
y
×
12
5
x
+
1
=
4
5
2
x
+
y
3^{11 x-2 y} \times 125^{x+1}=45^{2 x+y}
3
11
x
−
2
y
×
12
5
x
+
1
=
4
5
2
x
+
y
, 求
x
+
y
x+y
x
+
y
的值。
\newline
(A)
2
2
2
\newline
(B)
7
7
7
\newline
(C)
11
11
11
\newline
(D)
13
13
13
\newline
(E)
17
17
17
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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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- Tentukan
S
∝
S \propto
S
∝
dari deret tak hingga berikut:
\newline
-
∑
k
=
1
∞
k
−
5
k
+
2
\text { - } \sum_{k=1}^{\infty} \frac{k-5}{k+2}
-
k
=
1
∑
∞
k
+
2
k
−
5
Get tutor help
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}=\ldots
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}=\ldots
lim
x
→
∞
x
3
+
x
+
2
75
x
+
25
=
…
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∑
n
=
0
132
(
3
n
+
7.4
)
\sum_{n=0}^{132}(3 n+7.4)
∑
n
=
0
132
(
3
n
+
7.4
)
Get tutor help
Question
108
108
108
\newline
Factorise each of the following expressions completely:
\newline
(a)
5
a
2
−
25
a
b
5 a^{2}-25 a b
5
a
2
−
25
ab
\newline
(b)
8
a
b
+
8
a
c
+
b
+
c
=
8 a b+8 a c+b+c=
8
ab
+
8
a
c
+
b
+
c
=
\newline
Question
109
109
109
\newline
Factorise each of the following expressions completely:
\newline
(a)
24
y
2
z
−
3
y
x
24 y^{2} z-3 y x
24
y
2
z
−
3
y
x
\newline
(h)
m
x
−
11
m
v
+
x
−
11
v
m x-11 m v+x-11 v
m
x
−
11
m
v
+
x
−
11
v
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∑
k
=
1
50
(
5
k
−
161
)
=
\sum_{k=1}^{50}(5 k-161)=
∑
k
=
1
50
(
5
k
−
161
)
=
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∑
n
=
2
∞
(
n
−
1
n
+
1
)
n
(
n
−
5
)
\sum_{n=2}^{\infty}\left(\frac{n-1}{n+1}\right)^{n(n-5)}
∑
n
=
2
∞
(
n
+
1
n
−
1
)
n
(
n
−
5
)
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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x
2
+
y
2
−
16
x
+
26
y
+
208
=
0
x^{2}+y^{2}-16 x+26 y+208=0
x
2
+
y
2
−
16
x
+
26
y
+
208
=
0
Get tutor help
Khan Academy
\newline
Donate ca
\newline
Naymet
k
\mathrm{k}
k
\newline
Quiz
4
4
4
\newline
Factor
64
b
−
16
c
64 b-16 c
64
b
−
16
c
to identify the equivalent expressions.
\newline
Choose
2
2
2
answers:
\newline
A
16
(
4
b
−
c
)
16(4 b-c)
16
(
4
b
−
c
)
\newline
B
4
(
16
b
−
4
)
4(16 b-4)
4
(
16
b
−
4
)
\newline
c.
8
(
8
b
+
2
c
)
8(8 b+2 c)
8
(
8
b
+
2
c
)
\newline
D
2
(
32
b
−
8
c
)
2(32 b-8 c)
2
(
32
b
−
8
c
)
\newline
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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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F
(
x
)
=
20
x
−
40
4
x
2
−
16
F(x)= \frac{20x-40}{4x^{2}-16}
F
(
x
)
=
4
x
2
−
16
20
x
−
40
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3
z
−
4
y
=
2
z
−
4.5
3z-4y=2z-4.5
3
z
−
4
y
=
2
z
−
4.5
and
8
y
=
2
z
+
5
8y=2z+5
8
y
=
2
z
+
5
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32
32
32
. Erep
8
5
=
1
+
1
a
+
2
b
\frac{8}{5}=1+\frac{1}{a+\frac{2}{b}}
5
8
=
1
+
a
+
b
2
1
6
6
6
олея,
a
+
b
a+b
a
+
b
, табыныз
\newline
A)
5
5
5
\newline
B)
3
3
3
\newline
C)
4
4
4
\newline
D)
6
6
6
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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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∑
n
=
1
∞
(
1
+
1
n
)
2
e
−
n
\sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)^{2} e^{-n}
∑
n
=
1
∞
(
1
+
n
1
)
2
e
−
n
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ilus
1
1
1
: Exam
2
2
2
\newline
(
5
5
5
pts) Evaluate the limit
\newline
lim
x
→
5
4
x
−
4
−
4
4
x
−
20
\lim _{x \rightarrow 5} \frac{\sqrt{4 x-4}-4}{4 x-20}
x
→
5
lim
4
x
−
20
4
x
−
4
−
4
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1
1
1
. Change each to
log
\log
lo
g
base
2
2
2
\newline
(a)
log
3
81
\log _{3} 81
lo
g
3
81
\newline
(b)
log
4
16
\log _{4} 16
lo
g
4
16
\newline
(c)
log
100
\log 100
lo
g
100
\newline
(d)
ln
5
\ln 5
ln
5
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5
5
5
.
3
3
3
If
sin
A
=
p
\sin A=p
sin
A
=
p
and
cos
A
=
q
\cos A=q
cos
A
=
q
:
5
5
5
.
3
3
3
.
1
1
1
Write
tan
β
\tan \beta
tan
β
in terms of
p
p
p
and
q
q
q
5
5
5
.
3
3
3
.
2
2
2
Simplify
p
4
−
q
4
p^{4}-q^{4}
p
4
−
q
4
to a single trigonometric ratio
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5
5
5
.
3
3
3
If
sin
A
=
p
\sin A=p
sin
A
=
p
and
cos
A
=
q
\cos A=q
cos
A
=
q
:
\newline
5
5
5
.
3
3
3
.
1
1
1
Write
tan
A
\tan A
tan
A
in terms of
p
p
p
and
q
q
q
\newline
5
5
5
.
3
3
3
.
2
2
2
Simplify
p
4
−
q
4
p^{4}-q^{4}
p
4
−
q
4
to a single triganomeiric ratio
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