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Math Problems
Algebra 2
Csc, sec, and cot of special angles
baad m hogi
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Simplify.
\newline
Remove all perfect squares from inside the square roots. Assume
\newline
x
x
x
and
\newline
z
z
z
are positive.
\newline
72
x
3
z
3
=
□
\sqrt{72x^{3}z^{3}}=\square
72
x
3
z
3
=
□
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Oppgave
3
3
3
\newline
Under vises enhetssirkelen med en vinkelbue tegnet inn som tilsvarer
u
=
12
0
∘
u=120^{\circ}
u
=
12
0
∘
.
\newline
Bruk figuren til å bestemme en eksakt verdi for
sin
12
0
∘
\sin 120^{\circ}
sin
12
0
∘
når du får vite at
sin
3
0
∘
=
1
/
2
\sin 30^{\circ}=1 / 2
sin
3
0
∘
=
1/2
.
\newline
Husk å begrunne svaret ditt.
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(
sin
x
+
sinh
x
)
d
x
(\sin x+\sinh x) d x
(
sin
x
+
sinh
x
)
d
x
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If
x
2
−
a
=
(
x
−
3
)
(
x
+
3
)
x^{2}-a=(x-3)(x+3)
x
2
−
a
=
(
x
−
3
)
(
x
+
3
)
, then
a
=
a=
a
=
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If
x
2
−
y
2
=
x
+
y
x^{2}-y^{2}=x+y
x
2
−
y
2
=
x
+
y
, then
x
−
y
=
x-y=
x
−
y
=
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e
2
+
3
i
e^2+3i
e
2
+
3
i
can be written in the form of
a
+
b
i
a+bi
a
+
bi
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27
÷
3
2
×
3
2
×
2
−
3
=
27 \div 3^{2} \times 3^{2} \times 2-3=
27
÷
3
2
×
3
2
×
2
−
3
=
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Given
\newline
5
5
5
. that
△
A
B
C
≅
△
D
E
F
\triangle A B C \cong \triangle D E F
△
A
BC
≅
△
D
EF
. If
m
A
B
‾
=
2
x
+
5
m \overline{A B}=2 x+5
m
A
B
=
2
x
+
5
,
m
D
E
‾
=
3
(
6
+
y
)
m \overline{D E}=3(6+y)
m
D
E
=
3
(
6
+
y
)
,
m
E
F
‾
=
1
+
y
m \overline{E F}=1+y
m
EF
=
1
+
y
and
m
B
C
‾
=
3
y
−
x
m \overline{B C}=3 y-x
m
BC
=
3
y
−
x
what is the lenght of
m
E
F
‾
\overline{m E F}
m
EF
?
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Richard makes a number
7
7
7
by cutting out a rectangle and a parallelogram.
\newline
Find the area of the number
7
7
7
.
\newline
Area
=
=
=
□
\square
□
c
m
2
\mathrm{cm}^{2}
cm
2
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Evaluate. Write your answer as a whole number or as a simplified fraction.
\newline
2
9
2
7
=
\frac{2^{9}}{2^{7}}=
2
7
2
9
=
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Managed favorites
\newline
New folder
\newline
Whole Numbers and Integers
\newline
Multiplicative property of equally with integers
\newline
Solve for
y
y
y
.
\newline
−
y
5
=
−
30
-\frac{y}{5}=-30
−
5
y
=
−
30
\newline
Simplify your answer as much as possible.
\newline
y
=
y=
y
=
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. Find the perimeter in yards of a rectangle that is
18
y
d
.
1
f
t
18 \mathrm{yd} .1 \mathrm{ft}
18
yd
.1
ft
. by
27
y
d
27 \mathrm{yd}
27
yd
Reduce
36
100
\frac{36}{100}
100
36
to lowect terme
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Q
(
x
,
y
)
=
x
0
,
2
y
0
,
9
Q(x, y) = x^{0,2}y^{0,9}
Q
(
x
,
y
)
=
x
0
,
2
y
0
,
9
a) Essa fun\c{c}\~ao de produ\c{c}\~ao possui rendimentos crescentes, constantes ou decrescente de escala? Por qu\^e?
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19
19
19
.
(
8
+
12
)
(
48
+
18
)
=
(\sqrt{8}+\sqrt{12})(\sqrt{48}+\sqrt{18})=
(
8
+
12
)
(
48
+
18
)
=
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oblems.
\newline
7
7
7
.
−
19
+
(
7
+
4
)
3
=
-19+(7+4)^{3}=
−
19
+
(
7
+
4
)
3
=
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2
π
J
(
1
+
(
e
−
J
3
)
)
=
2\pi J\left(1+\left(\frac{e^{-J}}{3}\right)\right)=
2
π
J
(
1
+
(
3
e
−
J
)
)
=
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2
π
/
1
+
e
−
J
2
=
2 \pi / 1+\frac{e^{-J}}{2}=
2
π
/1
+
2
e
−
J
=
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u
−
4
=
−
u
+
34
u-4=\sqrt{-u+34}
u
−
4
=
−
u
+
34
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Find the following trigonometric values.
\newline
Express your answers exactly.
\newline
cos
(
22
5
∘
)
=
\cos \left(225^{\circ}\right)=
cos
(
22
5
∘
)
=
\newline
□
\square
□
\newline
sin
(
22
5
∘
)
=
\sin \left(225^{\circ}\right)=
sin
(
22
5
∘
)
=
\newline
□
\square
□
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(i)
f
(
t
)
=
1
+
t
1
−
t
f(t)=\frac{1+\sqrt{t}}{1-\sqrt{t}}
f
(
t
)
=
1
−
t
1
+
t
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1
1
1
.
1
−
sin
2
θ
1
+
cos
θ
=
cos
θ
1-\frac{\sin ^{2} \theta}{1+\cos \theta}=\cos \theta
1
−
1
+
c
o
s
θ
s
i
n
2
θ
=
cos
θ
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d
y
d
x
=
1
x
\frac{d y}{d x}=\frac{1}{x}
d
x
d
y
=
x
1
and
y
(
e
)
=
−
2
y(e)=-2
y
(
e
)
=
−
2
.
\newline
y
(
e
3
)
=
y\left(e^{3}\right)=
y
(
e
3
)
=
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Let
y
=
x
e
x
y=\sqrt{x} e^{x}
y
=
x
e
x
.
\newline
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
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d
d
x
[
cos
(
x
)
x
2
]
=
\frac{d}{d x}\left[\cos (x) x^{2}\right]=
d
x
d
[
cos
(
x
)
x
2
]
=
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d
d
x
(
e
x
cos
(
x
)
)
=
\frac{d}{d x}\left(e^{x} \cos (x)\right)=
d
x
d
(
e
x
cos
(
x
)
)
=
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If
2
a
=
2
4
9
2^{a}=\sqrt[9]{2^{4}}
2
a
=
9
2
4
, what is the value of
a
a
a
?
\newline
◻
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1
0
∘
,
∠
M
B
A
=
2
0
∘
,
∠
M
A
C
=
1
0
∘
10^{\circ}, \angle M B A=20^{\circ}, \angle M A C=10^{\circ}
1
0
∘
,
∠
MB
A
=
2
0
∘
,
∠
M
A
C
=
1
0
∘
si
∠
M
C
A
=
3
0
∘
\angle M C A=30^{\circ}
∠
MC
A
=
3
0
∘
. Denonstrati el triungtial ARC este inowert.
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Di sebuah toples terdapat
65
65
65
permen dengan rincian:
\newline
•
15
15
15
permen cokelat,
\newline
•
7
7
7
permen stroberi,
\newline
•
10
10
10
permen vanila,
\newline
•
8
8
8
permen jeruk,
\newline
•
10
10
10
permen kopi,
\newline
•
15
15
15
permen karamel.
\newline
Semua permen memiliki bungkus yang sama dan identik. Anda diminta untuk mengambil sejumlah permen dengan syarat setidaknya Anda memperoleh tiga permen dengan rasa yang sama (contohnya, Anda mungkin memperoleh:
3
3
3
permen cokelat; atau
3
3
3
permen stroberi; dan lain-lain). Paling sedikit, berapa banyak permen yang harus Anda ambil jika pengambilan dilakukan secara acak?
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arccos
(
θ
)
=
(
−
17
38
×
3
)
\arccos (\theta)=\left(\frac{-17}{\sqrt{38} \times 3}\right)
arccos
(
θ
)
=
(
38
×
3
−
17
)
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cot
x
+
sin
x
1
+
cos
x
=
csc
x
:
\cot x+\frac{\sin x}{1+\cos x}=\csc x:
cot
x
+
1
+
c
o
s
x
s
i
n
x
=
csc
x
:
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Evaluate. Write your answer as a whole number or as a simplified fraction.
\newline
3
7
3
5
=
\frac{3^{7}}{3^{5}}=
3
5
3
7
=
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Perhatikan gambar berikut ini! Dari gambar tersebut tentukan besar sudut
a
a
a
,
b
b
b
,
c
c
c
dan
d
d
d
!
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Terdapat dua kantong. Kantong pertama berisi
10
10
10
bola, terdiri atas
3
3
3
bola merah dan
7
7
7
bola biru. Kantong kedua berisi
10
10
10
bola, terdiri atas
4
4
4
bola merah dan
6
6
6
bola biru. Satu bola diambil dari kantong pertama, dicatat, dan disimpan pada kantong kedua. Setelah dikocok, satu bola diambil dari kantong kedua dan dikembalikan ke kantong pertama. a. Buatlah diagram pohonnya. b. Tentukan peluang memperoleh
1
1
1
bola merah dari kantong pertama dan
1
1
1
bola biru dari kantong kedua. c. Tentukan peluang memperoleh dua bola berwarna berbeda. d. Tentukan peluang memperoleh bola merah dari kantong kedua. Tentukan peluang bahwa kantong pertama tetap berisi
3
3
3
bola merah setelah pengambilan dari kantong kedua dan dikembalikan ke kantong pertama.
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lim
x
→
∞
(
x
2
−
2
x
−
x
)
=
\lim _{x \rightarrow \infty}\left(\sqrt{x^{2}-2 x}-x\right)=
lim
x
→
∞
(
x
2
−
2
x
−
x
)
=
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(
3
2
)
3
=
\left(\dfrac{3}{2}\right)^3=
(
2
3
)
3
=
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Dviženklio skaičiaus skaitmenų suma lygi mažiausiam dviženkliam skaičiui, o dešimčių skaitmuo – keturis kartus mažesnis už vienetų skaitmenį. Raskite tą skaičių.
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Distribute to create an equivalent expression with the fewest symbols possible.
1
2
(
10
x
+
20
y
+
10
z
)
=
\dfrac{1}{2}(10x + 20y +10z) =
2
1
(
10
x
+
20
y
+
10
z
)
=
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Distribute to create an equivalent expression with the fewest symbols possible.
1
2
(
10
x
+
20
y
+
10
z
)
=
\dfrac{1}{2}(10x + 20y +10z) =
2
1
(
10
x
+
20
y
+
10
z
)
=
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kalau proyek perumahan akan selesai
25
25
25
hari oleh
8
8
8
pekerja, jika proyek tersebut dikerjkan oleh
16
16
16
orang, maka proyek selesai berapa hari?
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lim
θ
→
0
+
(
sin
(
π
−
θ
)
)
θ
\lim _{\theta \rightarrow 0^{+}}(\sin (\pi-\theta))^{\theta}
lim
θ
→
0
+
(
sin
(
π
−
θ
)
)
θ
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12
⋅
y
3
⋅
6
y
=
\sqrt{12}\cdot\sqrt{y^3}\cdot\sqrt{6y}=
12
⋅
y
3
⋅
6
y
=
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LMS Universitas Teknologi Digita
x
x
x
\newline
E
\newline
1712375668373
1712375668373
1712375668373
Pert
1
1
1
_MatBis
\newline
easy viewing right in Microsoft Edge. Choose Download file if you want to use it later,
\newline
Download
\newline
FUNGSI KUADRAT
\newline
Os
\newline
a.
y
=
x
2
−
16
x
+
64
y=x^{2}-16 x+64
y
=
x
2
−
16
x
+
64
\newline
b.
y
=
p
2
−
10
p
+
25
y=p^{2}-10 p+25
y
=
p
2
−
10
p
+
25
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∫
x
arcsin
x
d
x
=
\int x \arcsin x d x=
∫
x
arcsin
x
d
x
=
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Diketahui
△
A
B
C
,
∠
A
=
6
0
∘
,
A
D
\triangle A B C, \angle A=60^{\circ}, A D
△
A
BC
,
∠
A
=
6
0
∘
,
A
D
garis bagi,
E
E
E
dan
F
F
F
pada garis bagi ini, sehingga
C
E
C E
CE
dan
B
F
⊥
B F \perp
BF
⊥
garis bagi ini.
\newline
Buktikan :
C
E
+
B
F
=
(
A
B
+
A
C
)
C E+B F=(A B+A C)
CE
+
BF
=
(
A
B
+
A
C
)
.
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(x)
lim
x
→
+
∞
x
x
+
1
=
\lim _{x \rightarrow+\infty} \frac{\sqrt{x}}{x+1}=
lim
x
→
+
∞
x
+
1
x
=
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Diberikan sebarang ring
R
R
R
dan
R
R
R
. Dibentak himpunan
a
R
=
{
a
r
∣
r
∈
R
}
aR= \{ar|r \in R\}
a
R
=
{
a
r
∣
r
∈
R
}
dan
R
a
=
{
r
a
∣
r
∈
R
}
Ra= \{ra|r \in R\}
R
a
=
{
r
a
∣
r
∈
R
}
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Evaluate. Write your answer as a whole number or as a simplified fraction.
\newline
5
5
5
2
=
\frac{5^{5}}{5^{2}}=
5
2
5
5
=
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y
−
3
=
y
−
3
\sqrt{y-3}=y-3
y
−
3
=
y
−
3
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Find all angles,
0
∘
≤
θ
<
36
0
∘
0^{\circ} \leq \theta<360^{\circ}
0
∘
≤
θ
<
36
0
∘
, that solve the following equation.
\newline
sin
θ
=
0
\sin \theta=0
sin
θ
=
0
\newline
Answer:
θ
=
\theta=
θ
=
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1
2
3
...
5
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