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Math Problems
Algebra 2
Reference angles
Wich is grater
7
−
3
\sqrt{7}-\sqrt{3}
7
−
3
or
5
−
1
\sqrt{5}-1
5
−
1
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The terms
w
w
w
,
x
x
x
,
y
y
y
and
z
z
z
are linked by the following relation:
x
2
y
=
3
w
z
1
3
x^2y = \frac{3w}{z^{\frac{1}{3}}}
x
2
y
=
z
3
1
3
w
. What is the correct expression for the term
z
z
z
?
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50
50
50
.
2
−
3
−
2
+
3
=
\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}=
2
−
3
−
2
+
3
=
?
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\qquad
33
33
33
. Given:
m
∠
I
F
G
=
12
8
∘
m \angle I F G=128^{\circ}
m
∠
I
FG
=
12
8
∘
\newline
m \overparen{I H F}=
\qquad
\newline
2
)
∘
2)^{\circ}
2
)
∘
, find m \overparen{D H} .
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lim
x
→
1
−
(
x
2
−
1
)
cos
(
(
π
x
)
/
2
)
=
\lim _{x \rightarrow 1^{-}}\left(x^{2}-1\right)^{\cos ((\pi x) / 2)}=
lim
x
→
1
−
(
x
2
−
1
)
c
o
s
((
π
x
)
/2
)
=
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80
80
80
.
4
x
5
−
x
x
3
\sqrt{4 x^{5}}-x \sqrt{x^{3}}
4
x
5
−
x
x
3
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If
7
a
=
7
3
8
7^a=\sqrt[8]{7^3}
7
a
=
8
7
3
, what is the value of
a
?
a?
a
?
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17
17
17
. If
m
∠
V
Y
U
=
6
7
∘
\mathrm{m} \angle V Y U=67^{\circ}
m
∠
VY
U
=
6
7
∘
, then what is
m
∠
T
Y
W
\mathrm{m} \angle T Y W
m
∠
T
YW
?
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My Drive - Google Dr...
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Computer science
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Classwork
\newline
Inbox - anton
\newline
ntonio.canas.
183
183
183
@k
12
12
12
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\newline
Indicates required question
\newline
Uuestion
1
1
1
\newline
1
1
1
. Which expression is equivalent to
60
m
−
2
n
6
5
m
−
4
n
−
2
\frac{60 m^{-2} n^{6}}{5 m^{-4} n^{-2}}
5
m
−
4
n
−
2
60
m
−
2
n
6
for all values of
m
m
m
and
n
n
n
where th expression is defined?
\newline
12
m
2
n
8
12
n
5
m
6
12 m^{2} n^{8} \quad \frac{12 n^{5}}{m^{6}}
12
m
2
n
8
m
6
12
n
5
\newline
A.
\newline
B.
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Shairah Kibler
\newline
Dilations of a
30
30
30
−
60
-60
−
60
−
90
-90
−
90
right triangle
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Php
[
(
144
)
2
−
48
sin
3
0
∘
]
\left[(\sqrt{144})^{2}-48 \sin 30^{\circ}\right]
[
(
144
)
2
−
48
sin
3
0
∘
]
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y. Show all your work
\newline
2
128
x
5
y
5
2 \sqrt{128 x^{5} y^{5}}
2
128
x
5
y
5
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11
11
11
. If
m
V
Q
undefined
=
(
y
+
7
)
∘
,
m
Q
R
undefined
=
(
x
+
11
)
∘
,
m
R
S
undefined
=
(
3
y
)
∘
m \widehat{V Q}=(y+7)^{\circ}, m \widehat{Q R}=(x+11)^{\circ}, m \widehat{R S}=(3 y)^{\circ}
m
V
Q
=
(
y
+
7
)
∘
,
m
QR
=
(
x
+
11
)
∘
,
m
RS
=
(
3
y
)
∘
, and
m
S
T
undefined
=
6
5
∘
m \widehat{S T}=65^{\circ}
m
ST
=
6
5
∘
, find the values of
x
x
x
and
y
y
y
.
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If
m
∠
3
=
2
6
∘
m \angle 3=26^{\circ}
m
∠3
=
2
6
∘
, find
m
∠
6
m \angle 6
m
∠6
. Justify your answer.
\newline
4
∘
4^{\circ}
4
∘
, alternate
\newline
2
6
∘
26^{\circ}
2
6
∘
, supplementary
\newline
26
26
26
\%. same side
\newline
15
4
∘
154^{\circ}
15
4
∘
, same side angles interior angles interior angles
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d
y
d
x
=
y
4
\frac{d y}{d x}=y^{4}
d
x
d
y
=
y
4
and
y
(
2
)
=
−
1
y(2)=-1
y
(
2
)
=
−
1
.
\newline
y
(
−
1
)
=
y(-1)=
y
(
−
1
)
=
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Scavenger Hunt
\newline
Name
\newline
\qquad
\newline
Task
\newline
Type
(
4
5
)
2
\left(4^{5}\right)^{2}
(
4
5
)
2
into the calculator.
\newline
Which does it match?
\newline
17
17
17
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Solwe for the wincing angle
\newline
A.
\newline
8
0
∘
80^{\circ}
8
0
∘
\newline
B.
\newline
24
5
∘
245^{\circ}
24
5
∘
\newline
C.
\newline
6
5
∘
65^{\circ}
6
5
∘
\newline
D.
\newline
10
0
∘
100^{\circ}
10
0
∘
\newline
Type here to search
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115
115
115
)
2
2
x
4
y
3
3
x
3
y
2
\frac{2 \sqrt{2 x^{4} y^{3}}}{\sqrt{3 x^{3} y^{2}}}
3
x
3
y
2
2
2
x
4
y
3
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simplify
(
11
−
7
)
(
11
+
7
)
(\sqrt{11}-\sqrt{7})(\sqrt{11}+\sqrt{7})
(
11
−
7
)
(
11
+
7
)
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y
−
1
+
3
=
y
\sqrt{y-1}+3=y
y
−
1
+
3
=
y
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588
5
(
@
)
=
\sqrt{5885^{(@)}}=
588
5
(
@
)
=
?
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e
5
−
4
x
=
3
e^{5-4 x}=3
e
5
−
4
x
=
3
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2
p
2
=
10
2 p^{2}=\sqrt{10}
2
p
2
=
10
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Home ws
\newline
(
3
+
7
)
(
4
−
13
)
(3+\sqrt{7})(4-\sqrt{13})
(
3
+
7
)
(
4
−
13
)
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m
∠
P
T
S
=
x
+
67
,
m
∠
U
T
P
=
x
+
74
m \angle P T S=x+67, m \angle U T P=x+74
m
∠
PTS
=
x
+
67
,
m
∠
U
TP
=
x
+
74
, and
m
∠
U
T
S
=
13
3
∘
m \angle U T S=133^{\circ}
m
∠
U
TS
=
13
3
∘
. Find
x
x
x
.
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Integrate
cosh
(
e
3
x
)
\cosh(e^{3x})
cosh
(
e
3
x
)
over the interval of
−
π
-\pi
−
π
to
π
\pi
π
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a Use a calculator to show that
2
×
32
\sqrt{2} \times \sqrt{32}
2
×
32
is a rational
\newline
b Find two irrational numbers with a product of
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Congruence OS \& real world application
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x
2
+
11
x
+
30
x
+
6
=
\frac{x^{2}+11 x+30}{x+6}=
x
+
6
x
2
+
11
x
+
30
=
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(
h
∘
g
)
(
−
6
)
=
h
(
g
(
−
6
)
)
(h \circ g)(-6)=h(g(-6))
(
h
∘
g
)
(
−
6
)
=
h
(
g
(
−
6
))
\newline
function
\newline
composition
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11
11
11
. Rotate rectangle
K
L
M
N
27
0
∘
K L M N 270^{\circ}
K
L
MN
27
0
∘
about the origin.
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Select all the expressions that are equivalent to
5
5
×
8
5
5^5 \times 8^5
5
5
×
8
5
.
\newline
Multi-select Choices:
\newline
(A)
4
0
5
40^5
4
0
5
\newline
(B)
1
4
0
5
\frac{1}{40^5}
4
0
5
1
\newline
(C)
1
4
0
−
5
\frac{1}{40^{-5}}
4
0
−
5
1
\newline
(D)
4
0
25
40^{25}
4
0
25
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g
(
x
)
=
(
2
3
x
+
1
)
3
+
(
2
x
)
3
g(x)=\sqrt{\left(2^{3 x}+1\right)^{3}+(2 x)^{3}}
g
(
x
)
=
(
2
3
x
+
1
)
3
+
(
2
x
)
3
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(
6
a
3
+
7
a
2
)
−
(
5
a
3
+
9
a
2
+
a
)
=
(6a^3+7a^2) -(5a^3+9a^2+a)=
(
6
a
3
+
7
a
2
)
−
(
5
a
3
+
9
a
2
+
a
)
=
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(
4
r
2
−
3
r
+
2
)
−
(
−
r
2
−
3
r
)
=
(4r^2-3r+2) -(-r^2-3r)=
(
4
r
2
−
3
r
+
2
)
−
(
−
r
2
−
3
r
)
=
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\(4^{
11
11
11
})(
4
4
4
^{
−
8
-8
−
8
})\
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lim
u
→
3
u
2
−
5
−
2
u
2
−
u
−
6
=
\lim _{u \rightarrow 3} \frac{\sqrt{u^{2}-5}-2}{u^{2}-u-6}=
lim
u
→
3
u
2
−
u
−
6
u
2
−
5
−
2
=
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cos
14
0
∘
−
cos
10
0
∘
sin
14
0
∘
−
sin
10
0
∘
=
\frac{\cos 140^{\circ}-\cos 100^{\circ}}{\sin 140^{\circ}-\sin 100^{\circ}}=
s
i
n
14
0
∘
−
s
i
n
10
0
∘
c
o
s
14
0
∘
−
c
o
s
10
0
∘
=
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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
=
1
\sin \theta=1
sin
θ
=
1
\newline
Answer:
θ
=
\theta=
θ
=
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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
tan
θ
=
3
\tan \theta=\sqrt{3}
tan
θ
=
3
\newline
Answer:
θ
=
\theta=
θ
=
Get tutor help
1
1
1
. Shade in the boxes of two numbers whose sum, when added, would be irrational.
\newline
\begin{tabular}{|c|c|c|c|c|}
\newline
\hline raturs & & Irrdt &
4
r
d
4^{r^{d}}
4
r
d
& त्रण \\
\newline
\hline
2
16
2 \sqrt{16}
2
16
& &
5
\sqrt{5}
5
&
4
3
\frac{4}{3}
3
4
&
4
10
4 \sqrt{10}
4
10
\\
\newline
\hline
\newline
\end{tabular}
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Write the expression as a single power of
9
9
9
.
\newline
(
9
−
3
)
(
9
12
)
=
(9^{-3})(9^{12})=
(
9
−
3
)
(
9
12
)
=
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Express as a function of a DIFFERENT angle,
0
∘
≤
θ
<
36
0
∘
0^{\circ} \leq \theta<360^{\circ}
0
∘
≤
θ
<
36
0
∘
.
\newline
tan
(
35
6
∘
)
\tan \left(356^{\circ}\right)
tan
(
35
6
∘
)
\newline
tan
(
□
∘
)
\tan \left(\square^{\circ}\right)
tan
(
□
∘
)
Get tutor help
Express as a function of a DIFFERENT angle,
0
∘
≤
θ
<
36
0
∘
0^{\circ} \leq \theta<360^{\circ}
0
∘
≤
θ
<
36
0
∘
.
\newline
tan
(
31
0
∘
)
\tan \left(310^{\circ}\right)
tan
(
31
0
∘
)
\newline
tan
(
□
∘
)
\tan \left(\square^{\circ}\right)
tan
(
□
∘
)
Get tutor help
Express as a function of a DIFFERENT angle,
0
∘
≤
θ
<
36
0
∘
0^{\circ} \leq \theta<360^{\circ}
0
∘
≤
θ
<
36
0
∘
.
\newline
sin
(
34
4
∘
)
\sin \left(344^{\circ}\right)
sin
(
34
4
∘
)
\newline
sin
(
□
∘
)
\sin \left(\square^{\circ}\right)
sin
(
□
∘
)
Get tutor help
Express as a function of a DIFFERENT angle,
0
∘
≤
θ
<
36
0
∘
0^{\circ} \leq \theta<360^{\circ}
0
∘
≤
θ
<
36
0
∘
.
\newline
tan
(
17
4
∘
)
\tan \left(174^{\circ}\right)
tan
(
17
4
∘
)
\newline
tan
(
□
∘
)
\tan \left(\square^{\circ}\right)
tan
(
□
∘
)
Get tutor help
Express as a function of a DIFFERENT angle,
0
∘
≤
θ
<
36
0
∘
0^{\circ} \leq \theta<360^{\circ}
0
∘
≤
θ
<
36
0
∘
.
\newline
tan
(
32
4
∘
)
\tan \left(324^{\circ}\right)
tan
(
32
4
∘
)
\newline
tan
(
□
∘
)
\tan \left(\square^{\circ}\right)
tan
(
□
∘
)
Get tutor help
Express as a function of a DIFFERENT angle,
0
∘
≤
θ
<
36
0
∘
0^{\circ} \leq \theta<360^{\circ}
0
∘
≤
θ
<
36
0
∘
.
\newline
sin
(
33
2
∘
)
\sin \left(332^{\circ}\right)
sin
(
33
2
∘
)
\newline
sin
(
□
∘
)
\sin \left(\square^{\circ}\right)
sin
(
□
∘
)
Get tutor help
Express as a function of a DIFFERENT angle,
0
∘
≤
θ
<
36
0
∘
0^{\circ} \leq \theta<360^{\circ}
0
∘
≤
θ
<
36
0
∘
.
\newline
sin
(
30
9
∘
)
\sin \left(309^{\circ}\right)
sin
(
30
9
∘
)
\newline
sin
(
□
∘
)
\sin \left(\square^{\circ}\right)
sin
(
□
∘
)
Get tutor help
Express as a function of a DIFFERENT angle,
0
∘
≤
θ
<
36
0
∘
0^{\circ} \leq \theta<360^{\circ}
0
∘
≤
θ
<
36
0
∘
.
\newline
tan
(
22
5
∘
)
\tan \left(225^{\circ}\right)
tan
(
22
5
∘
)
\newline
tan
(
□
∘
)
\tan \left(\square^{\circ}\right)
tan
(
□
∘
)
Get tutor help
1
2
3
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