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Let’s check out your problem:
Soient les matrices suivantes :
A
=
(
−
1
0
3
2
)
A = \begin{pmatrix} -1 & 0 \ 3 & 2 \end{pmatrix}
A
=
(
−
1
0
3
2
)
;
B
=
(
4
−
3
0
2
1
0
)
B = \begin{pmatrix} 4 & -3 \ 0 & 2 \ 1 & 0 \end{pmatrix}
B
=
(
4
−
3
0
2
1
0
)
;
C
=
(
1
0
2
4
0
−
3
)
C= \begin{pmatrix} 1 & 0 \ 2 & 4 \ 0 & -3 \end{pmatrix}
C
=
(
1
0
2
4
0
−
3
)
calcule
3
B
–
2
C
3B – 2 C
3
B
–2
C
View step-by-step help
Home
Math Problems
Algebra 1
Transformations of functions
Full solution
Q.
Soient les matrices suivantes :
A
=
(
−
1
0
3
2
)
A = \begin{pmatrix} -1 & 0 \ 3 & 2 \end{pmatrix}
A
=
(
−
1
0
3
2
)
;
B
=
(
4
−
3
0
2
1
0
)
B = \begin{pmatrix} 4 & -3 \ 0 & 2 \ 1 & 0 \end{pmatrix}
B
=
(
4
−
3
0
2
1
0
)
;
C
=
(
1
0
2
4
0
−
3
)
C= \begin{pmatrix} 1 & 0 \ 2 & 4 \ 0 & -3 \end{pmatrix}
C
=
(
1
0
2
4
0
−
3
)
calcule
3
B
–
2
C
3B – 2 C
3
B
–2
C
Multiply Matrix B by
3
3
3
:
Step
1
1
1
: Multiply matrix B by
3
3
3
.
\newline
Matrix B =
(
4
−
3
0
2
1
0
)
\begin{pmatrix} 4 & -3 \ 0 & 2 \ 1 & 0 \end{pmatrix}
(
4
−
3
0
2
1
0
)
\newline
3
B
=
3
×
(
4
−
3
0
2
1
0
)
3B = 3 \times \begin{pmatrix} 4 & -3 \ 0 & 2 \ 1 & 0 \end{pmatrix}
3
B
=
3
×
(
4
−
3
0
2
1
0
)
\newline
=
(
12
−
9
0
6
3
0
)
\begin{pmatrix} 12 & -9 \ 0 & 6 \ 3 & 0 \end{pmatrix}
(
12
−
9
0
6
3
0
)
Multiply Matrix C by
2
2
2
:
Step
2
2
2
: Multiply matrix C by
2
2
2
.
\newline
Matrix C =
(
1
0
2
4
0
−
3
)
\begin{pmatrix} 1 & 0 \ 2 & 4 \ 0 & -3 \end{pmatrix}
(
1
0
2
4
0
−
3
)
\newline
2
C
=
2
×
(
1
0
2
4
0
−
3
)
2C = 2 \times \begin{pmatrix} 1 & 0 \ 2 & 4 \ 0 & -3 \end{pmatrix}
2
C
=
2
×
(
1
0
2
4
0
−
3
)
\newline
=
(
2
0
4
8
0
−
6
)
\begin{pmatrix} 2 & 0 \ 4 & 8 \ 0 & -6 \end{pmatrix}
(
2
0
4
8
0
−
6
)
Subtract
2
2
2
C from
3
3
3
B:
Step
3
3
3
: Subtract
2
C
2C
2
C
from
3
B
3B
3
B
.
3
B
−
2
C
=
∣
12
−
9
06
30
∣
−
∣
20
48
0
−
6
∣
3B - 2C = \left| \begin{array}{cc} 12 -9 \ 0 6 \ 3 0 \end{array} \right| - \left| \begin{array}{cc} 2 0 \ 4 8 \ 0 -6 \end{array} \right|
3
B
−
2
C
=
∣
∣
12
−
9
06
30
∣
∣
−
∣
∣
20
48
0
−
6
∣
∣
=
∣
12
−
2
−
9
−
0
0
−
46
−
8
3
−
00
+
6
∣
= \left| \begin{array}{cc} 12-2 -9-0 \ 0-4 6-8 \ 3-0 0+6 \end{array} \right|
=
∣
∣
12
−
2
−
9
−
0
0
−
46
−
8
3
−
00
+
6
∣
∣
=
∣
10
−
9
−
4
−
2
36
∣
= \left| \begin{array}{cc} 10 -9 \ -4 -2 \ 3 6 \end{array} \right|
=
∣
∣
10
−
9
−
4
−
2
36
∣
∣
More problems from Transformations of functions
Question
What does the transformation
f
(
x
)
↦
f
(
x
)
−
6
f(x) \mapsto f(x) - 6
f
(
x
)
↦
f
(
x
)
−
6
do to the graph of
f
(
x
)
f(x)
f
(
x
)
?
\newline
Choices:
\newline
(A)
translates it
6
units down
\text{translates it } 6 \text{ units down}
translates it
6
units down
\newline
(B)
translates it
6
units right
\text{translates it } 6 \text{ units right}
translates it
6
units right
\newline
(C)
translates it
6
units left
\text{translates it } 6 \text{ units left}
translates it
6
units left
\newline
(D)
translates it
6
units up
\text{translates it } 6 \text{ units up}
translates it
6
units up
Get tutor help
Posted 3 months ago
Question
Find
g
(
x
)
g(x)
g
(
x
)
, where
g
(
x
)
g(x)
g
(
x
)
is the reflection across the
y
y
y
-axis of
f
(
x
)
=
−
4
x
+
1
f(x) = -4x + 1
f
(
x
)
=
−
4
x
+
1
.
\newline
Choices:
\newline
[g(x) = 4x - 1]
\text{[g(x) = 4x - 1]}
[g(x) = 4x - 1]
\newline
[g(x) = -4x + 1]
\text{[g(x) = -4x + 1]}
[g(x) = -4x + 1]
\newline
[g(x) = 4x + 1]
\text{[g(x) = 4x + 1]}
[g(x) = 4x + 1]
\newline
[g(x) = -4x - 1]
\text{[g(x) = -4x - 1]}
[g(x) = -4x - 1]
Get tutor help
Posted 3 months ago
Question
What kind of transformation converts the graph of
f
(
x
)
=
−
8
(
x
−
4
)
2
+
6
f(x) = -8(x - 4)^2 + 6
f
(
x
)
=
−
8
(
x
−
4
)
2
+
6
into the graph of
g
(
x
)
=
−
8
(
x
+
6
)
2
+
6
g(x) = -8(x + 6)^2 + 6
g
(
x
)
=
−
8
(
x
+
6
)
2
+
6
?
\newline
Choices:
\newline
[A]translation 10 units left
\text{[A]translation 10 units left}
[A]translation 10 units left
\newline
[B]translation 10 units right
\text{[B]translation 10 units right}
[B]translation 10 units right
\newline
[C]translation 10 units up
\text{[C]translation 10 units up}
[C]translation 10 units up
\newline
[D]translation 10 units down
\text{[D]translation 10 units down}
[D]translation 10 units down
Get tutor help
Posted 3 months ago
Question
Find
g
(
x
)
g(x)
g
(
x
)
, where
g
(
x
)
g(x)
g
(
x
)
is the reflection across the x-axis of
f
(
x
)
=
−
6
(
x
−
7
)
2
+
2
f(x) = -6(x - 7)^2 + 2
f
(
x
)
=
−
6
(
x
−
7
)
2
+
2
.
\newline
Choices:
\newline
(A)
g
(
x
)
=
6
(
x
−
7
)
2
−
2
g(x) = 6(x - 7)^2 - 2
g
(
x
)
=
6
(
x
−
7
)
2
−
2
\newline
(B)
g
(
x
)
=
−
6
(
x
+
7
)
2
+
2
g(x) = -6(x + 7)^2 + 2
g
(
x
)
=
−
6
(
x
+
7
)
2
+
2
\newline
(C)
g
(
x
)
=
−
6
(
x
−
7
)
2
−
2
g(x) = -6(x - 7)^2 - 2
g
(
x
)
=
−
6
(
x
−
7
)
2
−
2
\newline
(D)
g
(
x
)
=
6
(
x
+
7
)
2
−
2
g(x) = 6(x + 7)^2 - 2
g
(
x
)
=
6
(
x
+
7
)
2
−
2
Get tutor help
Posted 3 months ago
Question
lim
x
→
π
4
cos
(
x
)
=
?
\lim _{x \rightarrow \frac{\pi}{4}} \cos (x)=?
x
→
4
π
lim
cos
(
x
)
=
?
\newline
Choose
1
1
1
answer:
\newline
(A)
1
2
\frac{1}{2}
2
1
\newline
(B)
1
1
1
\newline
(C)
2
2
\frac{\sqrt{2}}{2}
2
2
\newline
(D) The limit doesn't exist.
Get tutor help
Posted 3 months ago
Question
Tess tried to solve the differential equation
d
y
d
x
=
2
x
y
−
6
x
\frac{d y}{d x}=2 x y-6 x
d
x
d
y
=
2
x
y
−
6
x
. This is her work:
\newline
d
y
d
x
=
2
x
y
−
6
x
\frac{d y}{d x}=2 x y-6 x
d
x
d
y
=
2
x
y
−
6
x
\newline
Step
1
1
1
:
d
y
d
x
=
(
y
−
3
)
2
x
\quad \frac{d y}{d x}=(y-3) 2 x
d
x
d
y
=
(
y
−
3
)
2
x
\newline
Step
2
2
2
:
∫
1
y
−
3
d
y
=
∫
2
x
d
x
\quad \int \frac{1}{y-3} d y=\int 2 x d x
∫
y
−
3
1
d
y
=
∫
2
x
d
x
\newline
Step
3
3
3
:
ln
∣
y
−
3
∣
=
x
2
+
C
1
\quad \ln |y-3|=x^{2}+C_{1}
ln
∣
y
−
3∣
=
x
2
+
C
1
\newline
Step
4
4
4
:
e
ln
∣
y
−
3
∣
=
e
x
2
+
C
1
\quad e^{\ln |y-3|}=e^{x^{2}}+C_{1}
e
l
n
∣
y
−
3∣
=
e
x
2
+
C
1
\newline
Step
5
5
5
:
∣
y
−
3
∣
=
e
x
2
+
C
1
\quad|y-3|=e^{x^{2}}+C_{1}
∣
y
−
3∣
=
e
x
2
+
C
1
\newline
Step
6
6
6
:
y
−
3
=
±
e
x
2
+
C
2
\quad y-3= \pm e^{x^{2}}+C_{2}
y
−
3
=
±
e
x
2
+
C
2
\newline
Step
7
7
7
:
y
=
±
e
x
2
+
C
\quad y= \pm e^{x^{2}}+C
y
=
±
e
x
2
+
C
\newline
Is Tess's work correct? If not, what is her mistake?
\newline
Choose
1
1
1
answer:
\newline
(A) Tess's work is correct.
\newline
(B) Step
1
1
1
is incorrect. We're not allowed to factor an expression when solving differential equations.
\newline
(C) Step
4
4
4
is incorrect. The right-hand side of the equation should be
e
x
2
+
C
1
e^{x^{2}+C_{1}}
e
x
2
+
C
1
.
\newline
(D) Step
7
7
7
is incorrect. Tess didn't account for adding
3
3
3
to the right side.
Get tutor help
Posted 2 months ago
Question
Aditya tried to solve the differential equation
d
y
d
x
=
6
x
−
30
\frac{d y}{d x}=6 x-30
d
x
d
y
=
6
x
−
30
. This is his work:
\newline
d
y
d
x
=
6
x
−
30
\frac{d y}{d x}=6 x-30
d
x
d
y
=
6
x
−
30
\newline
Step
1
1
1
:
∫
30
d
y
=
∫
6
x
d
x
\quad \int 30 d y=\int 6 x d x
∫
30
d
y
=
∫
6
x
d
x
\newline
Step
2
2
2
:
30
y
=
3
x
2
+
C
\quad 30 y=3 x^{2}+C
30
y
=
3
x
2
+
C
\newline
Step
3
3
3
:
y
=
x
2
10
+
C
\quad y=\frac{x^{2}}{10}+C
y
=
10
x
2
+
C
\newline
Is Aditya's work correct? If not, what is his mistake?
\newline
Choose
1
1
1
answer:
\newline
(A) Aditya's work is correct.
\newline
(B) Step
1
1
1
is incorrect. The separation of variables wasn't done correctly.
\newline
(C) Step
2
2
2
is incorrect. Aditya didn't integrate
30
30
30
correctly.
\newline
(D) Step
3
3
3
is incorrect. The right-hand side of the equation should be
30
3
x
2
+
C
\frac{30}{3 x^{2}+C}
3
x
2
+
C
30
.
Get tutor help
Posted 3 months ago
Question
Jonael dropped a sandbag from a hot air balloon at a target that is
250
250
250
meters below the balloon. At this moment, the sandbag is
75
75
75
meters below the balloon.
\newline
Which
2
2
2
of the following expressions represents the distance between the sandbag and the target?
\newline
Choose
2
2
2
answers:
\newline
(A)
∣
−
250
+
75
∣
|-250+75|
∣
−
250
+
75∣
\newline
(B)
∣
250
+
75
∣
|250+75|
∣250
+
75∣
\newline
(C)
∣
−
75
+
250
∣
|-75+250|
∣
−
75
+
250∣
Get tutor help
Posted 3 months ago
Question
f
(
x
)
=
−
(
x
+
1
)
(
x
+
7
)
f(x) = -(x+1)(x+7)
f
(
x
)
=
−
(
x
+
1
)
(
x
+
7
)
\newline
The function represents a parabola in the
x
y
xy
x
y
-plane. Which of the following is an equivalent form of
f
f
f
in which the
y
y
y
-intercept of the graph of
f
f
f
appears as a constant or coefficient?
\newline
Choose
1
1
1
answer:
\newline
(A)
f
(
x
)
=
−
(
x
+
4
)
2
+
9
f(x) = -(x+4)^{2}+9
f
(
x
)
=
−
(
x
+
4
)
2
+
9
\newline
(B)
f
(
x
)
=
−
(
x
+
4
)
2
−
9
f(x) = -(x+4)^{2}-9
f
(
x
)
=
−
(
x
+
4
)
2
−
9
\newline
(C)
f
(
x
)
=
−
x
2
+
8
x
+
7
f(x) = -x^{2}+8x+7
f
(
x
)
=
−
x
2
+
8
x
+
7
\newline
(D)
f
(
x
)
=
−
x
2
−
8
x
−
7
f(x) = -x^{2}-8x-7
f
(
x
)
=
−
x
2
−
8
x
−
7
Get tutor help
Posted 3 months ago
Question
Determine whether the function
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
−
3
x=-3
x
=
−
3
.
\newline
f
(
x
)
=
{
18
−
x
2
,
x
≤
−
3
15
+
3
x
,
x
>
−
3
f(x)=\left\{\begin{array}{ll} 18-x^{2}, & x \leq-3 \\ 15+3 x, & x>-3 \end{array}\right.
f
(
x
)
=
{
18
−
x
2
,
15
+
3
x
,
x
≤
−
3
x
>
−
3
\newline
f
(
x
)
f(x)
f
(
x
)
is discontinuous at
x
=
−
3
x=-3
x
=
−
3
\newline
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
−
3
x=-3
x
=
−
3
Get tutor help
Posted 3 months ago