Resources
Testimonials
Plans
Sign in
Sign up
Resources
Testimonials
Plans
Home
Math Problems
Algebra 2
Power rule
\newline
Find the vertical and horizontal asymptotes of the function:
\newline
f
(
x
)
=
(
2
x
−
1
)
(
3
x
+
1
)
(
x
−
2
)
(
x
+
4
)
f(x)=\frac{(2 x-1)(3 x+1)}{(x-2)(x+4)}
f
(
x
)
=
(
x
−
2
)
(
x
+
4
)
(
2
x
−
1
)
(
3
x
+
1
)
\newline
The fields below accept a list of numbers or formulas separated by semicolons (e.g.
2
;
4
;
6
2 ; 4 ; 6
2
;
4
;
6
or
x
+
1
;
x
−
1
)
x+1 ; x-1)
x
+
1
;
x
−
1
)
. The order of the list does not matter.
\newline
Vertical asymptotes:
\newline
x
=
x=
x
=
\newline
Horizontal asymptotes:
\newline
y
=
y=
y
=
\newline
Get tutor help
The graph of
\newline
y
=
500
(
0.6
)
x
y=500(0.6)^{x}
y
=
500
(
0.6
)
x
is shown. Which of the following statements about the graph is true?
\newline
Choose
1
1
1
answer:
\newline
A As
x
x
x
increases,
y
y
y
increases at an increasing rate.
\newline
B As
x
x
x
increases,
y
y
y
increases at a decreasing rate.
\newline
(C) As
x
x
x
increases,
y
y
y
decreases at an increasing rate.
\newline
(D) As
x
x
x
increases,
y
y
y
decreases at a decreasing rate.
Get tutor help
y
=
x
k
y=\frac{x}{k}
y
=
k
x
\newline
In the given equation,
k
k
k
is a constant. If
x
x
x
were multipled by
1
2
\frac{1}{2}
2
1
, how would the value of
y
y
y
change?
\newline
Choose
1
1
1
answer:
\newline
A The value of
y
y
y
would be multiplied by
1
2
k
\frac{1}{2 k}
2
k
1
.
\newline
B The value of
y
y
y
would be multiplied by
1
2
\frac{1}{2}
2
1
.
\newline
(c) The value of
y
y
y
would not change.
\newline
(D) The value of
y
y
y
would be multiplied by
2
2
2
.
Get tutor help
Question
1
1
1
: Which functions have
(
12
,
3
)
(12,3)
(
12
,
3
)
as a solution? (Select ALL that apply):
\newline
(Select all that apply.)
\newline
y
=
4
x
y=4 x
y
=
4
x
\newline
y
=
0.25
x
y=0.25 x
y
=
0.25
x
\newline
y
=
x
+
9
y=x+9
y
=
x
+
9
\newline
y
=
x
4
y=x^{4}
y
=
x
4
\newline
y
=
x
−
9
y=x-9
y
=
x
−
9
\newline
y
=
1
4
x
y=\frac{1}{4} x
y
=
4
1
x
\newline
y
=
(
x
÷
6
)
+
1
y=(x \div 6)+1
y
=
(
x
÷
6
)
+
1
\newline
y
=
x
÷
4
y=x \div 4
y
=
x
÷
4
\newline
Select all the functions whose graphs include the point
(
12
,
3
)
(12,3)
(
12
,
3
)
.
\newline
You may the whiteboard to work out any of your solutions / show your reasoning.
Get tutor help
4
4
4
y
−
3
-3
−
3
x=
40
40
40
\newline
4
4
4
y=
3
3
3
x
−
30
-30
−
30
\newline
Which of the following accurately describes all solutions to the system of equations shown?
\newline
Choose
1
1
1
answer:
\newline
(A)
x
=
0
x=0
x
=
0
and
y
=
0
y=0
y
=
0
\newline
(B)
x
=
5
3
x=\frac{5}{3}
x
=
3
5
and
y
=
45
4
y=\frac{45}{4}
y
=
4
45
\newline
(C) There are infinite solutions to the system.
\newline
(D) There are no solutions to the system.
Get tutor help
Simplify. Assume all variables are positive.
\newline
w
3
2
w
5
2
\frac{w^{\frac{3}{2}}}{w^{\frac{5}{2}}}
w
2
5
w
2
3
\newline
\newline
Write your answer in the form
A
A
A
or
A
B
\frac{A}{B}
B
A
, where
A
A
A
and
B
B
B
are constants or variable expressions that have no variables in common. All exponents in your answer should be positive.
\newline
______
\newline
Get tutor help
Rewrite the expression without using a negative exponent.
\newline
(
1
)
/
(
−
2
y
−
2
)
(1)/(-2y^{-2})
(
1
)
/
(
−
2
y
−
2
)
\newline
Simplify your answer as much as possible.
Get tutor help
Which of the following equations represent functions? Assume
x
x
x
is the input and
y
y
y
is the output.
\newline
Multi-select Choices:
\newline
(A)
x
=
−
1
x = -1
x
=
−
1
\newline
(B)
y
=
x
y = x
y
=
x
\newline
(C)
y
=
1
y = 1
y
=
1
\newline
(D)
x
+
y
=
1
x + y = 1
x
+
y
=
1
\newline
(E)
x
=
0
x = 0
x
=
0
Get tutor help
4
4
4
. Use the distributive property to write an expression that is equivalent to each given expression.
\newline
(
2
2
2
)
−
3
(
2
x
−
4
)
-3(2 x-4)
−
3
(
2
x
−
4
)
\newline
b.
0.1
(
−
90
+
50
a
)
0.1(-90+50 a)
0.1
(
−
90
+
50
a
)
\newline
c.
−
7
(
−
x
−
9
)
-7(-x-9)
−
7
(
−
x
−
9
)
\newline
d.
4
5
(
10
y
+
−
x
+
−
15
)
\frac{4}{5}(10 y+-x+-15)
5
4
(
10
y
+
−
x
+
−
15
)
Get tutor help
4
4
4
. Use the distributive property to write an expression that is equivalent to each given expression.
\newline
(a)
−
3
(
2
x
−
4
)
-3(2 x-4)
−
3
(
2
x
−
4
)
\newline
b.
0.1
(
−
90
+
50
a
)
0.1(-90+50 a)
0.1
(
−
90
+
50
a
)
\newline
c.
−
7
(
−
x
−
9
)
-7(-x-9)
−
7
(
−
x
−
9
)
\newline
d.
4
5
(
10
y
+
−
x
+
−
15
)
\frac{4}{5}(10 y+-x+-15)
5
4
(
10
y
+
−
x
+
−
15
)
Get tutor help
Solution
\qquad
9
9
9
. What is the total weight of the pieces of driftwood Soo collects most often?
\newline
(A)
11
3
4
11 \frac{3}{4}
11
4
3
pounds
\newline
(B)
22
1
4
22 \frac{1}{4}
22
4
1
pounds
\newline
(c)
33
3
8
33 \frac{3}{8}
33
8
3
pounds
\newline
(D)
33
3
4
33 \frac{3}{4}
33
4
3
pounds
\newline
66
66
66
Lesson
27
27
27
Make Line Plots and Interpret Data
Get tutor help
Factor the following binomial completely.
\newline
49
x
2
+
36
49 x^{2}+36
49
x
2
+
36
\newline
Select the correct choice below and, if necessary, fill in the answer box to complete your ch
\newline
A.
49
x
2
+
36
=
49 x^{2}+36=
49
x
2
+
36
=
□
\square
□
\newline
B.
49
x
2
+
36
49 x^{2}+36
49
x
2
+
36
is prime.
Get tutor help
Complete the table for the given rule.
\newline
Rule:
y
=
x
−
1
4
y=x-\frac{1}{4}
y
=
x
−
4
1
\newline
\begin{tabular}{cc}
\newline
x
x
x
&
y
y
y
\\
\newline
\hline
1
4
\frac{1}{4}
4
1
&
□
\square
□
\\
\newline
13
4
\frac{13}{4}
4
13
&
□
\square
□
\newline
\end{tabular}
\newline
2
2
2
Get tutor help
m
/
m
a
t
h
/
\mathrm{m} / \mathrm{math} /
m
/
math
/
grade
−
8
-8
−
8
/describe-transformations
\newline
Which of the following transformations maps VWXY onto
V
′
W
′
X
′
Y
′
V^{\prime} W^{\prime} X^{\prime} Y^{\prime}
V
′
W
′
X
′
Y
′
?
\newline
translation right
6
6
6
units and down
13
13
13
units
Get tutor help
i-Ready
\newline
Strategies to Add and Subtract Rationals - Instruction - Level G
\newline
Which equation matches the number line?
\newline
−
2
+
(
−
1
4
)
=
−
2
1
4
-2+\left(-\frac{1}{4}\right)=-2 \frac{1}{4}
−
2
+
(
−
4
1
)
=
−
2
4
1
\newline
−
2
+
1
4
=
−
1
3
4
-2+\frac{1}{4}=-1 \frac{3}{4}
−
2
+
4
1
=
−
1
4
3
\newline
−
2
+
3
4
=
−
1
1
4
-2+\frac{3}{4}=-1 \frac{1}{4}
−
2
+
4
3
=
−
1
4
1
\newline
−
2
+
(
−
3
4
)
=
2
3
4
-2+\left(-\frac{3}{4}\right)=2 \frac{3}{4}
−
2
+
(
−
4
3
)
=
2
4
3
Get tutor help
Girls'
H
\mathrm{H}
H
;
\newline
Exercis The graph of
y
=
1
2
x
2
+
1
x
−
5
y=\frac{1}{2} x^{2}+\frac{1}{x}-5
y
=
2
1
x
2
+
x
1
−
5
is partially shown below.
\newline
(a) State the asymptote of the curve.
\newline
(b) By drawing a tangent on the given graph, find the
x
x
x
-coordinate of a point on the curve at which the gradient is
−
1
-1
−
1
.
\newline
(e) (i) On the same graph, draw the line
2
y
=
x
+
2
2 y=x+2
2
y
=
x
+
2
.
\newline
(ii) Write down an equation in
x
x
x
which is satisfied by the
x
x
x
-coordinates of the points of intersection of the
2
2
2
graphs,
y
=
1
2
x
2
+
1
x
−
5
y=\frac{1}{2} x^{2}+\frac{1}{x}-5
y
=
2
1
x
2
+
x
1
−
5
and
2
y
=
x
+
2
2 y=x+2
2
y
=
x
+
2
. (It is not necessary to simplify the equation.)
\newline
(d) Using the graph(s),
\newline
(i) state the roots of the equation
1
2
x
2
+
1
x
=
5
\frac{1}{2} x^{2}+\frac{1}{x}=5
2
1
x
2
+
x
1
=
5
,
\newline
(ii) state the range of values of
x
x
x
for which the curve is decreasing,
\newline
(iii) find the range of values of
x
x
x
for which
y
=
1
2
x
2
+
1
x
−
5
y=\frac{1}{2} x^{2}+\frac{1}{x}-5
y
=
2
1
x
2
+
x
1
−
5
1
1
1
.
\newline
(v) draw by ensuing trat bofween
Get tutor help
Which of the equations are true identities?
\newline
A.
(
a
+
b
)
(
2
a
+
1
)
=
a
(
2
a
+
2
b
+
1
)
(a+b)(2 a+1)=a(2 a+2 b+1)
(
a
+
b
)
(
2
a
+
1
)
=
a
(
2
a
+
2
b
+
1
)
\newline
B.
(
n
+
2
)
2
−
n
2
=
4
(
n
+
1
)
(n+2)^{2}-n^{2}=4(n+1)
(
n
+
2
)
2
−
n
2
=
4
(
n
+
1
)
\newline
Choose
1
1
1
answer:
\newline
(A) Only A
\newline
(B) Only B
\newline
(C) Both A and B
\newline
(D) Neither A nor B
Get tutor help
Which of the equations are true identities?
\newline
A.
(
x
2
+
2
y
)
(
x
2
−
3
y
)
=
x
4
−
6
y
2
\left(x^{2}+2 y\right)\left(x^{2}-3 y\right)=x^{4}-6 y^{2}
(
x
2
+
2
y
)
(
x
2
−
3
y
)
=
x
4
−
6
y
2
\newline
B.
k
3
−
r
3
=
(
k
2
+
r
)
(
k
−
r
2
)
k^{3}-r^{3}=\left(k^{2}+r\right)\left(k-r^{2}\right)
k
3
−
r
3
=
(
k
2
+
r
)
(
k
−
r
2
)
\newline
Choose
1
1
1
answer:
\newline
(A) Only A
\newline
(B) Only B
\newline
(C) Both A and B
\newline
(D) Neither A nor B
Get tutor help
Which of the equations are true identities?
\newline
A.
n
(
n
−
2
)
(
n
+
2
)
=
n
3
−
4
n
n(n-2)(n+2)=n^{3}-4 n
n
(
n
−
2
)
(
n
+
2
)
=
n
3
−
4
n
\newline
B.
(
x
+
1
)
2
−
2
x
+
y
2
=
x
2
+
y
2
+
1
(x+1)^{2}-2 x+y^{2}=x^{2}+y^{2}+1
(
x
+
1
)
2
−
2
x
+
y
2
=
x
2
+
y
2
+
1
\newline
Choose
1
1
1
answer:
\newline
(A) Only A
\newline
(B) Only B
\newline
(C) Both A and B
\newline
(D) Neither A nor B
Get tutor help
How does
h
(
x
)
=
1
0
x
h(x) = 10^x
h
(
x
)
=
1
0
x
change over the interval from
x
=
5
x = 5
x
=
5
to
x
=
6
x = 6
x
=
6
?
\newline
Choices:
\newline
(A)
h
(
x
)
h(x)
h
(
x
)
increases by
10
%
10\%
10%
\newline
(B)
h
(
x
)
h(x)
h
(
x
)
increases by
900
%
900\%
900%
\newline
(C)
h
(
x
)
h(x)
h
(
x
)
increases by
10
10
10
\newline
(D)
h
(
x
)
h(x)
h
(
x
)
decreases by
10
10
10
Get tutor help
- Let
g
g
g
be a function such that
g
(
5
)
=
7
g(5)=7
g
(
5
)
=
7
and
g
′
(
5
)
=
−
2
g^{\prime}(5)=-2
g
′
(
5
)
=
−
2
.
\newline
- Let
h
h
h
be the function
h
(
x
)
=
x
h(x)=x
h
(
x
)
=
x
.
\newline
Evaluate
d
d
x
[
g
(
x
)
⋅
h
(
x
)
]
\frac{d}{d x}[g(x) \cdot h(x)]
d
x
d
[
g
(
x
)
⋅
h
(
x
)]
at
x
=
5
x=5
x
=
5
.
Get tutor help
Let
y
=
x
cos
(
x
)
y=\sqrt{x} \cos (x)
y
=
x
cos
(
x
)
.
\newline
Find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
Choose
1
1
1
answer:
\newline
(A)
−
1
x
+
sin
(
x
)
-\frac{1}{\sqrt{x}}+\sin (x)
−
x
1
+
sin
(
x
)
\newline
(B)
cos
(
x
)
2
x
−
x
sin
(
x
)
\frac{\cos (x)}{2 \sqrt{x}}-\sqrt{x} \sin (x)
2
x
c
o
s
(
x
)
−
x
sin
(
x
)
\newline
(C)
−
2
x
cos
(
x
)
−
x
sin
(
x
)
-2 \sqrt{x} \cos (x)-\sqrt{x} \sin (x)
−
2
x
cos
(
x
)
−
x
sin
(
x
)
\newline
(D)
−
sin
(
x
)
x
-\frac{\sin (x)}{\sqrt{x}}
−
x
s
i
n
(
x
)
Get tutor help
Which statement is true about the value of the expression below?
\newline
(
−
2
3
)
−
2
(-2^{3})^{-2}
(
−
2
3
)
−
2
\newline
(A) It is between
0
0
0
and
1
1
1
.
\newline
(B) It is less than
−
1
-1
−
1
.
\newline
(C) It is between
−
1
-1
−
1
and
0
0
0
.
\newline
(D) It is greater than
1
1
1
.
Get tutor help
Let it be known that
\newline
lim
x
→
1
f
(
x
)
=
+
∞
,
lim
x
→
1
g
(
x
)
=
−
∞
,
lim
x
→
1
v
(
x
)
=
0
,
\lim _{x \rightarrow 1} f(x)=+\infty, \lim _{x \rightarrow 1} g(x)=-\infty, \lim _{x \rightarrow 1} v(x)=0 \text {, }
x
→
1
lim
f
(
x
)
=
+
∞
,
x
→
1
lim
g
(
x
)
=
−
∞
,
x
→
1
lim
v
(
x
)
=
0
,
\newline
Find the values of the following limits:
\newline
3
3
3
)
lim
x
→
1
(
g
(
x
)
+
v
(
x
)
)
\lim _{x \rightarrow 1}(g(x)+v(x))
lim
x
→
1
(
g
(
x
)
+
v
(
x
))
;
\newline
4
4
4
)
lim
x
→
1
1
g
(
x
)
\lim _{x \rightarrow 1} \frac{1}{g(x)}
lim
x
→
1
g
(
x
)
1
;
\newline
5
5
5
)
lim
x
→
1
v
(
x
)
f
(
x
)
\lim _{x \rightarrow 1} \frac{v(x)}{f(x)}
lim
x
→
1
f
(
x
)
v
(
x
)
;
Get tutor help
A. Write the given mapping as (a) set of ordered pairs and (b) table of values,
\newline
1
1
1
.
\newline
2
2
2
.
\newline
3
3
3
.
\newline
4
4
4
.
\newline
5
5
5
.
Get tutor help
A balanced coin with one side heads
(
H
)
(H)
(
H
)
and the other side tails
(
T
)
(T)
(
T
)
is repeatedly flipped, and the results are recorded. The coefficients in the expansion of
(
H
+
T
)
n
(H+T)^{n}
(
H
+
T
)
n
give the number of ways to get a particular combination of heads and tails for
n
n
n
flips of the coin. For example, in the expansion of
(
H
+
T
)
7
(H+T)^{7}
(
H
+
T
)
7
, the term
21
H
2
T
5
21 H^{2} T^{5}
21
H
2
T
5
means there are
21
21
21
ways to get exactly
2
2
2
heads and
5
5
5
tails when flipping a coin
7
7
7
times. How many ways are there to get exactly
2
2
2
heads and
4
4
4
tails when flipping a balanced coin
6
6
6
times?
\newline
(A)
6
6
6
\newline
(B)
15
15
15
\newline
(C)
20
20
20
\newline
(D)
48
48
48
Get tutor help
Given that
y
=
e
2
x
x
y=\frac{e^{2 x}}{x}
y
=
x
e
2
x
for
x
>
0
x>0
x
>
0
, find the range of values of
x
x
x
for which
y
y
y
is decreasing.
Get tutor help
(
7
7
7
pts) Let
f
(
x
)
=
2
x
+
x
f(x)=2^{x}+x
f
(
x
)
=
2
x
+
x
. Use the theorem on derivatives of inverses to find
(
f
−
1
)
′
(
11
)
\left(f^{-1}\right)^{\prime}(11)
(
f
−
1
)
′
(
11
)
.
Get tutor help
7
7
7
. Which statement below best explains how to subtract
14
15
−
2
3
\frac{14}{15}-\frac{2}{3}
15
14
−
3
2
?
\newline
A. Subtract the numerators first then subtract the denominators.
\newline
B. Write an equivalent fraction for
14
15
\frac{14}{15}
15
14
with
3
3
3
as the new denominator. Then subtract only the numerators.
\newline
C. Write equivalent fractions for both fractions with
45
45
45
as the denominator. Then subtract the numerators and denominators separately.
\newline
D. Write an equivalent fraction for
2
3
\frac{2}{3}
3
2
with
15
15
15
as the new denominator. Then subtract only the numerators.
Get tutor help
(
2
x
+
5
)
(
−
m
x
+
9
)
=
0
(2 x+5)(-m x+9)=0
(
2
x
+
5
)
(
−
m
x
+
9
)
=
0
\newline
In the given equation,
m
m
m
is a constant. If the equation has the solutions
x
=
−
5
2
x=-\frac{5}{2}
x
=
−
2
5
and
x
=
3
2
x=\frac{3}{2}
x
=
2
3
, what is the value of
m
m
m
?
\newline
□
\square
□
Get tutor help
Simplify. Assume all variables are positive.
\newline
b
7
4
b
11
4
⋅
b
7
4
\frac{b^{\frac{7}{4}}}{b^{\frac{11}{4}} \cdot b^{\frac{7}{4}}}
b
4
11
⋅
b
4
7
b
4
7
\newline
Write your answer in the form
A
A
A
or
A
B
\frac{A}{B}
B
A
, where
A
A
A
and
B
B
B
are constants or variable expressions that have no variables in common. All exponents in your answer should be positive.
\newline
______
Get tutor help
Simplify. Assume all variables are positive.
\newline
b
8
3
b
4
3
⋅
b
2
3
\frac{b^{\frac{8}{3}}}{b^{\frac{4}{3}} \cdot b^{\frac{2}{3}}}
b
3
4
⋅
b
3
2
b
3
8
\newline
Write your answer in the form
A
A
A
or
A
B
\frac{A}{B}
B
A
, where
A
A
A
and
B
B
B
are constants or variable expressions that have no variables in common. All exponents in your answer should be positive.
\newline
______
Get tutor help
Simplify. Assume all variables are positive.
\newline
u
11
4
u
7
4
⋅
u
5
4
\frac{u^{\frac{11}{4}}}{u^{\frac{7}{4}} \cdot u^{\frac{5}{4}}}
u
4
7
⋅
u
4
5
u
4
11
\newline
Write your answer in the form
A
A
A
or
A
B
\frac{A}{B}
B
A
, where
A
A
A
and
B
B
B
are constants or variable expressions that have no variables in common. All exponents in your answer should be positive.
\newline
______
Get tutor help
Simplify. Assume all variables are positive.
\newline
r
3
4
r
5
4
\frac{r^{\frac{3}{4}}}{r^{\frac{5}{4}}}
r
4
5
r
4
3
\newline
Write your answer in the form
A
A
A
or
A
B
\frac{A}{B}
B
A
, where
A
A
A
and
B
B
B
are constants or variable expressions that have no variables in common. All exponents in your answer should be positive.
\newline
______
Get tutor help
Simplify. Assume all variables are positive.
\newline
r
3
4
r
11
4
\frac{r^{\frac{3}{4}}}{r^{\frac{11}{4}}}
r
4
11
r
4
3
\newline
Write your answer in the form
A
A
A
or
A
B
\frac{A}{B}
B
A
, where
A
A
A
and
B
B
B
are constants or variable expressions that have no variables in common. All exponents in your answer should be positive.
\newline
______
Get tutor help
Simplify.
\newline
(
2
m
2
3
m
−
1
)
2
\left(\frac{2 m^{2}}{3 m^{-1}}\right)^{2}
(
3
m
−
1
2
m
2
)
2
\newline
Write your answer using only positive exponents.
Get tutor help
Which of the following statements about the graph of
y
=
12
(
0.75
)
x
y=12(0.75)^{x}
y
=
12
(
0.75
)
x
is true?
\newline
Choose
1
1
1
answer:
\newline
(A) As
x
x
x
increases,
y
y
y
increases at an increasing rate.
\newline
(B) As
x
x
x
increases,
y
y
y
increases at a decreasing rate.
\newline
(C) As
x
x
x
increases,
y
y
y
decreases at an increasing rate.
\newline
(D) As
x
x
x
increases,
y
y
y
decreases at a decreasing rate.
Get tutor help
c)
\newline
h
Write ne equation in factored form.
.
h
=
−
5
(
t
2
+
3
t
−
10
)
\begin{array}{l} h^{\text {Write ne equation in factored form. }} \text {. } \\ h=-5\left(t^{2}+3 t-10\right) \end{array}
h
Write ne equation in factored form.
.
h
=
−
5
(
t
2
+
3
t
−
10
)
\newline
d) Find the roots. [
2
2
2
marks]
\newline
e) When will the rock hit the ground?
\newline
f) At what time will the rock reach its maxim
Get tutor help
Parts of algebraic expressions
\newline
Consider the expression
\newline
1
2
(
2
a
+
1
)
(
a
+
3
)
.
\frac{1}{2}(2 a+1)(a+3) \text {. }
2
1
(
2
a
+
1
)
(
a
+
3
)
.
\newline
Complete
2
2
2
descriptions of the parts of the expression.
\newline
1
1
1
. The entire expression is the product of
\newline
2
2
2
. On its own,
(
2
a
+
1
)
(2 a+1)
(
2
a
+
1
)
is a sum with
\newline
2
2
2
factors
\newline
3
3
3
factors
\newline
2
2
2
terms
\newline
3
3
3
terms
Get tutor help
rac{
6
6
6
}{
5
5
5
}p + kq = rac{
4
4
4
}{
5
5
5
} q = rac{
3
3
3
}{
5
5
5
}p - rac{
2
2
2
}{
5
5
5
} Consider the system of equations, where
k
k
k
is a constant. For which value of
k
k
k
is there no
(
p
,
q
)
(p,q)
(
p
,
q
)
solutions?
Get tutor help
Prove the identity.
\newline
cos
x
1
+
sin
x
=
sec
x
−
tan
x
\frac{\cos x}{1+\sin x}=\sec x-\tan x
1
+
sin
x
cos
x
=
sec
x
−
tan
x
\newline
Note that each Statement must be based on a Rule chosen from the Rule menu. To see a detail the right of the Rule.
Get tutor help
Which of the following relationships proves why
△
A
B
D
\triangle \mathrm{ABD}
△
ABD
and
△
C
B
D
\triangle \mathrm{CBD}
△
CBD
are congruent?
\newline
ASA
\newline
H
L
\mathrm{HL}
HL
\newline
SAS
\newline
Overlapping Triangles
\newline
Segments
A
B
A B
A
B
and
C
B
C B
CB
are congruent.
Get tutor help
f
(
x
)
=
2
x
5
+
x
4
−
18
x
3
−
17
x
2
+
20
x
+
12
f(x)=2 x^{5}+x^{4}-18 x^{3}-17 x^{2}+20 x+12
f
(
x
)
=
2
x
5
+
x
4
−
18
x
3
−
17
x
2
+
20
x
+
12
\newline
The function
f
f
f
is shown. If
x
−
3
x-3
x
−
3
is a factor of
f
f
f
, what is the value of
f
(
3
)
f(3)
f
(
3
)
Get tutor help
Calculus
1
1
1
: Exam
2
2
2
\newline
Name:
\newline
15
15
15
. (
8
8
8
pts) Use the first derivative test to find and classify all local extrema in the interval
(
−
4
,
5
)
(-4,5)
(
−
4
,
5
)
for the function
\newline
f
(
x
)
=
(
5
x
−
4
)
e
4
x
f(x)=(5 x-4) e^{4 x}
f
(
x
)
=
(
5
x
−
4
)
e
4
x
\newline
If there is more than one local maximum or minimum, list them all. If a local maximum or a local minimum does not occur on the interval, note that in your work. You may assume that
e
4
x
e^{4 x}
e
4
x
is a positive number for all values of
x
x
x
.
Get tutor help
Given:
P
(
x
)
=
x
4
+
x
3
−
11
х
2
−
9
x
+
18
P(x) = x^4 + x^3 - 11х^2 - 9x + 18
P
(
x
)
=
x
4
+
x
3
−
11
х
2
−
9
x
+
18
with factors
(
x
+
3
)
(x + 3)
(
x
+
3
)
and
(
x
−
1
)
(x - 1)
(
x
−
1
)
. A. Write
P
(
x
)
P(x)
P
(
x
)
in factored form. B. Find the zeros of the function.
Get tutor help
Use the reduction of order method to find the general solutio of the following equations. One solution of the homogeneous is shown alongside each equation.
\newline
13
13
13
.
2
x
2
y
′
′
+
3
x
y
′
−
y
=
1
x
,
y
1
=
x
1
/
2
2 x^{2} y^{\prime \prime}+3 x y^{\prime}-y=\frac{1}{x}, \quad y_{1}=x^{1 / 2}
2
x
2
y
′′
+
3
x
y
′
−
y
=
x
1
,
y
1
=
x
1/2
.
Get tutor help
The graph of
y
=
2
x
2
−
4
x
−
4
y=2x^2-4x-4
y
=
2
x
2
−
4
x
−
4
is shown in the
x
y
xy
x
y
plane. Which of the following characteristics of the graph is displayed as a constant or coefficient in the equation as written?
Get tutor help
Try Again
\newline
One or more of your answers are incorrect.
\newline
First, rewrite
\newline
4
5
\frac{4}{5}
5
4
and
\newline
9
11
\frac{9}{11}
11
9
so that they have a common denominator. Then, use
\newline
<
<
<
,
=
=
=
, or
\newline
>
>
>
to order
\newline
4
5
\frac{4}{5}
5
4
and
\newline
9
11
\frac{9}{11}
11
9
.
Get tutor help
Iestion
7
7
7
of
11
11
11
, Step
1
1
1
of
1
1
1
\newline
Correct
\newline
Find a formula for the inverse of the following function, if possible.
\newline
A
(
x
)
=
1
x
−
1
A(x)=\frac{1}{x-1}
A
(
x
)
=
x
−
1
1
\newline
Answer
\newline
How to enter your answer (opens in new window)
\newline
Selecting a radio button will replace the entered answer value(s) with the radio button value. If the radio button is not selected, the entered an
\newline
A
−
1
(
x
)
=
A^{-1}(x)=
A
−
1
(
x
)
=
\newline
does not have an inverse functi
Get tutor help
iestion
7
7
7
of
11
11
11
, Step
1
1
1
of
1
1
1
\newline
Correct
\newline
nd a formula for the inverse of the following function, if possible.
\newline
A
(
x
)
=
1
x
−
1
A(x)=\frac{1}{x-1}
A
(
x
)
=
x
−
1
1
\newline
nswer
\newline
How to enter your answer (opens in new window)
\newline
Selecting a radio button will replace the entered answer value(s) with the radio button value. If the radio button is not selected, the entered an
\newline
A
−
1
(
x
)
=
A^{-1}(x)=
A
−
1
(
x
)
=
\newline
does not have an inverse functi
Get tutor help
1
2
3
Next
