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Math Problems
Calculus
Evaluate definite integrals using the power rule
∫
x
2
−
5
x
+
6
2
x
−
6
d
x
=
\int \frac{x^{2}-5 x+6}{2 x-6} d x=
∫
2
x
−
6
x
2
−
5
x
+
6
d
x
=
Get tutor help
∫
2
x
d
x
=
\int 2^{x} d x=
∫
2
x
d
x
=
Get tutor help
∫
1
e
2
x
3
+
1
x
d
x
=
\int_{1}^{e^{2}} \frac{x^{3}+1}{x} d x=
∫
1
e
2
x
x
3
+
1
d
x
=
Get tutor help
∫
1
4
(
3
x
+
x
)
d
x
=
\int_{1}^{4}(3 x+\sqrt{x}) d x=
∫
1
4
(
3
x
+
x
)
d
x
=
Get tutor help
∫
(
sin
2
x
cos
4
2
x
+
1
(
2
x
+
1
)
ln
(
2
x
+
1
)
)
d
x
=
\int\left(\frac{\sin 2 x}{\cos ^{4} 2 x}+\frac{1}{(2 x+1) \ln (2 x+1)}\right) d x=
∫
(
c
o
s
4
2
x
s
i
n
2
x
+
(
2
x
+
1
)
l
n
(
2
x
+
1
)
1
)
d
x
=
Get tutor help
(
x
2
+
1
)
y
′
=
x
⋅
y
⋅
(
cos
(
7
⋅
π
/
2
)
⋅
y
2
−
7
2
7
⋅
y
+
sin
(
7
π
/
2
)
⋅
(
1
+
x
2
+
1
y
)
)
,
y
(
0
)
=
=
\left(x^{2}+1\right) y^{\prime}=x \cdot y \cdot\left(\cos (7 \cdot \pi / 2) \cdot \frac{y^{2}-7^{2}}{7 \cdot y}+\sin (7 \pi / 2) \cdot\left(1+\frac{x^{2}+1}{y}\right)\right), y(0)==
(
x
2
+
1
)
y
′
=
x
⋅
y
⋅
(
cos
(
7
⋅
π
/2
)
⋅
7
⋅
y
y
2
−
7
2
+
sin
(
7
π
/2
)
⋅
(
1
+
y
x
2
+
1
)
)
,
y
(
0
)
==
Get tutor help
Unit
6
⋅
6 \cdot
6
⋅
Lesson
9
9
9
\newline
Real World Problems
\newline
Write an equation. Then solve.
\newline
Shol
\newline
(
20
20
20
) Dimitri rode his bike
32
32
32
miles yesterday. He rode
12
4
5
12 \frac{4}{5}
12
5
4
miles before lunch and the rest of the distance after lunch. How far did he ride after lunch?
\qquad
\newline
21
21
21
Ms. Washington is taking an accounting class. Each class is
3
4
\frac{3}{4}
4
3
hour long. If there are
22
22
22
classes in all, how many hours will Ms. Washington spend in class?
\qquad
\newline
Elin bought a large watermelon at the farmers market.
\newline
She cut off a
5
5
8
5 \frac{5}{8}
5
8
5
-pound piece and gave it to her neighbor. She has
11
5
8
11 \frac{5}{8}
11
8
5
pounds of watermelon left. low much did the whole watermelon weigh?
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The population of a town grows at a rate of
r
(
t
)
r(t)
r
(
t
)
people per year (where
t
t
t
is time in years).
\newline
What does
∫
2
4
r
(
t
)
d
t
\int_{2}^{4}r(t)\,dt
∫
2
4
r
(
t
)
d
t
represent?
\newline
Choose
1
1
1
answer:
\newline
(A) The average rate at which the population grew between the second and the fourth year.
\newline
(B) The change in number of people between the second and the fourth year.
\newline
(C) The number of people in the town on the fourth year.
\newline
(D) The time it took for the town to grow from a population of
2
2
2
people to a population of
4
4
4
people.
Get tutor help
∫
3
t
−
4
(
2
+
4
t
−
3
)
7
d
t
=
\int 3 t^{-4}\left(2+4 t^{-3}\right)^{7} d t=
∫
3
t
−
4
(
2
+
4
t
−
3
)
7
d
t
=
Get tutor help
12
12
12
).
∫
cos
3
x
sin
2
x
−
2
d
x
=
\int \frac{\cos ^{3} x}{\sin ^{2} x-2} d x=
∫
s
i
n
2
x
−
2
c
o
s
3
x
d
x
=
Get tutor help
g
(
x
)
=
∫
0
x
(
5
t
2
−
t
)
d
t
g(x)=\int_{0}^{x}\left(5 t^{2}-t\right) d t
g
(
x
)
=
∫
0
x
(
5
t
2
−
t
)
d
t
\newline
g
′
(
2
)
=
g^{\prime}(2)=
g
′
(
2
)
=
Get tutor help
g
(
x
)
=
∫
−
12
x
(
4
t
−
7
)
d
t
g(x)=\int_{-12}^{x}(4 t-7) d t
g
(
x
)
=
∫
−
12
x
(
4
t
−
7
)
d
t
\newline
g
′
(
5
)
=
g^{\prime}(5)=
g
′
(
5
)
=
Get tutor help
g
(
x
)
=
∫
−
10
x
t
2
+
11
d
t
g(x)=\int_{-10}^{x} \sqrt{t^{2}+11} d t
g
(
x
)
=
∫
−
10
x
t
2
+
11
d
t
\newline
g
′
(
−
5
)
=
g^{\prime}(-5)=
g
′
(
−
5
)
=
Get tutor help
g
(
x
)
=
∫
−
10
x
(
10
−
3
t
)
d
t
g(x)=\int_{-10}^{x}(10-3 t) d t
g
(
x
)
=
∫
−
10
x
(
10
−
3
t
)
d
t
\newline
g
′
(
−
4
)
=
g^{\prime}(-4)=
g
′
(
−
4
)
=
Get tutor help
g
(
x
)
=
∫
−
7
x
(
t
−
t
2
)
d
t
g(x)=\int_{-7}^{x}\left(t-t^{2}\right) d t
g
(
x
)
=
∫
−
7
x
(
t
−
t
2
)
d
t
\newline
g
′
(
10
)
=
g^{\prime}(10)=
g
′
(
10
)
=
Get tutor help
F
(
x
)
=
∫
1
2
x
1
+
t
3
d
t
F(x)=\int_{1}^{2 x} \sqrt{1+t^{3}} d t
F
(
x
)
=
∫
1
2
x
1
+
t
3
d
t
\newline
F
′
(
x
)
=
F^{\prime}(x)=
F
′
(
x
)
=
Get tutor help
F
(
x
)
=
∫
−
2
2
x
(
3
t
2
+
2
t
)
d
t
F(x)=\int_{-2}^{2 x}\left(3 t^{2}+2 t\right) d t
F
(
x
)
=
∫
−
2
2
x
(
3
t
2
+
2
t
)
d
t
\newline
F
′
(
x
)
=
F^{\prime}(x)=
F
′
(
x
)
=
Get tutor help
F
(
x
)
=
∫
2
2
x
15
−
t
d
t
F(x)=\int_{2}^{2 x} \sqrt{15-t} d t
F
(
x
)
=
∫
2
2
x
15
−
t
d
t
\newline
F
′
(
x
)
=
F^{\prime}(x)=
F
′
(
x
)
=
Get tutor help
g
(
x
)
=
∫
2
x
(
10
t
+
2
)
d
t
g(x)=\int_{2}^{x}(10 t+2) d t
g
(
x
)
=
∫
2
x
(
10
t
+
2
)
d
t
\newline
g
′
(
3
)
=
g^{\prime}(3)=
g
′
(
3
)
=
Get tutor help
g
(
x
)
=
∫
0
x
5
+
4
cos
t
d
t
g(x)=\int_{0}^{x} \sqrt{5+4 \cos t} d t
g
(
x
)
=
∫
0
x
5
+
4
cos
t
d
t
\newline
g
′
(
π
)
=
g^{\prime}(\pi)=
g
′
(
π
)
=
Get tutor help
F
(
x
)
=
∫
1
2
x
t
2
t
2
+
1
d
t
F(x)=\int_{1}^{2 x} \frac{t^{2}}{t^{2}+1} d t
F
(
x
)
=
∫
1
2
x
t
2
+
1
t
2
d
t
\newline
F
′
(
x
)
=
F^{\prime}(x)=
F
′
(
x
)
=
Get tutor help
g
(
x
)
=
∫
0
x
5
+
4
tan
t
d
t
g(x)=\int_{0}^{x} \sqrt{5+4 \tan t} d t
g
(
x
)
=
∫
0
x
5
+
4
tan
t
d
t
\newline
g
′
(
π
4
)
=
g^{\prime}\left(\frac{\pi}{4}\right)=
g
′
(
4
π
)
=
Get tutor help
g
(
x
)
=
∫
0
x
t
2
−
t
t
d
t
g(x)=\int_{0}^{x} \frac{t^{2}-t}{\sqrt{t}} d t
g
(
x
)
=
∫
0
x
t
t
2
−
t
d
t
\newline
g
′
(
4
)
=
g^{\prime}(4)=
g
′
(
4
)
=
Get tutor help
F
(
x
)
=
∫
0
x
2
t
d
t
F(x)=\int_{0}^{\sqrt{x}} 2 t d t
F
(
x
)
=
∫
0
x
2
t
d
t
\newline
where
x
>
0
x>0
x
>
0
.
\newline
F
′
(
x
)
=
F^{\prime}(x)=
F
′
(
x
)
=
Get tutor help
g
(
x
)
=
∫
−
π
x
sin
(
t
)
d
t
g(x)=\int_{-\pi}^{x} \sin (t) d t
g
(
x
)
=
∫
−
π
x
sin
(
t
)
d
t
\newline
g
′
(
π
)
=
g^{\prime}(\pi)=
g
′
(
π
)
=
Get tutor help
F
(
x
)
=
x
+
7
F(x)=\sqrt{x+7}
F
(
x
)
=
x
+
7
\newline
f
(
x
)
=
F
′
(
x
)
f(x)=F^{\prime}(x)
f
(
x
)
=
F
′
(
x
)
\newline
∫
2
9
f
(
x
)
d
x
=
\int_{2}^{9} f(x) d x=
∫
2
9
f
(
x
)
d
x
=
Get tutor help
H
(
x
)
=
−
x
−
5
H(x)=-x-5
H
(
x
)
=
−
x
−
5
\newline
h
(
x
)
=
H
′
(
x
)
h(x)=H^{\prime}(x)
h
(
x
)
=
H
′
(
x
)
\newline
∫
−
3
6
h
(
x
)
d
x
=
\int_{-3}^{6} h(x) d x=
∫
−
3
6
h
(
x
)
d
x
=
Get tutor help
H
(
x
)
=
1
0
x
H(x)=10^{x}
H
(
x
)
=
1
0
x
\newline
h
(
x
)
=
H
′
(
x
)
h(x)=H^{\prime}(x)
h
(
x
)
=
H
′
(
x
)
\newline
∫
−
3
2
h
(
x
)
d
x
=
\int_{-3}^{2} h(x) d x=
∫
−
3
2
h
(
x
)
d
x
=
Get tutor help
G
(
x
)
=
2
x
3
−
4
G(x)=2 x^{3}-4
G
(
x
)
=
2
x
3
−
4
\newline
g
(
x
)
=
G
′
(
x
)
g(x)=G^{\prime}(x)
g
(
x
)
=
G
′
(
x
)
\newline
∫
1
2
g
(
x
)
d
x
=
\int_{1}^{2} g(x) d x=
∫
1
2
g
(
x
)
d
x
=
Get tutor help
G
(
x
)
=
3
x
G(x)=\sqrt{3 x}
G
(
x
)
=
3
x
\newline
g
(
x
)
=
G
′
(
x
)
g(x)=G^{\prime}(x)
g
(
x
)
=
G
′
(
x
)
\newline
∫
3
12
g
(
x
)
d
x
=
\int_{3}^{12} g(x) d x=
∫
3
12
g
(
x
)
d
x
=
Get tutor help
F
(
x
)
=
5
x
F(x)=5^{x}
F
(
x
)
=
5
x
\newline
f
(
x
)
=
F
′
(
x
)
f(x)=F^{\prime}(x)
f
(
x
)
=
F
′
(
x
)
\newline
∫
0
3
f
(
x
)
d
x
=
\int_{0}^{3} f(x) d x=
∫
0
3
f
(
x
)
d
x
=
Get tutor help
G
(
x
)
=
cos
(
3
x
)
G(x)=\cos (3 x)
G
(
x
)
=
cos
(
3
x
)
\newline
g
(
x
)
=
G
′
(
x
)
g(x)=G^{\prime}(x)
g
(
x
)
=
G
′
(
x
)
\newline
∫
0
π
g
(
x
)
d
x
=
\int_{0}^{\pi} g(x) d x=
∫
0
π
g
(
x
)
d
x
=
Get tutor help
∫
cos
3
(
x
)
sin
(
x
)
d
x
=
\int \cos ^{3}(x) \sin (x) d x=
∫
cos
3
(
x
)
sin
(
x
)
d
x
=
Get tutor help
LINEAR EQUATIONS
\newline
WARM-UP
\newline
1
1
1
. Kia's plant is
8
1
5
8 \frac{1}{5}
8
5
1
inches tall and is growing
2
3
\frac{2}{3}
3
2
inch each week. Lyric's plant is
10
7
10
10 \frac{7}{10}
10
10
7
inches tall and is growing
7
6
\frac{7}{6}
6
7
inch each week. Write and solve an equation to find how many weeks it will take for the height of the two plants to be the same.
\newline
Equation:
\qquad
\newline
Solution:
\qquad
Get tutor help
2
2
2
.
∫
x
3
+
4
x
2
−
x
+
3
x
⋅
d
x
=
\int \frac{x^{3}+4 x^{2}-x+3}{x} \cdot d x=
∫
x
x
3
+
4
x
2
−
x
+
3
⋅
d
x
=
Get tutor help
9
9
9
. Using distributivity, find
\newline
(i)
{
7
5
×
(
−
3
12
)
}
+
{
7
5
×
5
12
}
\left\{\frac{7}{5} \times\left(\frac{-3}{12}\right)\right\}+\left\{\frac{7}{5} \times \frac{5}{12}\right\}
{
5
7
×
(
12
−
3
)
}
+
{
5
7
×
12
5
}
Get tutor help
Three triangles of different sizes are drawn on the rectangular piece of paper below and painted grey. Given the rectangular paper has an area of
840
c
m
2
840 \mathrm{~cm}^{2}
840
cm
2
, find the area of the painted portions.
\newline
1
2
×
b
×
h
\frac{1}{2} \times b \times h
2
1
×
b
×
h
Get tutor help
Evaluate.
\newline
k
λ
4
R
∫
−
π
2
π
/
2
sec
θ
2
⋅
d
θ
\frac{k \lambda}{4 R} \int_{-\frac{\pi}{2}}^{\pi / 2} \sec \frac{\theta}{2} \cdot d \theta
4
R
kλ
∫
−
2
π
π
/2
sec
2
θ
⋅
d
θ
Get tutor help
Evaluate the limit:
\newline
lim
x
→
0
(
1
x
−
cot
x
)
=
\lim _{x \rightarrow 0}\left(\frac{1}{x}-\cot x\right)=
x
→
0
lim
(
x
1
−
cot
x
)
=
Get tutor help
∫
0
3
x
2
+
4
x
+
5
x
+
3
d
x
=
\int_{0}^{3} \frac{x^{2}+4 x+5}{x+3} d x=
∫
0
3
x
+
3
x
2
+
4
x
+
5
d
x
=
Get tutor help
a.
lim
n
→
0
sin
x
−
x
x
3
\lim _{n \rightarrow 0} \frac{\sin x-x}{x^{3}}
lim
n
→
0
x
3
s
i
n
x
−
x
\newline
b.
∫
0
1
/
2
ln
(
x
+
1
)
x
d
x
=
\int_{0}^{1 / 2} \frac{\ln (x+1)}{x} d x=
∫
0
1/2
x
l
n
(
x
+
1
)
d
x
=
Get tutor help
∫
(
4
x
2
−
6
)
(
5
x
2
+
3
)
d
x
=
…
\int\left(4 x^{2}-6\right)\left(5 x^{2}+3\right) d x=\ldots
∫
(
4
x
2
−
6
)
(
5
x
2
+
3
)
d
x
=
…
Get tutor help
∫
2
x
(
x
2
+
1
)
2
d
x
=
\int 2 x\left(x^{2}+1\right)^{2} d x=
∫
2
x
(
x
2
+
1
)
2
d
x
=
Get tutor help
Integrate.
\newline
∫
x
+
5
x
2
−
2
x
−
3
d
x
\int \frac{x+5}{x^{2}-2 x-3} d x
∫
x
2
−
2
x
−
3
x
+
5
d
x
Get tutor help
f
(
x
)
=
{
x
2
−
4
x
+
4
x
<
1
1
x
=
1
x
2
x
>
1
f(x)=\left\{\begin{array}{cc} x^{2}-4 x+4 & x<1 \\ 1 & x=1 \\ x^{2} & x>1 \end{array}\right.
f
(
x
)
=
⎩
⎨
⎧
x
2
−
4
x
+
4
1
x
2
x
<
1
x
=
1
x
>
1
\newline
a) Study Differentiabilitu
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Find the numerical value of the log expression.
\newline
log
a
=
10
log
b
=
12
log
c
=
−
3
log
a
2
b
9
c
5
3
\begin{array}{c} \log a=10 \quad \log b=12 \quad \log c=-3 \\ \log \frac{a^{2} b^{9}}{\sqrt[3]{c^{5}}} \end{array}
lo
g
a
=
10
lo
g
b
=
12
lo
g
c
=
−
3
lo
g
3
c
5
a
2
b
9
\newline
Answer:
Get tutor help
Find the numerical value of the log expression.
\newline
log
a
=
−
4
log
b
=
−
5
log
c
=
−
9
log
c
9
a
3
b
8
\begin{array}{c} \log a=-4 \quad \log b=-5 \quad \log c=-9 \\ \log \frac{c^{9}}{a^{3} b^{8}} \end{array}
lo
g
a
=
−
4
lo
g
b
=
−
5
lo
g
c
=
−
9
lo
g
a
3
b
8
c
9
\newline
Answer:
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Expand the logarithm fully using the properties of logs. Express the final answer in terms of
log
x
,
log
y
\log x, \log y
lo
g
x
,
lo
g
y
, and
log
z
\log z
lo
g
z
.
\newline
log
y
3
z
5
3
x
2
\log \frac{y^{3}}{\sqrt[3]{z^{5}} x^{2}}
lo
g
3
z
5
x
2
y
3
\newline
Answer:
Get tutor help
Expand the logarithm fully using the properties of logs. Express the final answer in terms of
log
x
,
log
y
\log x, \log y
lo
g
x
,
lo
g
y
, and
log
z
\log z
lo
g
z
.
\newline
log
x
5
z
3
y
2
\log \frac{\sqrt{x^{5}}}{z^{3} y^{2}}
lo
g
z
3
y
2
x
5
\newline
Answer:
Get tutor help
Expand the logarithm fully using the properties of logs. Express the final answer in terms of
log
x
,
log
y
\log x, \log y
lo
g
x
,
lo
g
y
, and
log
z
\log z
lo
g
z
.
\newline
log
x
5
y
3
z
4
\log \frac{\sqrt{x^{5}}}{y^{3} z^{4}}
lo
g
y
3
z
4
x
5
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Answer:
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