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Math Problems
Grade 8
Arithmetic sequences
What is the area of the region between the graphs of
f
(
x
)
=
2
x
2
+
5
x
f(x)=2 x^{2}+5 x
f
(
x
)
=
2
x
2
+
5
x
and
g
(
x
)
=
−
x
2
−
6
x
+
4
g(x)=-x^{2}-6 x+4
g
(
x
)
=
−
x
2
−
6
x
+
4
from
x
=
−
4
x=-4
x
=
−
4
to
x
=
0
x=0
x
=
0
?
\newline
Choose
1
1
1
answer:
\newline
(A)
40
40
40
\newline
(B)
355
12
\frac{355}{12}
12
355
\newline
(C)
8
8
8
\newline
(D)
128
3
\frac{128}{3}
3
128
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What is the area of the region between the graphs of
f
(
x
)
=
x
+
10
f(x)=\sqrt{x+10}
f
(
x
)
=
x
+
10
and
g
(
x
)
=
x
−
2
g(x)=x-2
g
(
x
)
=
x
−
2
from
x
=
−
10
x=-10
x
=
−
10
to
x
=
6
x=6
x
=
6
?
\newline
Choose
1
1
1
answer:
\newline
(A)
64
3
\frac{64}{3}
3
64
\newline
(B)
160
160
160
\newline
(C)
320
3
\frac{320}{3}
3
320
\newline
(D)
128
128
128
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What is the area of the region between the graphs of
f
(
x
)
=
x
+
1
f(x)=\sqrt{x+1}
f
(
x
)
=
x
+
1
and
g
(
x
)
=
2
x
−
4
g(x)=2 x-4
g
(
x
)
=
2
x
−
4
from
x
=
0
x=0
x
=
0
to
x
=
3
x=3
x
=
3
?
\newline
Choose
1
1
1
answer:
\newline
(A)
5
3
\frac{5}{3}
3
5
\newline
(B)
−
3
-3
−
3
\newline
(C)
14
3
\frac{14}{3}
3
14
\newline
(D)
23
3
\frac{23}{3}
3
23
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What is the area of the region between the graphs of
f
(
x
)
=
x
2
+
12
x
f(x)=x^{2}+12 x
f
(
x
)
=
x
2
+
12
x
and
g
(
x
)
=
3
x
2
+
10
g(x)=3 x^{2}+10
g
(
x
)
=
3
x
2
+
10
from
x
=
1
x=1
x
=
1
to
x
=
4
x=4
x
=
4
?
\newline
Choose
1
1
1
answer:
\newline
(A)
64
3
\frac{64}{3}
3
64
\newline
(B)
77
77
77
\newline
(C)
45
45
45
\newline
(D)
18
18
18
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Let
g
(
x
)
=
3
x
4
+
8
x
3
+
4
g(x)=3 x^{4}+8 x^{3}+4
g
(
x
)
=
3
x
4
+
8
x
3
+
4
.
\newline
What is the absolute maximum value of
g
g
g
?
\newline
Choose
1
1
1
answer:
\newline
(A)
4
4
4
\newline
(B)
36
36
36
\newline
(C)
116
116
116
\newline
(D)
g
g
g
has no maximum value
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The differentiable functions
x
x
x
and
y
y
y
are related by the following equation:
\newline
x
y
=
4
x y=4
x
y
=
4
\newline
We are also given that
d
y
d
t
=
1
\frac{d y}{d t}=1
d
t
d
y
=
1
.
\newline
Find
d
x
d
t
\frac{d x}{d t}
d
t
d
x
when
y
=
0.5
y=0.5
y
=
0.5
.
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Let
h
(
x
)
=
{
x
+
26
−
5
x
+
1
for
x
≥
−
26
,
x
≠
−
1
k
for
x
=
−
1
h(x)=\left\{\begin{array}{ll}\frac{\sqrt{x+26}-5}{x+1} & \text { for } x \geq-26, x \neq-1 \\ k & \text { for } x=-1\end{array}\right.
h
(
x
)
=
{
x
+
1
x
+
26
−
5
k
for
x
≥
−
26
,
x
=
−
1
for
x
=
−
1
\newline
h
h
h
is continuous for all
x
>
−
26
x>-26
x
>
−
26
.
\newline
What is the value of
k
k
k
?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
1
5
-\frac{1}{5}
−
5
1
\newline
(B)
1
1
1
\newline
(C)
1
10
\frac{1}{10}
10
1
\newline
(D)
0
0
0
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Let
f
(
x
)
=
x
−
1
−
2
x
−
5
f(x)=\frac{\sqrt{x-1}-2}{x-5}
f
(
x
)
=
x
−
5
x
−
1
−
2
when
x
≠
5
x \neq 5
x
=
5
.
\newline
f
f
f
is continuous for all
x
>
1
x>1
x
>
1
.
\newline
Find
f
(
5
)
f(5)
f
(
5
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
1
2
\frac{1}{2}
2
1
\newline
(B)
1
4
\frac{1}{4}
4
1
\newline
(C)
1
1
1
\newline
(D)
1
10
\frac{1}{10}
10
1
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f
(
x
)
=
{
sin
(
x
⋅
π
)
for
−
8
<
x
<
0
x
5
for
0
≤
x
≤
10
f(x)=\left\{\begin{array}{ll} \sin (x \cdot \pi) & \text { for }-8<x<0 \\ \frac{x}{5} & \text { for } 0 \leq x \leq 10 \end{array}\right.
f
(
x
)
=
{
sin
(
x
⋅
π
)
5
x
for
−
8
<
x
<
0
for
0
≤
x
≤
10
\newline
Find
lim
x
→
−
5
f
(
x
)
\lim _{x \rightarrow-5} f(x)
lim
x
→
−
5
f
(
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
−
1
-1
−
1
\newline
(B)
0
0
0
\newline
(C)
1
1
1
\newline
(D) The limit doesn't exist.
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