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Let’s check out your problem:
Complete the equation. Write the sum as a mixed number.
\newline
1
7
9
+
1
5
=
80
□
+
9
□
=
□
(
□
□
)
1\frac{7}{9}+\frac{1}{5}=\frac{80}{\Box}+\frac{9}{\Box}=\Box\left(\frac{\Box}{\Box}\right)
1
9
7
+
5
1
=
□
80
+
□
9
=
□
(
□
□
)
View step-by-step help
Home
Math Problems
Precalculus
Operations with rational exponents
Full solution
Q.
Complete the equation. Write the sum as a mixed number.
\newline
1
7
9
+
1
5
=
80
□
+
9
□
=
□
(
□
□
)
1\frac{7}{9}+\frac{1}{5}=\frac{80}{\Box}+\frac{9}{\Box}=\Box\left(\frac{\Box}{\Box}\right)
1
9
7
+
5
1
=
□
80
+
□
9
=
□
(
□
□
)
Convert to improper fraction:
Convert mixed number to improper
fraction
.
\newline
1
7
9
=
1
+
7
9
=
9
9
+
7
9
=
16
9
1\frac{7}{9} = 1 + \frac{7}{9} = \frac{9}{9} + \frac{7}{9} = \frac{16}{9}
1
9
7
=
1
+
9
7
=
9
9
+
9
7
=
9
16
.
Find common denominator:
Find a common denominator for
16
9
\frac{16}{9}
9
16
and
1
5
\frac{1}{5}
5
1
.
\newline
Common denominator of
9
9
9
and
5
5
5
is
45
45
45
.
\newline
16
9
=
(
16
×
5
)
(
9
×
5
)
=
80
45
\frac{16}{9} = \frac{(16\times5)}{(9\times5)} = \frac{80}{45}
9
16
=
(
9
×
5
)
(
16
×
5
)
=
45
80
.
\newline
1
5
=
(
1
×
9
)
(
5
×
9
)
=
9
45
\frac{1}{5} = \frac{(1\times9)}{(5\times9)} = \frac{9}{45}
5
1
=
(
5
×
9
)
(
1
×
9
)
=
45
9
.
Add fractions:
Add the
fractions
.
\newline
80
45
+
9
45
=
80
+
9
45
=
89
45
\frac{80}{45} + \frac{9}{45} = \frac{80 + 9}{45} = \frac{89}{45}
45
80
+
45
9
=
45
80
+
9
=
45
89
.
Convert to mixed number:
Convert improper fraction to mixed number.
89
45
=
1
\frac{89}{45} = 1
45
89
=
1
whole and
44
45
\frac{44}{45}
45
44
left over, so it's
1
(
44
45
)
1\left(\frac{44}{45}\right)
1
(
45
44
)
.
Complete the equation:
Complete the equation with the found values.
\newline
1
(
7
)
9
+
1
5
=
80
45
+
9
45
=
1
(
44
)
45
\frac{1(7)}{9} + \frac{1}{5} = \frac{80}{45} + \frac{9}{45} = \frac{1(44)}{45}
9
1
(
7
)
+
5
1
=
45
80
+
45
9
=
45
1
(
44
)
.
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Question
f
(
x
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=
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x
+
5
2
−
14
+
5
x
f(x)=\frac{6 x+5}{2-\sqrt{14+5 x}}
f
(
x
)
=
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+
5
x
6
x
+
5
\newline
We want to find
lim
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→
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f
(
x
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\lim _{x \rightarrow-2} f(x)
lim
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f
(
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)
.
\newline
What happens when we use direct substitution?
\newline
Choose
1
1
1
answer:
\newline
(A) The limit exists, and we found it!
\newline
(B) The limit doesn't exist (probably an asymptote).
\newline
(C) The result is indeterminate.
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What is the area of the region bound by the graphs of
f
(
x
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?
\newline
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1
1
answer:
\newline
(A)
19
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6
19
\newline
(B)
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\newline
(C)
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2
151
\newline
(D)
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2
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2
45
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\newline
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