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What is the sum of the sonutions to the equation
\newline
(
t
+
3
)
(
t
−
357
)
=
0
(t+3)(t-357)=0
(
t
+
3
)
(
t
−
357
)
=
0
?
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Math Problems
Algebra 2
Solve quadratic inequalities
Full solution
Q.
What is the sum of the sonutions to the equation
\newline
(
t
+
3
)
(
t
−
357
)
=
0
(t+3)(t-357)=0
(
t
+
3
)
(
t
−
357
)
=
0
?
Set
Factors
Equal to Zero:
Set each factor of the equation equal to zero to find the solutions.
\newline
(
t
+
3
)
=
0
(t + 3) = 0
(
t
+
3
)
=
0
implies
t
=
−
3
t = -3
t
=
−
3
.
\newline
(
t
−
357
)
=
0
(t - 357) = 0
(
t
−
357
)
=
0
implies
t
=
357
t = 357
t
=
357
.
Find Solutions:
Add the solutions together to find the sum.
\newline
Sum =
(
−
3
)
+
357
(-3) + 357
(
−
3
)
+
357
\newline
Sum =
354
354
354
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8
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x
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\newline
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\newline
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1
answer:
\newline
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Find
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x
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lim
x
→
4
x
−
4
2
−
4
x
−
12
.
\newline
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1
1
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\newline
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2
2
2
\newline
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1
1
1
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−
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Question
Find
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x
→
2
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4
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3
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x
−
2
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lim
x
→
2
x
−
2
x
4
−
4
x
3
+
4
x
2
.
\newline
Choose
1
1
1
answer:
\newline
(A)
−
4
-4
−
4
\newline
(B)
0
0
0
\newline
(C)
4
4
4
\newline
(D) The limit doesn't exist
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Posted 9 months ago
Question
Let
f
f
f
be a continuous function on the closed interval
[
−
5
,
0
]
[-5,0]
[
−
5
,
0
]
, where
f
(
−
5
)
=
0
f(-5)=0
f
(
−
5
)
=
0
and
f
(
0
)
=
5
f(0)=5
f
(
0
)
=
5
.
\newline
Which of the following is guaranteed by the Intermediate Value Theorem?
\newline
Choose
1
1
1
answer:
\newline
(A)
f
(
c
)
=
−
2
f(c)=-2
f
(
c
)
=
−
2
for at least one
c
c
c
between
0
0
0
and
5
5
5
\newline
(B)
f
(
c
)
=
2
f(c)=2
f
(
c
)
=
2
for at least one
c
c
c
between
0
0
0
and
5
5
5
\newline
(C)
f
(
c
)
=
−
2
f(c)=-2
f
(
c
)
=
−
2
for at least one
c
c
c
between
−
5
-5
−
5
and
0
0
0
\newline
(D)
f
(
c
)
=
2
f(c)=2
f
(
c
)
=
2
for at least one
c
c
c
between
−
5
-5
−
5
and
0
0
0
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Question
Let
g
(
x
)
=
tan
(
x
)
g(x)=\tan (x)
g
(
x
)
=
tan
(
x
)
.
\newline
Can we use the intermediate value theorem to say the equation
g
(
x
)
=
0
g(x)=0
g
(
x
)
=
0
has a solution where
π
4
≤
x
≤
3
π
4
\frac{\pi}{4} \leq x \leq \frac{3 \pi}{4}
4
π
≤
x
≤
4
3
π
?
\newline
Choose
1
1
1
answer:
\newline
(A) No, since the function is not continuous on that interval.
\newline
(B) No, since
0
0
0
is not between
g
(
π
4
)
g\left(\frac{\pi}{4}\right)
g
(
4
π
)
and
g
(
3
π
4
)
g\left(\frac{3 \pi}{4}\right)
g
(
4
3
π
)
.
\newline
(C) Yes, both conditions for using the intermediate value theorem have been met.
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Posted 9 months ago
Question
Let
g
(
x
)
=
cos
(
x
)
g(x)=\cos (x)
g
(
x
)
=
cos
(
x
)
.
\newline
Can we use the intermediate value theorem to say the equation
g
(
x
)
=
0.8
g(x)=0.8
g
(
x
)
=
0.8
has a solution where
0
≤
x
≤
π
2
0 \leq x \leq \frac{\pi}{2}
0
≤
x
≤
2
π
?
\newline
Choose
1
1
1
answer:
\newline
(A) No, since the function is not continuous on that interval.
\newline
(B) No, since
0
0
0
.
8
8
8
is not between
g
(
0
)
g(0)
g
(
0
)
and
g
(
π
2
)
g\left(\frac{\pi}{2}\right)
g
(
2
π
)
.
\newline
(C) Yes, both conditions for using the intermediate value theorem have been met.
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Posted 9 months ago
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