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Let’s check out your problem:
Complete the following statement. Use the
integers
that are closest to the number in middle.
\newline
□
\square
□
< \sqrt[3]{59} <
□
\square
□
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Math Problems
Geometry
Triangle Inequality Theorem
Full solution
Q.
Complete the following statement. Use the integers that are closest to the number in middle.
\newline
□
\square
□
<
59
3
<
< \sqrt[3]{59} <
<
3
59
<
□
\square
□
Understand the problem:
Step
1
1
1
: Understand the problem. We need to find the integers closest to the cube root of
59
59
59
.
Estimate cube root:
Step
2
2
2
: Estimate the cube root of
59
59
59
. Since
4
3
=
64
4^3 = 64
4
3
=
64
and
3
3
=
27
3^3 = 27
3
3
=
27
, the cube root of
59
59
59
is between
3
3
3
and
4
4
4
.
Calculate cube:
Step
3
3
3
: Calculate
3.
5
3
3.5^3
3.
5
3
to check if it's closer to
59
59
59
.
3.
5
3
=
42.875
3.5^3 = 42.875
3.
5
3
=
42.875
.
More problems from Triangle Inequality Theorem
Question
You pick a card at random, put it back, and then pick another card at random.
\newline
4
4
4
\newline
5
5
5
\newline
6
6
6
\newline
7
7
7
\newline
What is the probability of picking a number greater than
5
5
5
and then picking a
4
4
4
?
\newline
Write your answer as a percentage.
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Jeanne's bank account earns interest annually. The equation shows her starting balance of
$
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$500
and her balance at the end of three years,
$
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\$546.36
$546.36
. At what rate
r
r
r
did Jeanne earn interest?
\newline
546.36
=
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(
1
+
r
)
546.36 = 500(1 + r)
546.36
=
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(
1
+
r
)
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Question
Jeanne's bank account earns interest annually. The equation shows her starting balance of
$
350
\$ 350
$350
and her balance at the end of three years,
$
422.78
\$ 422.78
$422.78
. At what rate r did Jeanne earn interest?
\newline
422.78
=
350
(
1
+
r
)
3
422.78=350(1+r)^{3}
422.78
=
350
(
1
+
r
)
3
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Posted 1 month ago
Question
A leaky
10
-kg
10\text{-kg}
10
-kg
bucket is lifted from the ground to a height of
14
m
14\, \text{m}
14
m
at a constant speed with a rope that weighs
0.6
kg/m
0.6\, \text{kg/m}
0.6
kg/m
. Initially the bucket contains
42
kg
42\, \text{kg}
42
kg
of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the
14
m
14\, \text{m}
14
m
level. How much work is done? (Use
9.8
m/s
2
9.8\, \text{m/s}^2
9.8
m/s
2
for
g
g
g
.) Show how to approximate the required work (in
J
\text{J}
J
) by a Riemann sum. (Let
x
x
x
be the height in meters above the ground. Enter
x
i
∗
x_i^*
x
i
∗
as
x
i
x_i
x
i
.)
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Posted 1 month ago
Question
A leaky
10
−
k
g
10-\mathrm{kg}
10
−
kg
bucket is lifted from the ground to a height of
14
14
14
m at a constant speed with a rope that weighs
0.6
k
g
/
m
0.6 \mathrm{~kg} / \mathrm{m}
0.6
kg
/
m
. Initially the bucket contains
42
42
42
kg of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the
14
−
m
14-\mathrm{m}
14
−
m
level. How much work is done? (Use
9.8
m
/
s
2
9.8 \mathrm{~m} / \mathrm{s}^{2}
9.8
m
/
s
2
for g .)
\newline
Show how to approximate the required work (in J) by a Riemann sum. (Let
x
x
x
be the height in meters above the ground. Enter
x
i
∗
∗
x_{i}^{* *}
x
i
∗∗
as
x
i
x_{i}
x
i
)
\newline
lim
n
→
∞
∑
i
=
1
n
(
98
+
35.28
x
i
)
Δ
x
\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(\boxed{98+35.28 x_{i}}\right) \Delta x
lim
n
→
∞
∑
i
=
1
n
(
98
+
35.28
x
i
)
Δ
x
\newline
Express the work (in J) as an integral in terms of
x
x
x
(in m).
\newline
∫
0
14
(
98
+
35.28
x
\int_{0}^{14}(98+35.28 x
∫
0
14
(
98
+
35.28
x
\newline
Evaluate the integral (in J). (Round your answer to the nearest integer.)
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Posted 1 month ago
Question
Solve the equation.
\newline
6
w
7
=
98
,
304
6 w^{7}=98,304
6
w
7
=
98
,
304
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Posted 1 month ago
Question
A leaky
10
-kg
10\text{-kg}
10
-kg
bucket is lifted from the ground to a height of
14
m
14\text{ m}
14
m
at a constant speed with a rope that weighs
0.6
kg/m
0.6\text{ kg/m}
0.6
kg/m
. Initially the bucket contains
42
kg
42\text{ kg}
42
kg
of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the
14
-m
14\text{-m}
14
-m
level. How much work is done? (Use
9.8
m/s
2
9.8\text{ m/s}^2
9.8
m/s
2
for
g
g
g
.) Show how to approximate the required work (in
J
\text{J}
J
) by a Riemann sum. (Let
x
x
x
be the height in meters above the ground. Enter
x
i
∗
x_i^*
x
i
∗
as
x
i
x_i
x
i
.)
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Posted 1 month ago
Question
f
(
t
)
=
{
−
64
t
,
t
=
8
14
−
t
,
t
=
10
t
2
−
3
t
+
2
,
t
≠
8
,
10
f
(
2
)
=
□
\begin{array}{l}f(t)=\left\{\begin{array}{ll}-\frac{64}{t} & , \quad t=8 \\ 14-t & , \quad t=10 \\ t^{2}-3 t+2 & , \quad t \neq 8,10\end{array}\right. \\ f(2)=\square\end{array}
f
(
t
)
=
⎩
⎨
⎧
−
t
64
14
−
t
t
2
−
3
t
+
2
,
t
=
8
,
t
=
10
,
t
=
8
,
10
f
(
2
)
=
□
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Posted 1 month ago
Question
f
(
x
)
=
{
x
−
6
3
,
x
=
−
18
360
x
,
x
=
36
4
x
+
73
,
x
≠
−
18
,
36
f
(
−
18
)
=
\begin{array}{l}f(x)=\left\{\begin{array}{lll}\frac{x-6}{3} & , \quad x=-18 \\\frac{360}{x} & , \quad x=36 \\4 x+73 & , \quad x \neq-18,36\end{array}\right. \\f(-18)=\end{array}
f
(
x
)
=
⎩
⎨
⎧
3
x
−
6
x
360
4
x
+
73
,
x
=
−
18
,
x
=
36
,
x
=
−
18
,
36
f
(
−
18
)
=
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Posted 1 month ago
Question
c
(
n
)
=
4
9
(
−
3
)
n
−
1
c(n)=\frac{4}{9}(-3)^{n-1}
c
(
n
)
=
9
4
(
−
3
)
n
−
1
\newline
What is the
3
rd
3^{\text {rd }}
3
rd
term in the sequence?
\newline
□
\square
□
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Posted 1 month ago
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