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3 years ago

What is the volume of a box that will hold exactly 567of these cubes with 1.3 inch side

Mathematics

1 answer:

Bas_tet [7]3 years ago

5 0

Answer:

21 inch cube

Step-by-step explanation:

(1/3)(1/3)(1/3)=1/27 is the volume of the cube

the volume of the box : 567*(1/27)=567/27=21

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Divide and simplify. − 8 15 ÷ − 3 5 A) − 3 25 B) − 4 5 C) 13 15 D) 8 9 A 100 points

lubasha [3.4K]

Step-by-step explanation:

- will be cancel

then

815/35= 23.28

6 0

2 years ago

At which x-value does the cosine function attain a maximum value?.

Troyanec [42]

Answer:

cos(x) = ???

so the maximum cosine can be is 1 and that happens when x = 0

I'm pretty sure that's what you meant...

3 0

2 years ago

What sum will amount to rupees 4000 in 3 years at 6% p.a. compound interest

natali 33 [55]

Answer:

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Given:}}}}}}}\end{gathered}$

⇢ Principle = Rs.4000
⇢ Rate = 6%
⇢ Time = 3 year

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{To Find:}}}}}}}\end{gathered}$

⇢ Amount

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Using Formula:}}}}}}}\end{gathered}$

${\dag{\underline{\boxed{\sf{Amount ={P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}}}$

$\dag{\underline{\boxed{\sf{Compound \: Interest = Amount- Principle }}}}$

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Solution:}}}}}}}\end{gathered}$

${\bigstar \:{\underline{\pmb{\frak{\red{Firstly,Finding \: the \: Amount }}}}}}$

$\quad {:\implies{\sf{Amount = \bf{P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}$

Substituting the values

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg(1 + \dfrac{6}{100}{\bigg)}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg(1 \times 100 + \dfrac{6}{100}{\bigg)}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg( \dfrac{100 + 6}{100}{\bigg)}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg( \dfrac{106}{100}{\bigg)}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg({\cancel{\dfrac{106}{100}}{\bigg)}}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg( \dfrac{53}{50}{\bigg)}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg( \dfrac{53}{50} \times \dfrac{53}{50} \times \dfrac{53}{50}{\bigg)}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg( \dfrac{148877}{125000}{\bigg)}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000 \times \dfrac{148877}{125000}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4{\cancel{000}} \times \dfrac{148877}{125{\cancel{000}}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{\dfrac{148877 \times 4}{125}}}}}$

$\quad {:\implies{\sf{Amount = \bf{\dfrac{595508}{125}}}}}$

$\quad {:\implies{\sf{Amount = \bf{\cancel{\dfrac{595508}{125}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4764.064}}}}$

$\begin{gathered} \dag{\boxed{\textsf{\textbf{\underline{\color{green}{Amount = {Rs.4764.064}}}}}}}\end{gathered}$

Hence, The Amount is Rs.4764.064

$\begin{gathered}\end{gathered}$

${\bigstar \:{\underline{\pmb{\frak{\red{ Now,Finding \: The \: Compound \: Interest }}}}}}$

$\quad{: \implies{\sf{Compound \: Interest = \bf{Amount- Principle }}}}$

Substituting the values

$\quad{: \implies{\sf{Compound \: Interest = \bf{4764.064- 4000 }}}}$

$\quad{: \implies{\sf{Compound \: Interest =\bf{764.064}}}}$

$\begin{gathered} \dag{\boxed{\textsf{\textbf{\underline{\color{green}{Compound Interest = Rs.764.064}}}}}}\end{gathered}$

Henceforth,The Compound Interest is Rs.764064

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Learn More:}}}}}}}\end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered} \dag \: \underline{\bf{More \: Useful \: Formula}}\\ {\boxed{\begin{array}{cc}\dashrightarrow {\sf{Amount = Principle + Interest}} \\ \\ \dashrightarrow \sf{ P=Amount - Interest }\\ \\ \dashrightarrow \sf{ S.I = \dfrac{P \times R \times T}{100}} \\ \\ \dashrightarrow \sf{P = \dfrac{Interest \times 100 }{Time \times Rate}} \\ \\ \dashrightarrow \sf{P = \dfrac{Amount\times 100 }{100 + (Time \times Rate)}} \\ \end{array}}}\end{gathered}\end{gathered}\end{gathered}$

8 0

3 years ago

Consider that your profit-maximizing quantity of 500 can be produced at an average cost per unit of $9 and sold for a market pri

timama [110]

Based on the quantity sold and the price, the profit per unit is $5 and the total profit is $2,500

The profit made from each unit sold can be found as:

= Market price - Average cost

= 14 - 9

= $5

The total profit would therefore be:

= Profit per unit x Number of units sold

= 5 x 500

= $2,500

In conclusion, the profit per unit is $5 and the total profit is $2,500

Find out more about profit per unit at brainly.com/question/25531124.

4 0

2 years ago

Really need help on this timed

SSSSS [86.1K]

Answer:

10 units

Step-by-step explanation:

The distance from the point (0,0) to (10,0) is 10 units

4 0

3 years ago

Read 2 more answers