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never [62]
3 years ago
10

Find the volume of a rectangle box 12cm by 15cm by 10cm

Mathematics
2 answers:
olga nikolaevna [1]3 years ago
5 0

Answer:

Step-by-step explanation:

natka813 [3]3 years ago
5 0

Answer:

Volume of the rectangular box=1800cm³

Step-by-step explanation:

Volume of the rectangular box = length x breadth x height

Volume of the rectangular box=12cm x 15cm x 10cm =1800cm³

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Vlad1618 [11]

Answer:

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7 0
3 years ago
The function y = –0.296x2 + 2.7x models the length x and height y that your sister's pet rabbit can jump, in centimeters. How hi
Ne4ueva [31]
To figure out the solution in a problem related to calculate maximum or minimum of something, you've to derive the involved equation, and make it equal to 0, solving the x.

So:
y = -0'296x² + 2'7x
y' = -0'592x + 2'7
0 = <span>-0'592x + 2'7
</span><span>-0'592x = 2'7
</span>x = 4'56

And if we substitute 4'56 on the first equation:
y = -0'296·<span>4'56² + 2'7·4'56
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So the length of the jump is 4'56cm, and the height 6'16cm.</span>
8 0
3 years ago
Read 2 more answers
Write 405,000 in scientific notation
Luba_88 [7]
4.05 *10^5
I hope this helps
3 0
3 years ago
Read 2 more answers
Please help needed its due soon
sp2606 [1]

take the total surface area = 2πr(r+h)

2×22×3(3+10)

7

=1716

7

= 245.15cm² but i put 245.04 and got it right

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7 0
2 years ago
I dont want you to answer question for me, i have already answered it as shown in the picture. I want you to let me know if i ha
Mashutka [201]

Answer:

\begin{equation}&#10;\sqrt{3}-1,2 \sqrt{10} \div 5, \sqrt{14}, 3 \sqrt{2}, \sqrt{19}+1,6&#10;\end{equation}

Explanation:

Given the irrational numbers:

$3 \sqrt{2}, \sqrt{3}-1, \sqrt{19}+1,6$, $2 \sqrt{10} \div 5,\sqrt{14}$

In order to arrange the numbers from the least to the greatest, we convert each number into its decimal equivalent.

\begin{gathered} 3\sqrt{2}=3\times1.414\approx4.242 \\ \sqrt{3}-1\approx1.732-1=0.732 \\ \sqrt{19}+1\approx4.3589+1=5.3589 \\ 6=6 \\ 2\sqrt{10}\div5=2(3.1623)\div5=1.2649 \\ \sqrt{14}=3.7147 \end{gathered}

Finally, sort these numbers in ascending order..

\begin{gathered} \sqrt{3}-1\approx1.732-1=0.732 \\ 2\sqrt{10}\div5=2(3.1623)\div5=1.2649 \\ \sqrt{14}=3.7147 \\ 3\sqrt{2}=3\times1.414\approx4.242 \\ \sqrt{19}+1\approx4.3589+1=5.3589 \\ 6=6 \end{gathered}

The given numbers in ascending order is:

\begin{equation}&#10;\sqrt{3}-1,2 \sqrt{10} \div 5, \sqrt{14}, 3 \sqrt{2}, \sqrt{19}+1,6&#10;\end{equation}

Note: In your solution, you can make the conversion of each irrational begin on a new line.

8 0
1 year ago
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