I have 30 ones, 82 thousands, 4 hundred thousands, 60 tens, and 100 hundreds. What number am I?

Mathematics

2 answers:

Bad White [126]2 years ago

5 0

The number that you are is...
492,630

vladimir1956 [14]2 years ago

5 0

482163 and that’s your answer

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What is 0.1 litre : 500 ml simplified

Lesechka [4]

Answer:

Step-by-step explanation:

1 liter = 1000 ml

0.1 liter = 0.1*1000 = 100.0 = 100 ml

$\frac{0.1 litre}{500ml} = \frac{100}{500}\\\\=\frac{1}{5}$

0.1 liter : 500 ml = 1 : 5

3 0

2 years ago

In a certain year, the population of Smallburg was $4{,}000.$ In the following year, the population increased by $15\%.$ What wa

makkiz [27]

Answer:

4,600

Step-by-step explanation:

To solve this problem, we need to multiply add 15% of 4,000 back to 4,000.

First, change 15% into a decimal:

15% -> $\frac{15}{100}$ -> 0.15

Next, multiply 4,000 by 0.15:

$x= 4,000(0.15)$

$x=600$

The last step is to add the 600 back to 4,000:

$4,000 + 600 = 4,600$

The population of Smallburg the following year was 4,600.

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2 years ago

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Nikitich [7]

The acceleration of the particle is given by the formula mentioned below:

$a=\frac{d^2s}{dt^2}$

Differentiate the position vector with respect to t.

$\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}$

Differentiate both sides of the obtained equation with respect to t.

$\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}$

Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.