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IRINA_888 [86]
3 years ago
15

A rectangular prism with a length of 3cm, width of 4cm, and height of 5cm is packed with cubes whose side length is 1/2cm. how m

any 1/2cm cubes can fit inside the rectangular prism?
Mathematics
1 answer:
Dennis_Churaev [7]3 years ago
8 0
Length = 3cm 
1/2 cm cube = 3 ÷ 1/2 = 3 x 2 = 6

Width = 4cm
1/2 cm cube = 4 ÷ 1/2 = 4 x 2 = 8

Height = 5cm
1/2 cm cube = 5 ÷ 1/2 = 5 x 2 = 10

Number o 1/2 cubes = 6 x 8 x 10 = 480
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The distance between Lincoln, NE, and Boulder, CO, is about 500 miles. The distance between Boulder, CO, and a third city is 200
GenaCL600 [577]
We know that

The sum of the lengths of any two sides of a triangle is greater than the length of the third side (Triangle Inequality Theorem)

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A------> Lincoln, NE

B------> Boulder, CO

C------> third city


we know that

in the triangle ABC

AB=500 miles

BC=200 miles

AC=x


Applying the Triangle Inequality Theorem

1) 500+200 > x------> 700 > x------> x < 700 miles

2) 200+x > 500----> x > 500-200------> x > 300 miles


the solution for x is 

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Find the differential coefficient of <br><img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^
Gemiola [76]

Answer:

\rm \displaystyle y' =   2 {e}^{2x}   +    \frac{1}{x}  {e}^{2x}  + 2 \ln(x) {e}^{2x}

Step-by-step explanation:

we would like to figure out the differential coefficient of e^{2x}(1+\ln(x))

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

\displaystyle y =  {e}^{2x}  \cdot (1 +   \ln(x) )

to do so distribute:

\displaystyle y =  {e}^{2x}  +   \ln(x)  \cdot  {e}^{2x}

take derivative in both sides which yields:

\displaystyle y' =  \frac{d}{dx} ( {e}^{2x}  +   \ln(x)  \cdot  {e}^{2x} )

by sum derivation rule we acquire:

\rm \displaystyle y' =  \frac{d}{dx}  {e}^{2x}  +  \frac{d}{dx}   \ln(x)  \cdot  {e}^{2x}

Part-A: differentiating $e^{2x}$

\displaystyle \frac{d}{dx}  {e}^{2x}

the rule of composite function derivation is given by:

\rm\displaystyle  \frac{d}{dx} f(g(x)) =  \frac{d}{dg} f(g(x)) \times  \frac{d}{dx} g(x)

so let g(x) [2x] be u and transform it:

\displaystyle \frac{d}{du}  {e}^{u}  \cdot \frac{d}{dx} 2x

differentiate:

\displaystyle   {e}^{u}  \cdot 2

substitute back:

\displaystyle    \boxed{2{e}^{2x}  }

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

\displaystyle  \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)

let

  • f(x) \implies   \ln(x)
  • g(x) \implies    {e}^{2x}

substitute

\rm\displaystyle  \frac{d}{dx}  \ln(x)  \cdot  {e}^{2x}  =  \frac{d}{dx}( \ln(x) ) {e}^{2x}  +  \ln(x) \frac{d}{dx}  {e}^{2x}

differentiate:

\rm\displaystyle  \frac{d}{dx}  \ln(x)  \cdot  {e}^{2x}  =   \boxed{\frac{1}{x} {e}^{2x}  +  2\ln(x)  {e}^{2x} }

Final part:

substitute what we got:

\rm \displaystyle y' =   \boxed{2 {e}^{2x}   +    \frac{1}{x}  {e}^{2x}  + 2 \ln(x) {e}^{2x} }

and we're done!

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Answer:

Step-by-step explanation:

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