All Answers 358095 Answers

pretty nice bro :) you??/

Volume of one cube  = (.25in)^3  = .015625 in^3

Volume of prism  = ( 4.5 * 5 * 3.75 )   = 84.375 in^3

So to completely fill the box we have :

84.375 / .015625   =  5400 cubes

Lets see. So we have a 4-digit palindrome. that means the first digit and the last digit are the same. it also means the second digit and the third digit are the same. lets call the first one a, the second- b, thhrd- c, and fourth- d

a=d

b=c

we know that a*1000+b*100+c*10+d=a*1000+b*100+b*10+a=a*1001+b*110 is divisible by 7.

we know that 1001 is divisible by 7 (1001/7=143), and that 110 is NOT divisible by 7. that means we need b to be divisible by 7 (because a number divisible by 7+a number NOT divisible by7=a number NOT divisible by7

and

a number divisible by 7+a number divisible by 7=a number divisible by 7

the first digit of the number has to be between 1 and 9, and the other digits have to be betwee 0 and 9. meaning 1<=a<=9 and 0<=b<=9. if we need b to be divisible by 7 there are 2 options for b- 0 or 7. there are 9 options for a because a*1001 will always be divisible by 7 and 1<=a<=9 leaving 9 options for a.

2*9=18

so there are 18 4-digit palindromic numbers are divisible by 7

17 1/2%  =  .175  ....so..

84 + 84(.175)   =

£98.7

5400

My proffesor gave me this to problem to resolve it, but I really do not know how to do this 

8444...4...455= divisible to 19

Between the 8 and the two 5 are 2017 fours

8444...4...455 is divisible to 19, if 8444...4...455 modulo 19 = 0

• We calculate 8444...4...455 modulo 19:
  We divide 8444...4...455 in parts of 3-digits. ( The partition is arbitrary ).
  But the first part is 8444.

So the Number is:

\(\begin{array}{lcll} \underbrace{8444}_{\text{ Fist Part }}\ \underbrace{444}_{\text{ Second Part }}\ 444\ 444\ 444\ 444\ 444\ 444\ 444\ \ldots \underbrace{444}_{\text{ Part 672 }} \ \underbrace{455}_{\text{ Last Part }} \qquad \text{2017 fours} \\ \end{array} \)

\(\begin{array}{|rcll|} \hline && \text{first part} \pmod {19} \\ &\equiv& 8444\pmod {19}\\ &=& \color{red}8 \color{black}\qquad (8444=444\cdot 19 + \color{red}8 \color{black}) \\\\ && \text{The last remainder } = \color{red}8 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline && \text{Second part} \pmod {19} \\ &\equiv& \underbrace{8}_{\text{ The last remainder }}\underbrace{444}_{\text{ Second part }}\pmod {19}\\ &=& \color{red}8 \color{black}\qquad (8444=444\cdot 19 + \color{red}8 \color{black}) \\\\ && \text{The new last remainder } = 8 \\ \hline \end{array} \)

\(\cdots \)

\(\begin{array}{|rcll|} \hline && \text{Part 672} \pmod {19} \\ &\equiv& \underbrace{8}_{\text{ The last remainder }}\underbrace{444}_{\text{ Part 672 }}\pmod {19}\\ &=& \color{red}8 \color{black}\qquad (8444=444\cdot 19 + \color{red}8 \color{black}) \\\\ && \text{The new last remainder } = 8 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \text{Last part} \pmod {19} \\ &\equiv& \underbrace{8}_{\text{ The last remainder }}\underbrace{455}_{\text{ Last part }}\pmod {19}\\ &=& \color{red}0 \color{black}\qquad (8455=445\cdot 19 + \color{red}0 \color{black}) \\\\ && \text{The new last remainder } = 0 \\ \hline \end{array} \)

8444...4...455 is divisible to 19, because 8444...4...455 modulo 19 = 0

Whoops!! Sorry Alan!

180 - 130 = 50, for the 2 sides. So, 50/2=25.

Thx Mel!

You should try the log button. ;)

n=5*(10!)

no of zeros does n end with?

Legendre's Theorem - The Prime Factorization of Factorials

In mathematics,

Legendre's formula gives an expression for the exponent of the largest power of a prime p that divides the factorial 

\({\displaystyle n!}\)

1. The exponent of 2^x

We calculate 10 in base 2:

\(\begin{array}{rcll} 10_{10} &=& 1010_2 \\ \end{array}\)

The sum of the digits in base 2 is \(\begin{array}{rcll} 1+0+1+0 = 2 \end{array}\)

The exponent is: \(\begin{array}{|rcll|} \hline x &=&\frac{10-(\text{sum of the standard base-}\mathbf{2}\text{ digits of 10})}{\mathbf{2}-1} \\ x &=&\frac{10-2}{2} \\ x &=&4\\ \hline \end{array}\)

2. The exponent of 5^x

We calculate 10 in base 5:

\(\begin{array}{rcll} 10_{10} &=& 20_5 \\ \end{array}\)

The sum of the digits in base 5 is \(\begin{array}{rcll} 2+0 = 2 \end{array}\)

The exponent is: \(\begin{array}{|rcll|} \hline x &=&\frac{10-(\text{sum of the standard base-}\mathbf{5}\text{ digits of 10})}{\mathbf{5}-1} \\ x &=&\frac{10-2}{4} \\ x &=&2\\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 10! &=& 2^4\cdot 5^2\ldots \\ n=5\cdot 10! &=& 2^4\cdot 5^3\ldots \\ \hline \end{array}\)

Number of zeros does n end with ?:

\(\begin{array}{|rcll|} \hline && 2^4\cdot 5^3 \\ &=& 2^3\cdot5^3\cdot 2 \\ &=& (2\cdot 5)^3 \cdot 2 \\ &=& 2\cdot 10^\mathbf{\color{red}3} \\ \hline \end{array}\)

n ends with 3 zeros.

You forgot Alan...

Here's an alternative approach:

Find the length of a pendulum that oscillates with a frequency of 0.19 Hz.

The acceleration due to gravity is 9.81 m/s 2 .

Formula:

For small amplitudes, the period of such a pendulum can be approximated by:

\(\begin{array}{|lrcll|} \hline & T &=& 2\pi\sqrt{\frac{L}{g}} \\ \qquad L~ \text{expresses the pendulum length in meters } \\ \qquad g~ \text{expresses the acceleration of gravity}\approx 9.81 \frac{m}{s^2} \\ & T &=& \frac{1}{f} \\ \qquad f~ \text{expresses the frequency in Hz} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \frac{1}{f} &=& 2\pi\sqrt{\frac{L}{g}} \\ \frac{1}{2\pi f} &=& \sqrt{\frac{L}{g}} \\ \Big(\frac{1}{2\pi f}\Big)^2 &=& \frac{L}{g} \\ L &=& g\cdot \Big(\frac{1}{2\pi f}\Big)^2 \\ L &=& 9.81\cdot \left(\frac{1}{2\pi\cdot 0.19}\right)^2 \\ L &=& 9.81\cdot \left(0.83765759522\right)^2 \\ L &=& 9.81\cdot 0.70167024683 \\ \mathbf{L} &\mathbf{=}& \mathbf{6.88338512141\ m} \\ \hline \end{array}\)

 The length of the pendulum is 6.88 m

(cosx / sinx) + (2sinx / cosx) + (sin^3x / cos^3x)

How to simplify the expression with the LCM?

\(\begin{array}{|rcll|} \hline && \frac{\cos(x)}{\sin(x)} + \frac{2\sin(x) }{ \cos(x) } + \frac{ \sin^3(x) } { \cos^3(x) } \\ &=& \frac{\cos(x)}{\sin(x)}\cdot \frac{\cos^3(x)}{\cos^3(x)} + \frac{2\sin(x) }{ \cos(x) } \cdot \frac{\sin(x)\cos^2(x)}{\sin(x)\cos^2(x)} + \frac{ \sin^3(x) } { \cos^3(x) }\cdot \frac{ \sin(x) } { \sin(x) } \\ &=& \frac{\cos(x)\cos^3(x)+ 2\sin(x)\sin(x)\cos^2(x) + \sin^3(x)\sin(x) } {\sin(x)\cos^3(x) } \\ &=& \frac{\cos^4(x)+ 2\sin^2(x)\cos^2(x) + \sin^4(x) } {\sin(x)\cos^3(x) } \\ &=& \frac{\sin^4(x)+2\sin^2(x)\cos^2(x)+\cos^4(x) } {\sin(x)\cos^3(x) } \\ &=& \frac{\Big( \sin^2(x)+\cos^2(x) \Big)^2 } {\sin(x)\cos^3(x) } \quad & | \quad \sin^2(x)+\cos^2(x) = 1 \\ &=& \frac{ 1^2 } {\sin(x)\cos^3(x) } \\ &=& \frac{ 1 } {\sin(x)\cos^3(x) } \\ \hline \end{array} \)

The lateral area of a cylinder  =

2 * pi * (radius of the cylinder) * (height of the cylinder )

It's easier to find the x intercepts by factoring x^2 - 2x - 8   and setting  y = 0

0 = ( x - 4) (x + 2)

And setting both of the factors to 0 and solving for x, we have that x = 4  and x = -2

To find the vertex :

 y=x^2-2x-8   complete the square on x

y = x^2 - 2x + 1 - 8 - 1      factor the first three terms

y = (x - 1)^2  - 9

And the vertex  is  ( 1, - 9)

Here's the graph : https://www.desmos.com/calculator/pqqn4ceovc

I think you must mean "inscribed" if the circle is inside the regular pentagon...

Look at this picture to get a good idea of the situation :

AE  =  28/5 inches  =   5.6 inches....so AM is 1/2 of this  =  2.8 inches

Angle AFE  =  360/5 = 72°.....so.....angle AFM is 1/2 of this  = 36°

And angle BAE = 108°.....so......angle FAM is 1/2 of this  = 54°

And FM is the radius of the circle which can be found using The Law of Sines

AM / sin AFM  = FM / sin FAM

2.8 / sin 36° = FM /  sin 54°

Rearrange  as

FM  = sin 54° * 2.8 / sin 36°  ≈ 3.854 inches

cylinder volume = π * radius² * height

cylinder volume = π * 3² * 19

cylinder volume = 171π cubic inches

sphere volume = 4/3 * π * radius³

sphere volume = 4/3 * π * 3³

sphere volume = 36π cubic inches

cylinder volume - sphere volume

= 171π cubic inches - 36π cubic inches

= 135π cubic inches

≈ 424.1 cubic inches

Mmm

the equation c^2 = p can have zero, one, or two solutions. What can you say about the number of solutions of the equation c^3 = p?

\(c^2=p\\so\\c=\pm\sqrt p\\ \text{no real solutions if }p<0\\ \text{one real solutions if }p=0\\ \text{2 real solutions if }p>0\\\\~\\ If\\ c^3=p \;\;then\;\; c=\sqrt[3]{p}\\ \text{If p=0 then c=0}\\ \text{If p<0 then }c=-\sqrt[3]{|p|}\\ \text{If p>0 then }c=+\sqrt[3]{|p|}\\~\\ \text{asinus is correct, no matter what real value p has, there is exactly on solution for c} \)

Thanks asinus :)

I suppose you have this ???

 √6 / 4  +  √2 / 4  =

[  √6 +  √2 ] / 4   ≈ .966

As to your other question,   √6 *  √2   =  √12  =  √4 * √3   =  2 √3

The answer to this is just.....  log (6)  ≈ 0.77815

how do you find the height of a rectangle?

Area content of the rectangle divided by width.

  !

length x width