All Answers 359318 Answers

Do you mean this problem:

Insert three fractions between $\frac14$ and $\frac12$ so the five fractions form an arithmetic sequence. What is the sum, expressed as a common fraction, of these three new fractions?

If you do then the answer is 9/8

this is the reasoning:

The term in the middle is the arithmetic mean of $\dfrac{1}{4}$ and $\dfrac{1}{2}$, or $\dfrac{3}{8}$, so the arithmetic sequence looks like $$\dfrac{1}{4}, \underline{\qquad} , \dfrac{3}{8}, \underline{\qquad} , \dfrac{1}{2}.$$The fraction $\dfrac{3}{8}$ is also the arithmetic mean of the two missing terms, and thus $\dfrac{3}{8}$ is the average of the $3$ new fractions. So, their sum is $3\cdot\dfrac{3}{8}=\boxed{\dfrac{9}{8}}$.

Graphical solution:

Let $a_n$ be the $n$th term of the sequence. ($a_n=a_1+n*(d-1)$ where $d$ is the common difference, $a_1$ is the first term, and $n$ is the number of the term you want to find out)

We are given that $a_5=9$ and $a_{32}=84$.

From those, we can get the equations $9=a_1+4d$ and $84=a_1+31d$.

(Math time!)

$a_1+31d = 84$

$a_1+4d   = 9$

---------------------

        $27d =  75$

            $d = \frac{25}{9}$

Since $9=a_1+4d$:

$9=a_1+\frac{100}{9}$

$a_1=\frac{-19}{9}$

Finally, $a_23=a_1+22*d$, so:

$a_{23}=\frac{-19}{9}+22*\frac{25}{9}$

$a_{23}=\frac{550-19}{9}$

$a_{23}=\frac{531}{9}$

$a_{23}=\boxed{59}$.

...BUT you CAN change it to  

         5 ( 12 - x)

Not so sure you can do that simplification of a math EXPRESSION......

just plug into the given equation

4 x 10⁸  m = 4 x 10⁵  km

(4/3)(pi)(4x 10⁵)³

(4/3)(π)(16 x 10¹⁵ ) = .................  km³

discriminant of the quadratic equation,    b² -4ac ,   must be > zero 

                              45² - 4(8)(c) >0

                             2025 - 32c > 0

                                  32c<2025

                                      c < 63.28         the product of 1 thru 63   is a big number.....did you enter the correct numbers in your Q ?

63! = 1.982608315 x 10⁸⁷

Hmm. . . I'm not sure why you changed your question to \(60-5x\). Anyways, you can't solve for \(x\) in this situation; as the equation isn't complete. But you can simplify it into: 

\(\frac{60-5x}{5}= 12-x\)

 guest.exe

You can plug it into a calculator or and get:

\(\frac{1500}{75}\)\(=20\)

Answer: 

75 can go into 1500 20 times.

 guest.exe

1500 ÷ 75 = 1500/75 = ........  use calculator

(x-4)^2     - 16 +1

(x-4)² -15

If \(50x = 500\), then: 

Divide both sides by \(50\):

\(\frac{50x}{50} = \frac{500}{50}\)

This simplifies to:

\( x = 10\)

 guest.exe

She has 3/2 kilos left.

3.141592653589793238262643383279502884196179393751

Thanks Alan,

That makes sense.  

deleted  - it was wrong

The foci of an ellipse have to be inside the ellipse.  (0, 4) and (0, -4) are not in your ellipse.

Part b can be done as follows:

In triangle \(ABC\), \(\angle ABC = 2 \alpha\) and \(\angle ACB = 140 ^\circ - \alpha\).
\(D\) is a point on line segment \(AB\) such that \(CB = BD\).
What is the measure (in degrees) of \(\angle DCA\)?

\(\text{Let $\angle DCA=x$} \\ \text{Let $\angle BDC=\angle BCD= \beta$} \)

\(\begin{array}{|rcll|} \hline \mathbf{2\alpha + 2\beta} &=& \mathbf{180^\circ} \\ \alpha + \beta &=& 90^\circ \\ \mathbf{\beta} &=& \mathbf{90^\circ -\alpha} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{140^\circ-\alpha} &=& \mathbf{x + \beta} \quad | \quad \mathbf{\beta=90^\circ -\alpha} \\ 140^\circ-\alpha &=& x + 90^\circ -\alpha \\ 140^\circ&=& x + 90^\circ \\ x &=& 140^\circ-90^\circ \\ \mathbf{x} &=& \mathbf{50^\circ} \\ \hline \end{array}\)

you are making a rectangular box out of a 12 in by 8 in piece of cardboard. The box will be formed by making two slits on each corner and folding up the sides. A) Draw the  3D solids with the dimentions b) find the volume of the rectangular box c) Find the surface area.

Hello Guest!

The slots are a long.

A)

You can do that yourself.

B)

\(V=(12\ in-2a)\cdot (8\ in-2a)\cdot a=96a\ in^2-24\ in\cdot a^2-16\ in\cdot a^2+4a^3\)

\(V=4a^3-40a^2\cdot in+96a\cdot in^2\)

C)

\(S=12\cdot 8\cdot \ in^2-4a^2\)

\(S=96\ in^2-4a^2\)

  !

\(\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{b}-\frac{1}{a}\right)\\ =\left(\frac{1}{b}+\frac{1}{a}\right)\left(\frac{1}{b}-\frac{1}{a}\right)\\ =\frac{1}{b^2}-\frac{1}{a^2}\\ smallest - biggest=smallest\\ \text{so I need the biggest }b^2 \text { and the smallest }a^2\\ biggest \;\;b^2=9\qquad smallest\;\;a^2=1\\ \text{smallest value = }\frac{1}{9}-1 = -\frac{8}{9}\\ \)

(z + 1/z)^2   = 1

z^2  + 2z (1/z) + 1/z^2  = 1

z^2 + 2 + 1/z^2  =  1                 subtract 2 from both sides

z^2 + 1/z^2  =  -1

(z^2 + 1/z^2)^2  =  1

z^4 + 2 z^2(1/z^2) + 1/z^4  =  1

z^4 + 2 + 1/z^4  =  1               subtract 2 from  both sides

z^4 + 1/z^4  =  -1

(z^4 + 1/z^4)^3    = ( -1 )^3            binomial expansion on the left side

(z^4)^3  + 3(z^4)^2(1/z^4)  + 3(z^4)(1/z^4)^2  + (1/z^4)^3  =  -1

z^12  + 3(z^8/z^4) + 3(z^4/z^8 + 1/z^12   =  -1

z^12 + 3z^4 + 3/z^4  + 1/z^12  =  -1

z^12 + 3 (z^4 + 1/z^4)  + 1/z^12  =  -1

z^12 + 3(-1) +1/z^12  =  -1

z^12 + 1/z^12  - 3  =  -1          add 3 to  both sides

z^12 + 1/z^12  =  2

A little different from jugoslav's method  :

C = (5.0)

Construct circles  x^2 + y^2  =25      and  (x -5)^2 + y^2 = 1.25^2

Find the coordinates of B

Subtract the  first equation  from the  second

(x-5)^2  - x^2  = 1.25^2 - 25

x^2  -10x + 25  - x^2  =   1.25^2 -25 

-10x =   1.25^2 - 50

-10x =  25/16  - 800/16

-10x =  -775/16

x= 775/160  = 155/32

And

(155/32)^2  + y^2 = 25

y ^2 =  25 - (155/32)^2    

y ^2  =  1575/1024       take the negative  root

y  =  sqrt  (1575/1024)=  - (15/32)sqrt (7)

So B  =( 155/32,  -(15/32) sqrt (7)  )

Slope of line segment BC =   ( 0  + 15sqrt 7 (32) )  /( 5 - 155/32)  =  [ (15sqrt (7)) / 32 ] /  (5/32)   =

3sqrt (7) =   sqrt ( 63)

The equation of  a line through (0,0)  with this slope has the equation

y= sqrt (63)x

The intersection of this line with   x^2 + y^2  = 25    has the x coordinate

x^2 +  63x^2   =  25

64x^2  = 25

x^2  = 25/64       take the negative root

x = - 5/8

And 

y = -sqrt  (25 - 25/64)  = - sqrt (1575/64)  = -( 15/8) sqrt (7)  

H = ( -5/8, -(15/8) sqrt (7) )

So  CH  = sqrt  [ (5  + 5/8)^2  + ( (15/8)sqrt (7) )^2 ] =

sqrt  [( 45/8)^2  +  1575/64  ]  =

sqrt  [ 45^2 + 1575 ] / 8

sqrt  [ 3600] / 8  =  

60/8   =

7.5