All Answers 356785 Answers

x^3  -3x^2  - 10x

First....take out the GCF  = x

x ( x^2 - 3x  - 10)

Now....to factor the polynomial inside the parentheses, we need two numbers that multiply to -10  and sum to -3......  -5  and +2 are what we need....so we have......

x ( x - 5) ( x + 2)

jeez i am going nuts with these odd answers...sorry if im being a bother

actually melody is at 91 thousand

Thanks, rebecca.....!!!

The inverse will have reverse coordinates.....so the points on the inverse  are

(-8, -2),  (9, 0)

f(x)  = 4x + 3

For f(x), write y

y = 4x + 3     subtract 3 from both sides

y - 3  =  4x    divide both sides by   4

[ y - 3 ] / 4  =  x     swap x and y

[x - 3] / 4  = y     for y, write f^-1(x)

f^-1 (x)  =  (1/4)x - 3/4

f(x)  = (1/4)x  + 10

y = (1/4)x + 10

y - 10  =  (1/4)x     multiply both sides by 4

4y - 40  =  x

4x - 40  = y

f^-1(x)  = 4x - 40

We are using these  log properties

log a +  log b  =  log(a * b)

log a - log b  =  log (a /b)

alog b  =  log b^a

So we have

log x  + log y - 2 log z    =

log  (x * y)  - log z^2

log  [ (x * y) / z^2 ]

now its edited

is that me? how rude

I represent that remark!

Solution (and precipitates):

\(\text {Solve and balance the equation to determine the reaction ratio. }\\ \text{ }\\ Ni(NO_3)_2 + K_2CO_3 \Rightarrow NiCO_3 + 2KNO_3 \small \text { Note that the ratio is 1:1}\\ \small \text {Determine the number of moles in the solution of 85ml with a 0.55 molar concentration of potassium carbonate. }\\ 0.25 * 85*10^{-3} L = \small \text{moles of potassium carbonate }\\ \small \text{Divide this by the 0.55 molar concentration of nickel (II) nitrate }\\ 0.02125/0.55 = 0.0386 L \\ 38.6 \text{ml} \small \text { of 0.55 molar nickel (II) nitrate } \)

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Easy as pealing a high potassium banana

GA

A molasses can is made up with a cylindrical base having a radius of 6 cm that is topped by a smaller cylindrical pouring spout with a radius 2 cm. When upright the top of the molasses is 14 cm above the base of the can but when turned over the top of the molasses is 19 1/3 cm above the base. Find the total height of the can in cm.

Do you mean that when the can is turned over the top of the molasses is 19 1/3 cm BELOW the base?

Afterall, if the container is upside down then the base is at the top.

ALSO, what has the pouring spout to do with it?   When the container is upside down does the pouring spout fill up?  We would need to know how long the pouring spout is to accurately answer this question.

Hi Old Timer    

Sum of n terms for -3n^2+2n+5

\(\displaystyle\sum_{m=1}^n\; (-3m^2+2m+5)\\ =\displaystyle\sum_{m=1}^n\; -3m^2\quad+\displaystyle\sum_{m=1}^n2m\quad +5n\\ =\displaystyle\sum_{m=1}^n\; -3m^2\quad+\displaystyle\sum_{m=1}^n2m\quad +5n\\ \qquad\qquad \displaystyle\sum_{m=1}^n2m=\frac{n}{2}(a+L) = \frac{n}{2}(2+2n) = n(1+n)=n^2+n\\ =-3\left[\displaystyle\sum_{m=1}^n\; m^2\right]\quad+n^2+n\quad +5n\\ \text{** the next line was taken from a khan academy video which I will reference at the end.}\\ =-3\left[\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}\right]\quad+n^2+n\quad +5n\\ =\left[-n^3-\frac{3n^2}{2}-\frac{n}{2}\right]\quad+n^2+n\quad +5n\\ =\frac{-2n^3}{2}-\frac{3n^2}{2}-\frac{n}{2}\quad+\frac{2n^2+2n\quad +10n}{2}\\~\\ =\dfrac{ -2n^3-n^2 +11n }{2} \)

Khan Academy videos:

You should post it as an equation instead, so that it's easier to understand your question.

Find the remainder when r^13 + 1 is divided by r-1.

Geometric sequence:

\(\begin{array}{|rcll|} \hline \displaystyle 1+r+r^2+r^3+r^4+\ldots +r^{12} &=& \displaystyle\frac{r^{13}-1}{r-1} \\\\ \displaystyle 1+r+r^2+r^3+r^4+\ldots +r^{12} &=& \displaystyle\frac{r^{13}}{r-1}- \frac{1}{r-1} \quad & | \quad + \frac{2}{r-1} \\\\ \displaystyle 1+r+r^2+r^3+r^4+\ldots +r^{12} +\frac{2}{r-1} &=& \displaystyle\frac{r^{13}}{r-1}- \frac{1}{r-1} + \frac{2}{r-1} \\\\ \displaystyle 1+r+r^2+r^3+r^4+\ldots +r^{12} +\frac{2}{r-1} &=& \displaystyle\frac{r^{13}}{r-1}+ \frac{1}{r-1} \\\\ \displaystyle 1+r+r^2+r^3+r^4+\ldots +r^{12} +\frac{\color{red}2}{r-1} &=& \displaystyle\frac{r^{13}+1}{r-1} \\ \hline \end{array}\)

The remainder is 2.

The multiples of 12 and of 21 are written in order.

What is the smallest positive difference between a multiple of 12 and a multiple of 21?

The difference is \(n\cdot 21 -m \cdot 12 = d \) with \( n,m,d \in Z\) has a solution,

if the \(gcd(21,12)\) is a divider of \(d\). Or \(gcd(21,12) | d\).

The greatest common divisor  \(gcd(21,12) = 3.\)

The difference is \(n\cdot gcd(21,12)\)

so the difference is a multiple of 3

The differences are: \(0, 3, 6, 9, 12, \ldots\)

The smallest positive difference is 3.

-218   + 262    =   44° C

 t^4-t^2+3t-7 is divided by t^2-3t+8

                                 t^2     +   3t    

t^2  -  3t   +  8     [     t^4     +  0t^3     -  t^2     +   3t      -   7    ]

                                 t^4     -  3t^3     +8t^2

                                 _________________________________

                                             3t^3     -  9t^2    +   3t

                                             3t^3    -  9t^2    +   24t

                                            _______________________

                                                                     -  21t      - 7

Quotient   t^2  + 3t      Remainder     -21t  - 7

x^4  - 3x^3  + 6x^2  - 12x  + 8

Rewrite as  :

x^4  - 3x^3  + 2x^2  +  4x^2 - 12x + 8

x^2 (x^2- 3x + 2) +  4 (x^2 - 3x + 2)

(x^2 + 4)  ( x^2 - 3x  + 2)

(x^2  + 4)  (x -2) ( x - 1)

The roots are    x  = 1, x = 2   and  x = ±2i

If this is divisible by  x^2  - 1.....it must be divisible by  (x - 1)  and ( x + 1)

So....using synthetic division

1  [   1     a         b  + 2                    1       ]

               1        a  +  1          a+ b + 3 

       ______________________________

        1   a + 1     a + b + 3     a +  b + 4

-1  [   1       a           b +   2                   1          ]

                  -1          -a +  1               a  -  b  - 3   

         ______________________________

          1     a - 1      b - a  + 3           a  -  b  - 2

And  for 1 and -1 to be roots.....the following system must be true

a  + b  +   4   =  0

a  -  b  -   2    =  0      add  these

2a   +   2   =    0        ⇒   a  =  -1

And    a  + b  + 4  = 0

          -1   + b  +  4  = 0

                b + 3   =  0

                 b  = -3   

So  (a,b)   =  ( -1, - 3) 

Using synthetic division, we have

-4  [  9    1    0     - 12          21  ]

            -36  140   -560   2288

        ___________________

        9  -35  140   -572   2309

So......the quotient is    9x^3 - 35x^2  + 140x - 572   remainder  2309    

2/3  =  (4/3) / 2

So he needs 4/3 scoops. That is 1 and 1/3 scoops.

Performing synthetic division, we have

1  [   1    0    0    0    0    0   0    0   0   0   0  0   0    1 ]

              1    1    1    1    1   1    1   1   1   1  1   1    1

          ____________________________________

        1    1    1    1    1     1   1   1   1    1   1  1   1   2

The remainder is 2