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I am sorry but this is incorrect

Here's another way to solve this......

The area  of the triangle  =   

Integer coordinates on the boundary / 2  +  integer coordinates in the interior   - 1

The area  of the  triangle  is   20 * 5 / 2 =    50   units ^2

The line  joining   (-10,0)  and (0, 5)   has the equation     y  = x/2  + 5

So we will have integer  coodinates on this line when x=  -10, x = -8, x = -6,  x = -4,  x  = -2 and  x  = 0

Using symmety....we will also have integer coordinates when  x = 2, x = 4 ,  x = 6  x  = 8  and x  = 10

And we have  19 additional  integer coordinates on the  base

So.....the number of  integer coordinates on the  boundary  = 11 + 19  = 30

So

50 =  30 / 2  +   integer  coordinates in the interior  - 1

50  = 30/2  +  integer coordinates in the interior   - 1

50  =  15  - 1   + integer coordinates in the  interior

50 =  14   +  integer coodinates in the interior

36   =    integer coodinates in the interior

Even plaindromes

Starts  with                        Ends with

    2                                         2                 

    4                                         4

    6                                         6

    8                                         8

We have 10  choices for the middle two numbers  in each case   = 40  possibiliities

We have   90   4 digit palindromes   (the other 50  come from starting and ending the number with 1,3,5,7, or 9....and for each of these we have  10 possibiliites)

So....the  probability of an even palindrome  =   40  / 90    = 4 / 9

A 3rd degree polynomial has the  form      ax^3  + bx^2 + cx + d

Since  f(0)  = 0     then   x = 0   is one of the y intercepts  and    d  = 0

So....we can solve  for  a, b and  c     as follows :

a(-1)^3  + b(-1)^2  + c(-1) =  15  ⇒      -a   +  b - c  = 15     (1)

a(1)^3  + b(1)^2  + c(1)=  -5  ⇒              a    +  b  + c =  -5   (2)

a(2)^3  + b(2)^2  + c(2)  =  12  ⇒           8a   + 4b  + 2c  =  12    (3)

Adding  (1)  and (2)   we have that     2b  =  10 ⇒    b  = 5

Sub this into  each equation for b and we  have that

-a +5 - c  =  15 ⇒    -a  - c =  10    (4)

a +5  + c  = -5  ⇒   a  + c  = -10        (5)

8a + 4(5) + 2c =  12  ⇒   8a  + 2c = -8   (6)

Multiply (5)  by -2  and add it to (6)   and we get that     6a  =12 ⇒   a  = 2

Using (5)  to find c  we have that    2 + c  = -10 ⇒    c  =-12

So....the polynomial is   f(x)  =   2x^3 + 5x^2  - 12x      set to 0 and factor and we  get  that

(2x  - 3) (x + 4 )  = 0

Setting each factor to  0 and  solving  for  x  and  we  get the  other  two x intercepts  of   x  = 3/2    and x =  -4

See the graph here  : https://www.desmos.com/calculator/ef2xuaesrm

Suppose a is directly proportional to b, but inversely proportional to c. If a=2 when b=5 and c=9, then what is c when b=3?

\(a:b=2:5\\ a:c=9:2\)                    Nonsense. How blamable 
 

\(b=\frac{5a}{2}\\ c=\frac{2a}{9}\)

b=3

\(a=\frac{2b}{5 }\\ a=\frac{2\cdot 3}{5}\\ a=\frac{6}{5}\\ c=\frac{2a}{9}=\frac{2\cdot6}{9\cdot 5}\\ \color{blue}c=\frac{4}{15}\ \color{black}Not\ correct.\)      

  !

\(a = p b\\ a = \dfrac q c\\ c = \dfrac{q}{pb}\)

\(2=5p \Rightarrow p=\dfrac 2 5\\ 2 = \dfrac q 9 \Rightarrow q = 18\\~\\ c = \dfrac{18}{\frac 2 5 \cdot 3} = 15 \)

well basically you are looking at the set of output from each matrix when fed all possible vectors.

The matrix being invertible should immediately suggest one of the 3 options.

Think about the output of matrix A.  It tells you it maps every vector to the zero vector.

Is the zero vector a point, line, plane, or all of 3D space?

Simlarly with C.  Every output it produces lies along the vector i+j.  That suggests that the output grid covers a .. ?

Let the small size cost  = C   and let the  volume of the  small  size be V

So....the unit cost for the  small size is  just  cost / volume  =   C/ V

The medium size  costs 50% more than the  small  size....so it  cost is  1.5C

The large size holds twice as  much as the  small size =  2V   and  costs 30% more than the medium size = (1.5)C (1.3)  = 1.95C

So...the unit  cost  for the large  size =  1.95C / 2V =   =   (195/200)C/V  =  (39/40)C/V =  .975C/V

Finally....the  medium size contains 20% less detergent than the large  size=   2V (.80)   = 1.6V

So....the uint  cost  for the  meduim  size  is  just   ( 1.5)C / (1.6 V) = (15/16)C/V  = .9375C/V

So.....best to worse we have

.9375C/V       .975C/V           C / V 

    M                   L                    S          ⇒        "E"

z^2  =  2i

(a + bi)^2  =   0 + 2i

a^2  + 2abi  - b^2   =   0 + 2i              

(a^2 - b^2)   + 2abi  =  0 + 2i        equate  coefficients

a^2 - b^2  = 0        (1)                     and             ab   = 1         (2)

(a - b) ( a + b)  =0 

(a - b)  = 0     or  ( a + b)  = 0

a = b          or         a  =  - b      (reject     because  a and  b will  have opposite signs and   ab  will be  negative if this is true  )

If   a  = b    then    (2)  becomes     a^2  =1        ⇒  a  = ±1

And if a   = 1   then b  = 1

And if a  = -1, then b  =-1

So....the solutions are       a  = 1, b  = 1     ⇒     z  = 1  + i        or    a =- 1 , b = -1   ⇒   z  = -1 - i 

We can geneate  a  liunear regression  approximation   using this website  : :   https://www.socscistatistics.com/tests/regression/default.aspx

The equation is

y = 1.58095 x - 1.12975

When x  = -3.....the approximate value fo y  is   -5.8726  ≈  -5.9

When x  = -20    .....the  approximate  value  for  y   is   -32.7488  ≈  -32.7

f ( √ [ x + 1) )   =  1  / x

f(2)   implies that

2  = √ [x + 1 ]        square both sides

4  = x + 1      subtract 1 from  both sides

3  = x

So

f(2)    =     1 / 3

We have a vertical asymptote at  x  = -1   because this value makes the  denominator undefined....

So....the  domain  iss (-inf, -1 ) U (-1, inf)

See my previous answer  as to the range of this rational function for the  graph

f(x)   =  3x - 7

            ______

              x +  1

We have  vertical  asymptote at  x = -1      and a horizontal asymptote at  y =  3

So......when x  < -1      y  is  >  3

And when  x  > -1       y is < 3

So....the range  is   ( -infinity , 3)  U (3 , inf)

Here is a graph : https://www.desmos.com/calculator/r9bpootve2

Let me ask YOU some questions.  Just to help you learn.

One of them can be seen immediately.  What one is that?

How many (at most) can there be?

Thank you, Melody !

ok that is a pity. I'll see you later :)

432=  2*2*2*2*3*3*3*1   (times 1 as many times as you want)

So the single digit numbers that go into 432 are   1, 2, 3, 4, 6, 8, 9

To add 4 numbers to 20 I need an even number of even numbers.  ie, 0,2 or 4 even numbers

Lets see if i can split 432 into 4 odd single digit factors ...odd, odd, odd, odd      This is impossible.

Ok

Lets see if i can split 432 into 2 odd single digit factors and 2 even digit factors ...odd, odd, even, even  

9 * 3 * 8 * 2  = 432       But these add to 22  So they are no good

9 * 3 * 4 * 4   = 432   These add to 20   

9 * 1 * 8 * 6   = 432     But these add to 24 so they are no good.

3 * 3 * 6 * 8  = 432   These add to 20    

AND LASTLY

Lets see if i can split 432 into 4 odd single digit factors ...even, even, even, even

6 * 6 * 6 * 2 = 432     These add to 20    

So there are only 3 combinations of digits possible

9 * 3 * 4 * 4     This has the biggest digit so the largest number will be    9443

3 * 3 * 6 * 8 

6 * 6 * 6 * 2     This has the smallest digit so the smallest number will be  2666

9443-2666 = 6777

Note: I did not use a calculator but I would not get many of this level question done in 40 minutes either.

Still not all the ones you have been asking have been so time consuming.

Hi Melody,

okay nothing is working for some reason. We are restricted to downloading apps on our work pc's, I will make a plan to get my home computer connected to our wifi at home, and just work from there. So, these questions will just have to wait.......

Chat again later...

Hi Melody,

thank you for that...I remember now how I used to do it in the past, I used gladwin printscreen app. I will try that again now, and if no luck, will look into your suggestion. Thanx a lot...

I did not realize it was going to be so complicated.

Nicely done Heureka!

Here is the graph from Desmos graphing calculator

Hi Juriemagic,

Have you looked at how I post pics?  I have not had any poblems at all.  

Can you give it a go?

Find all solutions to the inequality \(\large{\dfrac{(2x-7)(x-3)}{x} \ge 0}\) in interval notation

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{(2x-7)(x-3)}{x}} &\ge& \mathbf{0} \quad | \quad \boxed{x\neq 0 !} \\\\ \dfrac{(2x-7)(x-3)}{x} &\ge& 0 \quad | \quad \times x^2 \\ \dfrac{(2x-7)(x-3)x^2}{x} &\ge& 0\times x^2 \\ (2x-7)(x-3)x &\ge& 0 \\ 2\left(x-\dfrac{7}{2}\right)(x-3)x &\ge& 0 \quad | \quad : 2 \\ \left(x-\dfrac{7}{2}\right)(x-3)x &\ge& 0 \\ \mathbf{ \left(x-\dfrac{7}{2}\right)(x-3)(x-0) } &\ge& \mathbf{0} \\ \hline \end{array}\)

We create a sign table:

\(\begin{array}{|l|c|c|c|c|c|c|c|} \hline \text{Interval or position} : & (-\infty,0) & 0 & (0,3) & 3 & \left(3,\dfrac{7}{2}\right) &\dfrac{7}{2} & \left(\dfrac{7}{2},\infty\right) \\ \hline \text{sign of } (x-0): & - & 0 & + & + & + & + & + \\ \hline \text{sign of } (x-3): & - & - & - & 0 & + & + & + \\ \hline \text{sign of } \left(x-\dfrac{7}{2}\right): & - & - & - & - & - & 0 & + \\ \hline \text{sign of }\left(x-\dfrac{7}{2}\right)(x-3)(x-0): & - & \color{red}0 & \color{red}+ & \color{red}0 & - & \color{red}0 & \color{red}+ \\ \hline \end{array}\)

We can read the result:

in the interval \([0,3] \)and \(\Big[\dfrac{7}{2},\infty\Big)\) the left side of the inequality is positive or zero, and thus the inequality is true there.

But \(\boxed{x\neq 0 !}\) and the solution is: \(\Big(0,3\Big]\) and \(\Big[\dfrac{7}{2},\infty\Big)\)

Two companies working together can clear a parcel of land in 26 hours.  Working alone, it would take Company A 10 hours longer to clear the land than it would Company B.  How long would it take Company B to clear the parcel of land alone?

Thanks guests.

I struggle with these myself.  I decided to have a go.  I got the same as the last guest.    

Speed of A + Speed of B = speed of both together

\(speed=\frac{\text{Land to be cleared}}{time \;taken}\)

Let the land to be cleared be L

\(S_B=\frac{L}{t_B}\qquad \qquad S_A=\frac{L}{t_B+10} \qquad \qquad \\~\\ S_B+S_A=\frac{L}{26}\\ \frac{L}{t_B}+\frac{L}{t_B+10} =\frac{L}{26}\\ \frac{1}{t_B}+\frac{1}{t_B+10} =\frac{1}{26}\\ \text{It will be easier to follow if I let }Z=T_B\\ \qquad \frac{1}{Z}+\frac{1}{Z+10} =\frac{1}{26}\\ \qquad \frac{Z}{Z(Z+10)}+\frac{Z+10}{Z(Z+10)} =\frac{1}{26}\\ \qquad\frac{2Z+10}{Z(Z+10)} =\frac{1}{26}\\ \qquad 52Z+260=Z^2+10Z\\ \qquad Z^2-42Z-260=0\\ \qquad \text{Using the quadrativ euqation the answer is}\\ \qquad Z=21+\sqrt{701}\approx 47.48 \)

so   On their own B will take 47.48 hours to clear the land.

Guest above has already checked this answer.

latex interpreter fails

If we restrict the domain from the vertex to positive infinity then f(x) will be a bijection.

We can read the vertex off as x=-2, so we restict the domain as -2 <= x < +infinity