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I think you mean,  do you smell color when you are falling into water?

If it is regular water it is most likely no, but if it is pool water it depends on the person.

So the question is do you smell color falling into pool water? Or Do you smell color falling into water.Try it and see, if you can't get to a pool at least try it with regular water.

By simple calculation, the smallest n that works is n = 6.

Take 1/2 of the 'x' coefficient.....square it and use that value in the blank      (1/2 * 5/4)² = ............

Monday morning    4 1/4 hrs

            afternoon    4 1/4 hrs       total hours   8 1/2

if no overtime wages     8.5 hr x $13 /hr =   8.5 * 13 = $ ............

2.5 is correct

Graph the line equation to see the x and y-axis intercepts (or realize by subbing in x = 0     and y = 0  that the intercepts are both 8)

right triangle with legs = 8

area = 1/2 b * h =   1/2 *(8)(8)  32 units²

Thank you!!!

Are you sure?

Same question already answered here:

https://web2.0calc.com/questions/please-help_33982

The lowest point shown in the graph is y = 2.5

2 tsp /  (1/4 tsp/qt) =  2 * 4/1 = 8 qts

Carpet is usually sold by the yd²      your question does not include units for the $ 8 per.......

IF it is square yards then     15/3 x12/3 = 5 x 4 = 20 sq yds       20 yd²  * $ 8/ yd² =  $ 160

.030303.....  =       (3  - 0)  / (99)     =   3/99  =   1/33

.003003..... =       (3 - 0)  / ( 999)  =  3/999 =   1/333

(1/33)   / ( 1/333)   =

333 / 33   =   111 / 11  =    10 + 1/11

angle BFD = Angle CED

15 * 12   * $ 8/ sq ft   =    $1440

Okay ty what is ur teachers name?

A full  bowl  must contain  11.5 + 1.25 + 2  = 14.75  cups

So

14.75   -  12 =   n

2.75  cups  =  n

Regarded as a set of four equations in the four unknowns a, b, c and d, (k being a parameter),

the equations will possess non-zero solutions only if the determinant of coefficients is equal to zero.

That is,

$\displaystyle \begin{array} {| cccc |}1 & 1 & -k & 0 \\ 0 & 1 & 1 & -k \\ -k & 0 & 1 & 1\\ 1 & -k & 0 & 1 \\ \end{array}=0$

Expanding, that gets you 

$\displaystyle k^{4}-2k^{2} -4k = 0, \\ \text{ or, }\\ k(k-2)(k^{2}+2k+2)=0, \\ \text{ so } \\ k=0 \text{ or } k=2.$

k = 0 produces the solution a = t, b = -t, c = t, d = -t , t a parameter, from which a/b + b/c + c/d + d/a = -4.

k = 2 produces a = b = c = d = t, t a parameter so a/b + b/c + c/d + d/a = 4.

So the maximum would appear to be 4.

101 - 72 = 29.

$\displaystyle 29 \equiv -72 \bmod101,\\ 100 \equiv -1 \bmod 101, \\ \text{so}\\ 2900 \equiv 72 \bmod 101.$

Let  the original  fraction be  x  /  y

After  the  decreases we have

.15x  / .75 y   =   ( 1/5) (x/y)    =   .20 (x / y) 

So  .....         .20(x/y)   =  (1 -.80) (x/y)   =    80%  decrease

A polynomial is of degree 3 if it is of the form $h(x)=ax^3+bx^2+cx+d$.

$f(x)+cg(x)=1-12x+3x^2-4x^3+2x^4+c(3-2x-6x^3+9x^4)\\ f(x)+cg(x)=1-12x+3x^2-4x^3+2x^4+3c-2cx-6cx^3+9cx^4\\ f(x)+cg(x)=(9c+2)x^4+(-6c-4)x^3+3x^2+(-2c-12)x+(3c+1)$

Expanding completely really was not necessary, but I did for completeness sake. In order to make this polynomial a degree 3 polynomial, the coefficient of x^4 should be 0.

$9c+2=0\\ c=-\frac{2}{9}$

That's it!

This question gives more information than necessary because writing the equation of a line really only requires two points. The third one is not really necessary.

Let's find the slope of the line. Since the slope of any line is constant, the two points used to compute the slope is immaterial. Just pick any two. I will pick thw first two in the table, $(-1,\frac{7}{2})\text{ and } (1,-\frac{1}{2})$.

$m=\frac{\frac{7}{2}-(-\frac{1}{2})}{-1-1}\\ m=-\frac{4}{2}\\ m=-2$

Because we can write the equation in any form, I will opt for point-slope form, as it is probably the easiest one to construct in this particular situation.

Point-slope Form: $y-y_1=m(x-x_1)$ where m is the slope and $(x_1,y_1)$ is any point on the line.

Equation of Line: $y-\frac{7}{2}=-2(x+1)$

10.5  =   10 + 5/10  =   10  +  1/2   =    (2 *10  +  1)  / 2  =   21  / 2

There are a few ways to approach this question. Here was my approach.

A point lies on an arbitrary line if the slope between any two points always remains constant. We already have two points, so we can determine the slope of this arbitrary line in space. Be careful with your signs!

$m=\frac{\text{change in y}}{\text{change in x}}\\ m=\frac{-2-(-7)}{3-(-1)}\\ m=\frac{5}{4}$

Now, let's find the slope between $(-6,y)\text{ and }(-1,-7)$.

$m=\frac{\text{change in y}}{\text{change in x}}\\ m=\frac{y-(-7)}{-6-(-1)}\\ m=\frac{y+7}{-5}\\ $

As I already stated, the slopes must be the same in order for this point to be on this line. It is now possible to solve for y.

$\frac{y+7}{-5}=\frac{5}{4}\\ 4(y+7)=-5*5\\ 4y+28=-25\\ 4y=-53\\ y=-\frac{53}{4}$

We can write

2  +  log (2a)   / log 2    =  3   +  log (2b)  / log 4    +  log (a + b)  / log 6

2  +  (log 2 + log a)  / (log 2)  =  3   +  (log 2) (log b) / log (2)^2   +   log (a + b) / log 6

2 +  1   + log a  / log 2     =   3  +  (log 2) (log b) / (2 log 2)    +  log (a + b)  / log 6

log a / log 2  =  log b /2    +  log (a + b)  / log 6

( 1/2)  (log a  - log b)  =  log ( a +b) / log 6

log a  -  log b  = 2 log (a + b)/log 6

log 6 * log (a/ b)  = 2 log ( a + b)

log 6 / 2  =   log (a + b)  / (log (a/b))

Note that this will  be true  when   a = 4   and  b =   2      

Check  :

2 +  log₂ 8   =  3  + log ₄ 4  +  log ₆ 6

2 +  3  =  3  +  1   +   1

5   =  5

So    (a + b)  /(ab)   =        6   /  8    =   3  / 4

is that

$log_2(2a)\qquad or \qquad log_{10}(2*2a)$

I assume it is the first one?