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z= kxy/t^2

5 = k*2*(-3)/(1/4).    Hence k = -5/24

z = -(5/24)xy/t^2

When x = -12/(5z)^2 then z = -(5/24)(-12)y/(5zt)^2 or z^3 = y/(10t^2)

With z = 2/5 we have 8/125 = y/(10t^2) so that t = (5/4)y^(1/2)

Hence if f(y) = (5/4)y^(1/2) then f(y) = t when the above conditions hold.

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Let C be the original cost......and we have

[C * .90] * .80   = 1.08   simplify

.72C   = 1.08      divide both sides by .72

1.08/.72 = C =  $1.50

If this is even, then f(x)   = f(-x)

f(x)  = (x^3)/(2x^2-2)      f(-x)   = (-x)^3 / [ 2(-x)^2 - 2] = -x^3 / [ 2x^2 - 2]   and these are not the same, so it's not even

If it's odd, then  f(-x)  = -f(x)

f(-x)  = -x^3 / [ 2(-x)x^2 - 2]    and     -f(x) =  -x^3 / [2x^2 - 2]     and these are the same......so it's odd

The vertical asymptotes occur at the values that make the denominator 0, i.e., x = -1 and x = 1

There are no horizontal asymptotes......there is a "slant" asymptote of y = x/2

And this function has the same end behavior as y  = x^3 / [2x^2]  →  y = x/2

Here's the graph : https://www.desmos.com/calculator/dgix5c99wx

The marks scored by 20 students are as under

Marks      no of students       

0-6   -              4           

6-12 -              x

12-18-             5       let a+b=5  

16-24-             y

24-30-             1           

If 10 students scored below 14.4 marks and remaining 10 students scored above 14.4 marks,find the value of x and y

Since the 5 students in  the middle group could have all scorred below 14.4  OR they could of all scored above 14.4

there are mulpiple answers

If    x=1   then  y=9         All below

If    x=2   then  y=8

If    x=3   then  y=7

If    x=4   then  y=6

If     x=5  then  y=5

If     x=6  then  y=4        All above.

{y = x^3 - x^2 + 2

 y = x^2 + 15x + 2

For "y" in the first equation, substitute the second....and we have

x^3- x^2 + 2 = x^2 + 15x + 2     subtract the terms on the right side

x^3 - 2x^2 -15x = 0   factor

x [ x^2 -2x - 15] = 0

x (x - 5) (x + 3)  = 0   and setting each factor to 0, x = 0 or x = 5 or x = -3

Subbing these values back into either equation  gives the values {-3, -34}, {0, 2} and {5, 102}

Answer A is correct

You can see immediately that neither a nor d can be correct as they give negative heights for all values of t, whereas the question specifies positive heights.

You can also rule out b because this could only give a height of 161 at time t = 0, and for greater times, including t= 1, the height would be less than 161, which would conflict with the information given in the question.

Substitute one of the given times into c and e to see which produces the corresponding specified height.

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{y = x^3 - x^2 + 2

 y = x^2 + 15x + 2

A. (5, 102), (-3, -34), (0,2)

B. (5, 102), (-1, -12)

C. (5, 102), (3, 56), (1, 18)

D. no real solution

E. )-3, -34), (1, 18)

\(y_1 = x^3 - x^2 + 2\\  y_2 = x^2 + 15x + 2\\\)

Sub in (0,2) becasue it is the easiest.  Yes that makes both statements true

If x=-1

\(y_1=-1-1+2=0 \qquad \mbox{So if x=-1, y=0 So B is not correct}\)

If x=1

\(y_1=1-1+2=2 \qquad \mbox{not 18 so C and E are no good} \)

So that leaves A and D

Check the other two point in A and if they both work then that is the answer otherwise it must be D

(3/5)total = 12

total = 12*5/3

total = 20

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3000*1.03=3090

ans:3090

what is 1 (1/8)+(1/3)?

\(1\frac{1}{8} + \frac{1}{3}= \frac{1\times 3\times 8+3+8}{3\times 8} \) = \(\frac{35}{24} = 1\frac{11}{24} \)

Let  X  =  amount of water

Final Amount  =  X + 30

Percentage of sulfuric acid in water  =  0%

Final Amount x Final Percentage  =  (First Amount x First Percentage) + (Second Amount x Second Percentage)

        (X + 30) x (0.23)                    =  (                30 x 0.36                   ) + (                       X x 0.00                         )

           0.23X + 6.90                       =                      10.80                        +                            0.00

                                0.23X + 6.90  =  10.80

                                         0.23X  =  3.90

                                                 X  =  16.96 ml

Let  W  =  width of the plot.

Let  L  =  length of the plot.

L  =  2W - 3

Total perimeter =  L + W + L + W

                 510  =  (2W - 3) + (W) + (2W - 3) + (W)

                  510  =  6W - 6

                  516  =  6W

                    86  =  W

L  =  2W - 3     --->     L  =  2(86) - 3  =  169

3[8+6x+4]=162x+132

3[12+6x]=162x+132

12+6x=(162x+132)/3

 12+6x=54x+44

12-44=54x-6x

-32=48x

x=-32/48

   = - 2/3

4{ 8z - [9 - 7(7z - 5)] }

Working from the inside out:  -7(7z - 5)  =  -49z + 35

4{ 8z - [9 - 49z + 35] }

Simplifying:  9 - 49z + 35  =  44 - 49z

4{ 8z - [44 - 49z] }

Again, working from the inside out:  8z - [44 - 49z]  =  8z - 44 + 49z  =  57z - 44

4{ 57z - 44 }

228z - 176

3 (6 X + 12) = 162 X + 132

4 (8 z + 7 (7 z - 5) -9)

He must get a 94% on his 4th math exam to get an 88% on the class

(90 + 76 + 92 + 94) 352

352 divided by 4 = 88%

3. N=6 because clearly 6 == 6 for any mod, and no number under 6 satisfies the conditions.

24 = 2 x 2 x 2 x 3

x=2.8473221018630726395189199142257

x^3 - 6x - 6=0

(2.8473221)^3 - 6(2.8473221) - 6 =0

23.0839326 - 17.0839326 - 6 = 0

                    6 - 6 = 0

x  =  2^1/3 + 2^2/3     --->     x³ - 6x - 6  =  0

First, let's evaluate:  [ 2^1/3 + 2^2/3 ]³

 Notice that  [ a + b ]³  =  a³ + 3a²b + 3ab² + b³

The x³ part:     [ 2^1/3 + 2^2/3 ]³  =  [2^1/3]³ + 3·[2^1/3]²·[2^2/3] + 3·[2^1/3]·[2^2/3]² + [2^2/3]³

                                              =  2^3/3 + 3·[2^2/3]·[2^2/3] + 3·[2^1/3]·[2^4/3] + 2^6/3

                                              =  2 + 3·2^4/3 + 3·2^5/3 + 4

                                              =  6 + 3·2·2^1/3 + 3·2·2^2/3 

                                              =  6 + 6·2^1/3 + 6·2^2/3 

The 6x part:    6·[ 2^1/3 + 2^2/3 ]  =  6·2^1/3 + 6·2^2/3 

Thus:  x³ - 6x - 6  =  [ 6 + 6·2^1/3 + 6·2^2/3  ] - [ 6·2^1/3 + 6·2^2/3  ] - 6

                             =  6 + 6·2^1/3 + 6·2^2/3 - 6·2^1/3 - 6·2^2/3 - 6

                             =  0