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a + b (2)  = 15

a + b (5) = 3      subtract the second equation  from the first   and we have

     -3b  = 12      divide both sides by -3

         b =  -4

Sub this into the first equation and we have

a + -4(2)  = 15

a - 8  =  15          add 8 to both sides

a  =  23

So    a + b  =   23 + (-4)  =   19

Since this is a right triangle, we can use the Pythagorean theorem to find the missing side.

Let's call the missing side length  " x " .

6² + x²  =  12.3²        Subtract  6²  from both sides.

x²  =  12.3² - 6²         Simplify

x²  =  151.29 - 36

x²  =  115.29             Take the positive square root of both sides.

x  =  √115.29   ≈   10.737

But there is a small disagreement here... the side opposite the 30° must be the side with length  6 .

However,  6  is not exactly half of 12.3 . In a true 30-60-90 triangle, the side opposite the 30° angle is half the hypotenuse.

So this can't be exactly a 30-60-90 triangle, but it is close.

You can either use a fractions calculator, or use the division sign. Three fourth can written as 3 / 4. 

Not sure about this.....but I think it occurs twice [ in row 41 ]

This is just the first two.

1.

So we get  x² - 2x + 4 - 12/(x + 2)  as the final result.

2.

We get  2x² - 9x + 10  as the final result.

In the form ax^2 + bx + c.....the x coordinate of the max height occurs at

-b / [ 2a]  =  -32 / [ 2 (-16) ]  =  1

So.....the max height is   -16(1)^2 + 32(1) + 15  =  31 ft 

Here's the graph :  https://www.desmos.com/calculator/tp0ornpndr

 9x^2+nx+36=0

If this has only one [ double root ] solution......the discriminant will = 0

So

n^2  - 4(9)(36) =  0

n^2 - 36^2  = 0

n^2  = 36^2

n = 36

I'm  assuming this is

3/x+3/y=11/10

5/x+18/y = 4

Multiply the first equation by  -5    and the second by 3  ....  this gives

-15/x  - 15/x  =  -5.5

 15/x + 54/y = 12          add these

          39/y  = 6.5

          39/y  = 13/2    and we can write

           y / 39 = 2/13   multiply by 39

           y =  39 * 2 / 13  =  78/13   =  6

And putting this into the second equation for, y....we have

5/x + 18/6  =  4

5/x + 3  = 4       subtract 3 from both sides

5/x  = 1      which implies that x = 5    

WOW, Mathhemathh....I've not seen that method before.....but....I like it  !!!!

Kudos.....!!!

This parabola will turn downward since the directrix is above the foucus

The vertex will   be  at   ( -1, [ sum of directix and y coordinate of the vertex]/2 )  =

 (-1, [ 6 + 5 ] / 2 )  =  (-1, 11/2)  

And  the distance between the focus and the vertex  = p = -.5 = -1/2

And the equation  is

4p ( y - 11/2)  =  ( x + 1)^2

4 (-1/2) ( y - 11/2)  = (x + 1)^2

-2 ( y - 11/2)  = ( x + 1)^2

y - 11/2   =  (-1/2)( + 1)^2

y =  (-1/2)(x + 1)^2 + 11/2

Here's the graph : https://www.desmos.com/calculator/zfnw68cp1f

1)3x+3y=10 

2)(-9x - 9y)/-3 = -30/-3 <=> 3x + 3y = 10        so 1) = 2) so have solution for every x,y 

Both coins tossed are independent of each other, so each coin you toss will always have a $\frac{1}{2}$ chance of being either heads or tails.

Using distance formula, the distance of the parabola from its focus is $\sqrt{(x+1)^2+(y-5)^2}$ and the distance from the directrix is $\sqrt{(y-6)^2}$. On aparabola, these distances are always equal, so:

$y-6=\sqrt{(x+1)^2+(y-5)^2}$

$y^2-12y+36=(x+1)^2+y^2-10y+25$

$-2y+11=(x+1)^2$

$y=-{(x+1)^2-11\over2}$

And that is your equation.

OK.....

Put  (0,8)  into  3x + 2y  =  3(0) + 2(8)  = 0 + 16  = 16

       ( 5, 4) into  3x + 2y  =  3(5)  + 2(4)  =  15 + 8  = 23

       (9, 0) into  3x + 2y =  3(9) + 2(0)  =  27 + 0  =  27

So.....  (9,0)  maximizes the objective function....and the max value is 27

The first two  constraints just mean that our feasible region is found in the first quadrant

So....look at the graph of the other two, here : 

https://www.desmos.com/calculator/5dmzwfn5cj

A max will occur at a corner point.....there is only one at (1.5, 1.5)....so this point will max the objective function

I don't know how that's the thing...

The second answer is correct.

You begin by placing the opposite of the number (of the divisor) into the box on the top left.

Then, you bring down the coefficient of the squared term (the '2').

Now, you multiply the number in the box (the '5') times the coefficient of the squared term (the '2') and you place the answer (a '10') into the second column. This column now contains a '-1' and a '10; add these together, getting '9'.

Finally, multiply the number in the box (the '5') times the result of the second column (the '9'), getting '45'. Place this number into the third column and add down, getting '49'.

The division results in a quotient of '2x + 9' and a remainder of '49'.

We have 12 in total if we have to use at least two of the flavors......if we can have 4 scoops of only one flavor....then just add 3 to this giving 15 in all 

3x + x  + 5 - 2x      rearrange

3x + x - 2x  + 5      simplify

4x  - 2x  +  5

2x + 5

You can have plain flavors without mixing them together so that gives 3 plus all the mixtures.

Is your answer 9 or 12 or something else? 

NSS....Just sub the x and y value of each point into 3x + 2y .....you should have 3 results....the largest result is the max value

For instance..... if (3,4)  were on the graph....we have  3(3) + 2(4)  = 9 + 8 = 17

We have to take at least two scoops of one flavor if we use 4 scoops

I'm also assuming that we don't necessarily have to use all the flavors at once, but we must use at least two of the flavors

Here's the way I would do this.......but....maybe not the fastest

Flavors     Vanilla      Chocolate   Strawberry

Scoops      2                 1                 1 

                  1                 2                 1

                  1                 1                 2

                  1                 3                 0

                  2                 2                 0

                  3                 1                 0

                  1                 0                 3

                  2                 0                 2

                  3                 0                 1

                  0                 1                 3

                  0                 2                 2

                  0                 3                 1

If we have to use all 3 flavors......there are 3 new flavors......using my assumption adds 9 more

:D  👍

3x + 2y +  z  = 7     (1)

5x + 5y + 4z  = 3    (2)

3x + 2y + 3z  = 1    (3)

The object, NSS, is to eliminate a variable and end up with 2 equations with two unknowns

We can choose any variable that we want......here....z seems easiest

Multiply the first equation by -4  and add it to equation 2

-12x - 8y -  4z = -28

  5x  + 5y + 4z =   3

________________

  -7x - 3y          =  -25         →  7x + 3y = 25      (4)

Multiply the first equation by -3 and add it to to the 3rd equation

  -9x - 6y - 3z = -21

   3x  + 2y + 3z = 1

_________________

 - 6x - 4y      =  -20     →  6x + 4y  =  20    (5)

Multiply (4) by  4   and (5) by  -3

   28x +12 y = 100

 -18x -  12y  =  -60         add these together

10x =  40      divide both sides by 10

x = 4

Using (5) to find y, we have

6(4) + 4y = 20

24 + 4y  = 20    subtract24 from both sides

4y  = -4      divide both sides by 4

y = - 1

And using   3x  + 2y + 3z = 1  to find z, we have

3 (4) + 2 (-1) + 3z  = 1

12 - 2 + 3z  = 1

10 + 3z  = 1    subtract 10 from both sides

3z = -9       divide both sides by 

z = -3

So....{ x , y, z }  =  { 4, -1, -3 }

Just so you know, the 2 other coordinates were right, heureka. They are (2, 4) and (4, 2) because 4^2 = 2^4 = 8. However, the intersection is (e,e) to the limits of Desmos's calculations.