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Len [333]
3 years ago
8

Advanced order of operations 5-2•[3-{5+2}]=

Mathematics
1 answer:
irina [24]3 years ago
5 0

5 - 2 * (3 - [5 + 2])

5 + 2 = 7

5 - 2 * (3 - 7)

3 - 7 = -4

5 - 2 * (-4)

- 2 * - 4 = 8

5 + 8 = 13

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Mr. Kohl has a breaker containing n milliliters of solution to distribute to the students in his chemistry class. If he gives ea
77julia77 [94]

Answer:

There were 26 students in his class and the teacher had 83 ml of the solution.

Step-by-step explanation:

Mr. Kohl has a "x" amount of solution, if he divides it by the number of students "n" he'll give each student 3 milliliters  and have a left over of 5 milliliters. If the amount of solution Mr. Kohl had was "x + 21" then he'd be able to give each student 4 milliliters of the solution. From these informations we have:

x = 3*n + 5

(x + 21)/n = 4

x + 21 = 4*n

x = 4*n - 21

Now that we have two equations and two variables we can solve the system of equations, as seen bellow:

3*n + 5 = 4*n - 21

3*n - 4*n = -21 - 5

-n = -26

n = 26

x = 4*26 - 21 = 83 ml

There were 26 students in his class and the teacher had 83 ml of the solution.

3 0
3 years ago
What are the solutions to the equation 2(x-3)^2=54 ?
erastova [34]

Answer:

      x =(6-√108)/2=3-3√ 3 = -2.196

 x =(6+√108)/2=3+3√ 3 = 8.196

Step-by-step explanation:

Step  1  :

Equation at the end of step  1  :

2 • (x - 3)2 - 54 = 0

Step  2  :

 2.1    Evaluate :  (x-3)2   =  x2-6x+9 

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   2x2 - 12x - 36  =   2 • (x2 - 6x - 18) 

Adding  9  has completed the left hand side into a perfect square :

   x2-6x+9  =

   (x-3) • (x-3)  =

  (x-3)2  (x-3)1 =

   x-3

Now, applying the Square Root Principle to  Eq. #4.3.1  we get:

   x-3 = √ 27

Add  3  to both sides to obtain:

   x = 3 + √ 27

Since a square root has two values, one positive and the other negative

   x2 - 6x - 18 = 0

   has two solutions:

  x = 3 + √ 27

   or

  x = 3 - √ 27

Solve Quadratic Equation using the Quadratic Formula

 4.4     Solving    x2-6x-18 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  x  , the solution for   Ax2+Bx+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :

                                     

            - B  ±  √ B2-4AC

  x =   ————————

                      2A

  In our case,  A   =     1

                      B   =    -6

                      C   =  -18

Accordingly,  B2  -  4AC   =

                     36 - (-72) =

                     108

Applying the quadratic formula :

               6 ± √ 108

   x  =    —————

                    2

Can  √ 108 be simplified ?

Yes!   The prime factorization of  108   is

   2•2•3•3•3 

To be able to remove something from under the radical, there have to be  2  instances of it (because we are taking a square i.e. second root).

√ 108   =  √ 2•2•3•3•3   =2•3•√ 3   =

                ±  6 • √ 3

  √ 3   , rounded to 4 decimal digits, is   1.7321

 So now we are looking at:

           x  =  ( 6 ± 6 •  1.732 ) / 2

Two real solutions:

 x =(6+√108)/2=3+3√ 3 = 8.196

or:

 x =(6-√108)/2=3-3√ 3 = -2.196

7 0
3 years ago
How to solve to system y=x^2-2 y=2x+1
vovikov84 [41]

Answer:

the solutions are (3, 7) and (-1, -1)

Step-by-step explanation:

Insert the " = " symbol between the two equations, obtaining:

y = x^2 - 2 = y = 2x + 1

Then x^2 - 2 = 2x + 1, or

x^2 - 2x - 3 = 0, and this can be factored into (x - 3)(x + 1) = 0.

Thus, the x values that satisfy this system are {-1, 3}.

Use y = 2x + 1 to find the corresponding y values:

y = 2(-1) + 1 = -1

and

y = 2(3) + 1 = 7

Then the solutions are (3, 7) and (-1, -1)

7 0
4 years ago
2 5 x y + = <br> Solve for Y<br> Please hellppppppppp
Setler79 [48]
I do not think all of the question is there.

25 X y = / y would have to
7 0
4 years ago
How do I find X when the triangle is a right angle?
qwelly [4]

1:Replace the variables in the theorem with the values of the known sides.

2:Square the measures, and subtract from each side

3: find the square root of each side

6 0
4 years ago
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