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Gelneren [198K]
3 years ago
11

Find x 5[x^2]+5[x]-x^2-x=2004

Mathematics
2 answers:
Vsevolod [243]3 years ago
5 0

Answer:

-22.89, 21.89   to the nearest hundredth.

Step-by-step explanation:

5[x^2]+5[x]-x^2-x=2004

5x^2 - x^2 + 5x - x - 2004 = 0

4x^2 + 4x - 2004 = 0

Divide through by 4:

x^2 + x - 501 = 0

Using  the quadratic formula:

x = -1 +/- √(1^2 - 4*1*-501) / 2

x = -0.5 +/- 44.777/2

= -0.5 +/- 22.389

= -22.89, 21.89.

a_sh-v [17]3 years ago
4 0

The answer is either x=\frac{-1}{2}+\frac{1}{2}\sqrt{2005}\\ or x=\frac{-1}{2}+\frac{-1}{2}\sqrt{2005}

To solve: Let's solve your equation step-by-step.

5x2+5x−x2−x=2004

Step 1: Simplify both sides of the equation.

4x2+4x=2004

Step 2: Subtract 2004 from both sides.

4x2+4x−2004=2004−2004

4x2+4x−2004=0

Step 3: Use quadratic formula with a=4, b=4, c=-2004.

x= −b±√b2−4ac /2a

x= −4±√32080 /8

x=\frac{-1}{2}+\frac{1}{2}\sqrt{2005}\\ or x=\frac{-1}{2}+\frac{-1}{2}\sqrt{2005}

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{4x-2y+5z=6 <br> {3x+3y+8z=4 <br> {x-5y-3z=5
lesya692 [45]

There are three possible outcomes that you may encounter when working with these system of equations:


  •    one solution
  •    no solution
  •    infinite solutions

We are going to try and find values of x, y, and z that will satisfy all three equations at the same time. The following are the equations:

  1. 4x-2y+5z = 6
  2. 3x+3y+8z = 4
  3. x-5y-3z = 5

We are going to use elimination(or addition) method

Step 1: Choose to eliminate any one of the variables from any pair of equations.

In this case it looks like if we multiply the third equation by 4 and  subtracting it from equation 1, it will be fairly simple to eliminate the x term from the first and third equation.

So multiplying Left Hand Side(L.H.S) and Right Hand Side(R.H.S) of 3rd equation with 4 gives us a new equation 4.:

4. 4x-20y-12z = 20      

Subtracting eq. 4 from Eq. 1:

(L.HS) : 4x-2y+5z-(4x-20y-12z) = 18y+17z

(R.H.S) : 20 - 6 = 14

5. 18y+17z=14

Step 2:  Eliminate the SAME variable chosen in step 2 from any other pair of equations, creating a system of two equations and 2 unknowns.

Similarly if we multiply 3rd equation with 3 and then subtract it from eq. 2 we get:

(L.HS) : 3x+3y+8z-(3x-15y-9z) = 18y+17z

(R.H.S) : 4 - 15 = -11

6. 18y+17z = -11

Step 3:  Solve the remaining system of equations 6 and 5 found in step 2 and 1.

Now if we try to solve equations 5 and 6 for the variables y and z. Subtracting eq 6 from eq. 5 we get:

(L.HS) : 18y+17z-(18y+17z) = 0

(R.HS) : 14-(-11) = 25

0 = 25

which is false, hence no solution exists



3 0
2 years ago
Solve for w. 5w + 9z = 2z + 3w w = w equals negative StartFraction 7 Over 2 EndFraction z.Z w = w equals negative StartFraction
TEA [102]

Answer:

w = -7z/2

Step-by-step explanation:

We need to solve for w in terms of z.

The equation given is:

5w + 9z = 2z + 3w

Collect like terms:

5w - 3w = 2z - 9z

2w = -7z

w = -7z/2

5 0
3 years ago
Read 2 more answers
2a +5 - 3a -12 please
chubhunter [2.5K]

Answer:

-7-a

Step-by-step explanation:

2a-3a +5-12

-a - 7

or -7-a

7 0
2 years ago
Read 2 more answers
A deli offers 2 platters of sandwiches. Platter A has 2 roast beef and 3 turkey sandwiches. Platter B has 3 roast beef and 2 tur
erica [24]

Answer:

2 x+3 y and 3 x+ 2 y

Step-by-step explanation:

Let x represents the number of roast beef and y represents the number of turkey sandwiches.

Let r and t represent cost of the roast beef and sandwich.

Platter A:

Platter A has 2 roast beef and 3 turkey sandwiches. So, equation is 2 x+3 y.

Total cost of the platter = 2 r x+3 t y

Platter B:

Platter B has 3 roast beef and 2 turkey sandwiches. So, equation is 3 x+2 y.

Total cost of the platter = 3 r x+2 t y

So, the two equations are 2 x+3 y and 3 x+ 2 y.

7 0
3 years ago
Help needed quick!!!!!!!!!1
sleet_krkn [62]

Answer:

\cos(D) = 0.2195

Step-by-step explanation:

Given

The attached triangle

Required

\cos(D)

This is calculated as:

\cos(D) = \frac{Adjacent}{Hypotenuse}

So, we have:

\cos(D) = \frac{1.8}{8.2}

\cos(D) = 0.2195

5 0
3 years ago
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