How do you write "the difference between twenty and a number and four is six" as an algebraic expression?

. If α and β are the roots of

Lostsunrise [7]

Answer:

$18x^2+85x+18 = 0$

Step-by-step explanation:

Given Equation is

=> $2x^2+7x-9=0$

Comparing it with $ax^2+bx+c = 0$ , we get

=> a = 2, b = 7 and c = -9

So,

Sum of roots = α+β = $-\frac{b}{a}$

α+β = -7/2

Product of roots = αβ = c/a

αβ = -9/2

Now, Finding the equation whose roots are:

α/β ,β/α

Sum of Roots = $\frac{\alpha }{\beta } + \frac{\beta }{\alpha }$

Sum of Roots = $\frac{\alpha^2+\beta^2 }{\alpha \beta }$

Sum of Roots = $\frac{(\alpha+\beta )^2-2\alpha\beta }{\alpha\beta }$

Sum of roots = $(\frac{-7}{2} )^2-2(\frac{-9}{2} ) / \frac{-9}{2}$

Sum of roots = $\frac{49}{4} + 9 /\frac{-9}{2}$

Sum of Roots = $\frac{49+36}{4} / \frac{-9}{2}$

Sum of roots = $\frac{85}{4} * \frac{2}{-9}$

Sum of roots = S = $-\frac{85}{18}$

Product of Roots = $\frac{\alpha }{\beta } \frac{\beta }{\alpha }$

Product of Roots = P = 1

The Quadratic Equation is:

=> $x^2-Sx+P = 0$

=> $x^2 - (-\frac{85}{18} )x+1 = 0$

=> $x^2 + \frac{85}{18}x + 1 = 0$

=> $18x^2+85x+18 = 0$

This is the required quadratic equation.

5 0

3 years ago

Read 2 more answers

A product can be made in sizes huge, average and tiny which yield a net unit profit of $14, $10, and$5, respectively. Three cent

navik [9.2K]

Answer:

The model is:

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃ to maximize

Subject to:

First center X₁₁ + X₂₁ + X₃₁ ≤ 550

Second center X₁₂ + X₂₂ + X₃₂ ≤ 750

Third center X₁₃ + X₂₃ + X₃₃ ≤ 275

22* X₁₁ + 16* X₂₁ + 9*X₃₁ ≤ 11000

22* X₁₂ + 16* X₂₂ + 9*X₃₂ ≤ 2700

22*X₁₃ + 16* X₂₃ + 9*X₃₃ ≤ 3400

X₁₁ + X₁₂ + X₁₃ ≤ 710

X₂₁ + X₂₂ + X₂₃ ≤ 900

X₃₁ + X₃₂ + X₃₃ ≤ 350

2700*(X₁₁ + X ₂₁ + X₃₁) - 11000*(X₁₂ + X₂₂ + X₃₂) = 0

3400*(X₁₁ + X ₂₁ + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

Xij >= 0

Step-by-step explanation:

Let´s call Xij product size i produced in center j

According to this, we get the following set of variable

X₁₁ product size huge produced in center 1

X₁₂ product size huge produced in center 2

X₁₃ product size huge produced in center 3

X₂₁ product size average produced in center 1

X₂₂ product size average produced in center 2

X₂₃ product size average produced in center 3

X₃₁ product size-tiny produced in center 1

X₃₂ product size-tiny produced in center 2

X₃₃ product size-tiny produced in center 3

Then Objective function is

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃

Constrains

Center capacity

1.- First center X₁₁ + X₂₁ + X₃₁ ≤ 550

2.- Second center X₁₂ + X₂₂ + X₃₂ ≤ 750

3.- Third center X₁₃ + X₂₃ + X₃₃ ≤ 275

Water available

1.- 22* X₁₁ + 16* X₂₁ + 9*X₃₁ ≤ 11000

2.- 22* X₁₂ + 16* X₂₂ + 9*X₃₂ ≤ 2700

3.- 22*X₁₃ + 16* X₂₃ + 9*X₃₃ ≤ 3400

Demand constrain

Product huge

X₁₁ + X₁₂ + X₁₃ ≤ 710

Product average

X₂₁ + X₂₂ + X₂₃ ≤ 900

Product tiny

X₃₁ + X₃₂ + X₃₃ ≤ 350

Fraction SP/CC must be the same

First and second centers fraction SP/CC

(X₁₁ + X ₂₁ + X₃₁)/ 11000 = (X₁₂ + X₂₂ + X₃₂)/ 2700

2700*(X₁₁ + X ₂₁ + X₃₁) - 11000*(X₁₂ + X₂₂ + X₃₂) = 0

First and third centers fraction SP/CC

(X₁₁ + X ₂₁ + X₃₁)/ 11000 = ( X₁₃ + X₂₃ + X₃₃)/ 3400

3400*(X₁₁ + X ₂₁ + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

The model is:

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃

Subject to:

First center X₁₁ + X₂₁ + X₃₁ ≤ 550

Second center X₁₂ + X₂₂ + X₃₂ ≤ 750

Third center X₁₃ + X₂₃ + X₃₃ ≤ 275

22* X₁₁ + 16* X₂₁ + 9*X₃₁ ≤ 11000

22* X₁₂ + 16* X₂₂ + 9*X₃₂ ≤ 2700

22*X₁₃ + 16* X₂₃ + 9*X₃₃ ≤ 3400

X₁₁ + X₁₂ + X₁₃ ≤ 710

X₂₁ + X₂₂ + X₂₃ ≤ 900

X₃₁ + X₃₂ + X₃₃ ≤ 350

2700*(X₁₁ + X ₂₁ + X₃₁) - 11000*(X₁₂ + X₂₂ + X₃₂) = 0

3400*(X₁₁ + X ₂₁ + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

Xij >= 0

6 0

3 years ago

Sarahs craft project uses peices of yarn that are 1/8 yard long.She has a peice of yarn that is 3 yards ling.How many 1/8 yard p

motikmotik

Pieces of yarn .... 1/8 yard long (each)
a piece of yarn ... n = 3 yards long
remaining ... 1 1/4 yards = 5/4 yards

If you would like to calculate how many 1/8 yard pieces can Sarah cut and still have 1 1/4 yards left, you can do this using the following steps:

n - 1 1/4 = 3 - 1 1/4 = 3 - 5/4 = 12/4 - 5/4 = 7/4 = 1 3/4 yards

7/4 / 1/8 = 7/4 * 8/1 = 14 pieces

Result: Sarah can cut 14 1/8 yard pieces and still have 1 1/4 yards left.

3 0

4 years ago

Read 2 more answers

an engineer has a 54 inch wire that she has to cut into two pieces so that one piece is 8 inches shorter than the other find the

Brilliant_brown [7]

Hey there!!

Let's take the length of the bigger piece as x inches

Then, the length of the smaller piece would be x - 8 inches

The sum of these lengths would be 54 inches

... x + x - 8 = 54

... 2x - 8 = 54

... Add 8 on both sides

... 2x = 62

... Divide 2 on both sides

... x = 31

Hence, the length of the bigger piece = 31 inches

The length of the smaller piece would be = 31 - 8 = 23 inches

Hope my answer helps!!

4 0

3 years ago

Solve for x & y solve x = 3 + yy = -2x + 9

sergey [27]

x = 3 + y Eqn(1)

y = -2x + 9 Eqn(2)

Let us solve the system of equations with the substitution method

x - 3 = y (Subtracting 3 from both sides of the Eqn(1))

Replacing y = x - 3 in Eqn (2), we have:

x - 3 = -2x + 9

x = -2x + 9 + 3 (Adding 3 to both sides of the equation)

x + 2x = 9 + 3 (Adding 2x to both sides of the equation)

3x = 12 ( Adding like terms)

x = 12/3 (Dividing by 3 on both sides of the equation)

x = 4

Replacing x=4 in Eqn(1), we have:

4 = 3 + y

4 - 3 = y (Subtracting 3 from both sides of the equation)

y=1

The answers are:

x= 4 and y=1

6 0

1 year ago