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Alik [6]
2 years ago
15

Solve for x by completing the square: x2 − 8x + 13 = 0. show all work

Mathematics
1 answer:
Murljashka [212]2 years ago
5 0

There are two <em>real</em> roots for the <em>quadratic</em> equation x² - 8 · x + 13 = 0, contained in the number x = 4 ± √2.

<h3>How to find the roots of a polynomial by completing the square</h3>

In this question we must apply algebraic handling to simplify a <em>quadratic</em> equation and find the roots that satisfy the expression. Completing the square consists in transforming part of the equation into a <em>perfect square</em> trinomial, and then we clear for x:

x² - 8 · x + 13 = 0

x² - 8 · x + 16 = 3

(x - 4)² = 3

x - 4 = ± √2

x = 4 ± √2

There are two <em>real</em> roots for the <em>quadratic</em> equation x² - 8 · x + 13 = 0, contained in the number x = 4 ± √2.

To learn more on quadratic equations: brainly.com/question/2263981

#SPJ1

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a person sold 100 shares of stock at a loss of 40%. If the selling price for the 100 shares was 3000 which of the following come
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3 0
3 years ago
The price of one share of a company’s stock declined $13.89 every day for 2.5 days. What was the stock’s total decline in price?
castortr0y [4]

Answer:

-$14(2.5) & -$34.73

Step-by-step explanation:

Which estimate uses compatible numbers to estimate the problem?  

✔ -<u>$14(2.5) </u>

What is the exact answer?  

✔ -<u>$34.73 </u>

<u></u>

6 0
4 years ago
Read 2 more answers
The length of a rectangle is three times its width. if the area of the rectangle is 75 in2, find its perimeter.
UkoKoshka [18]

Answer:

  40 in

Step-by-step explanation:

For a width of w, the length is 3w and the area and perimeter are ...

  A = LW = (3w)(w) = 3w^2

  P = 2(L+W) = 2(3w +w) = 8w

We are given the area, so we can find w to be ...

  75 in^2 = 3w^2

  25 in^2 = w^2 . . . . . divide by 3

  5 in = w . . . . . . . . . square root

Then the perimeter is ...

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7 0
3 years ago
The maintenance department at the main campus of a large state university receives daily requests to replace fluorecent lightbul
pshichka [43]

Answer:

P₂   [ 32  ≤ X ≤ 46 ] = 47,71 %

Step-by-step explanation:

The rule:

68-95-99.7

establishes

P₁  [ μ₀ - σ ≤ X ≤ μ₀ + σ] ≈ 0,683

P₂   [ μ₀ - 2σ ≤ X ≤ μ₀ + 2σ] ≈ 0,954

P₃  [ μ₀ - 3σ ≤ X ≤ μ₀ + 3σ] ≈ 0,997

For our paticular case we have

μ₀   =  46

 σ    = 7

Then we get:

μ₀ - σ  = 39            μ₀ +  σ  =   53

μ₀ - 2σ  =  32        μ₀ +  2σ = 60

μ₀ - 3σ  =  25          μ₀ +  3σ  = 67

We were asked by % of replacement request numbering between 32 and 46.

Normal curve is symmetric therefore μ₀ - 2σ  =  32  is just half of the interval  P₂   [ μ₀ - 2σ ≤ X ≤ μ₀ + 2σ] ≈ 0,954, then between 32 and 46 the porcentage of lightbulb is 0,954/2  , is 0,4771 or 47,71 %

P₂   [ μ₀ - 2σ ≤ X ≤ μ₀ ]    = P₂   [ 32  ≤ X ≤ 46 ] = 47,71 %

6 0
4 years ago
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