By applying algebraic handling on the two equations, we find the following three <em>solution</em> pairs: x₁ ≈ 5.693 ,y₁ ≈ 10.693; x₂ ≈ 1.430, y₂ ≈ 6.430; x₃ ≈ - 0.737, y₃ ≈ 4.263.
<h3>How to solve a system of equations</h3>
In this question we have a system formed by a <em>linear</em> equation and a <em>non-linear</em> equation, both with no <em>trascendent</em> elements and whose solution can be found easily by algebraic handling:
x - y = 5 (1)
x² · y = 5 · x + 6 (2)
By (1):
y = x + 5
By substituting on (2):
x² · (x + 5) = 5 · x + 6
x³ + 5 · x² - 5 · x - 6 = 0
(x + 5.693) · (x - 1.430) · (x + 0.737) = 0
There are three solutions: x₁ ≈ 5.693, x₂ ≈ 1.430, x₃ ≈ - 0.737
And the y-values are found by evaluating on (1):
y = x + 5
x₁ ≈ 5.693
y₁ ≈ 10.693
x₂ ≈ 1.430
y₂ ≈ 6.430
x₃ ≈ - 0.737
y₃ ≈ 4.263
By applying algebraic handling on the two equations, we find the following three <em>solution</em> pairs: x₁ ≈ 5.693 ,y₁ ≈ 10.693; x₂ ≈ 1.430, y₂ ≈ 6.430; x₃ ≈ - 0.737, y₃ ≈ 4.263.
To learn more on nonlinear equations: brainly.com/question/20242917
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72 3 plus 62 is 65 plus 7 is 72
Answer:
The answer is: 181.5 square centimeters.
Step-by-step explanation:
The cube has six faces, and each face is a square with sides of length 5.5, so the area of one of this faces is
, adding the six faces, we get the answer: 
The probablility is 1/2 or 0.50 % chance.
Answer:
The minimum value of f(x) is 2
Step-by-step explanation:
- To find the minimum value of the function f(x), you should find the value of x which has the minimum value of y, so we will use the differentiation to find it
- Differentiate f(x) with respect to x and equate it by 0 to find x, then substitute the value of x in f(x) to find the minimum value of f(x)
∵ f(x) = 2x² - 4x + 4
→ Find f'(x)
∵ f'(x) = 2(2)
- 4(1)
+ 0
∴ f'(x) = 4x - 4
→ Equate f'(x) by 0
∵ f'(x) = 0
∴ 4x - 4 = 0
→ Add 4 to both sides
∵ 4x - 4 + 4 = 0 + 4
∴ 4x = 4
→ Divide both sides by 4
∴ x = 1
→ The minimum value is f(1)
∵ f(1) = 2(1)² - 4(1) + 4
∴ f(1) = 2 - 4 + 4
∴ f(1) = 2
∴ The minimum value of f(x) is 2