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
#SPJ1
Four $5 bills and two $1 bills
In total it’s six bills
5(4)=20 and 1(2)=2
20+2=22
Answer:
3) f(x) = 3(x + 2)^2 - 1.
Step-by-step explanation:
f(x) = 3x² + 12x +11
f(x) = 3(x^2 + 4x) + 11 Completing the square:
f(x) = 3[(x + 2)^2 - 4] + 11
f(x) = 3(x + 2)^2 -+3*-4 + 11
f(x) = 3(x + 2)^2 - 12 + 11
f(x) = 3(x + 2)^2 - 1.
Answer:
the answer to this question 2 cups
Step-by-step explanation:
2 cups is equal to a pint
The base would be 22 m squared since it is 22 x 10 = 220 and then you divide it by 2 to get 110 m squared.
