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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obtuse because it all adds up to 180 degrees
Answer:
Step-by-step explanation:
She must write an email to the new employees explaining how to access documents that need editing and what to do with new versions of the documents.
Consider this option:
1. if the point (4;6) is the centre of the circle and the point (2;5) is the first endpoint of its diameter, then point (4;6) is the middle point of the diameter (it means that is the middle between the 1st and the 2d endpoints of diameter).
2. using the property described above:
for x of the 2d endpoint of the diameter: x=4*2-2=6;
for y of the 2d endpoint of the diameter: y=6*2-5=7.
answer: (6;7)
Answer:
x=4.8
This here you use Pythagorean Theorem.
Step-by-step explanation:
a^2+b^2=c^2
The longest side is c. And the two legs are a and b.
So it'd be 11^2+x^2=12^2
121+x^2=144
Subtract 121 from both sides.
x^2=23
Then take the √ of x^2 and 23
So √x^2=√23
x=4.8
Hope this helps
