It is definitely the letter B because of the liner parent function
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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Peta has 10 plums. The second sentence is unnecessary.
<h3>Answer to Question 1:</h3>
AB= 24cm
BC = 7cm
<B = 90°
AC = ?
<h3>Using Pythagoras theorem :-</h3>
AC^2 = AB^2 + BC ^ 2
AC^2 = 24^2 + 7^2
AC^2 = 576 + 49
AC^2 = √625
AC = 25
<h3>Answer to Question 2 :-</h3>
sin A = 3/4
CosA = ?
TanA = ?
<h3>SinA = Opp. side/Hypotenuse</h3><h3> = 3/4</h3>
(Construct a triangle right angled at B with one side BC of 3cm and hypotenuse AC of 4cm.)
<h3>Using Pythagoras theorem :-</h3>
AC^2 = AB^2 + BC ^ 2
4² = AB² + 3²
16 = AB + 9
AB = √7cm
<h3>CosA = Adjacent side/Hypotenuse</h3>
= AB/AC
= √7/4
<h3>TanA= Opp. side/Adjacent side</h3>
=BC/AB
= 3/√7
Answer:
3y^2 - (y + 2) (y - 2) = 0
<=> 3y^2 - (y^2 - 4) = 0
<=> 2y^2 + 4 =0
<=>y^2 + 2 = 0
=> Because y^2 is always equal or larger than 0, there is no real solution.
Hope this helps!
:)
