HURRY!!!! ∠A and \angle B∠B are supplementary angles. If m\angle A=(6x+10)^{\circ}∠A=(6x+10)
2 answers:
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There is a multiple zero at 0 (which means that it touches there), and there are single zeros at -2 and 2 (which means that they cross). There is also 2 imaginary zeros at i and -i.
You can find this by factoring. Start by pulling out the greatest common factor, which in this case is -x^2.
-x^6 + 3x^4 + 4x^2
-x^2(x^4 - 3x^2 - 4)
Now we can factor the inside of the parenthesis. You do this by finding factors of the last number that add up to the middle number.
-x^2(x^4 - 3x^2 - 4)
-x^2(x^2 - 4)(x^2 + 1)
Now we can use the factors of two perfect squares rule to factor the middle parenthesis.
-x^2(x^2 - 4)(x^2 + 1)
-x^2(x - 2)(x + 2)(x^2 + 1)
We would also want to split the term in the front.
-x^2(x - 2)(x + 2)(x^2 + 1)
(x)(-x)(x - 2)(x + 2)(x^2 + 1)
Now we would set each portion equal to 0 and solve.
First root
x = 0 ---> no work needed
Second root
-x = 0 ---> divide by -1
x = 0
Third root
x - 2 = 0
x = 2
Forth root
x + 2 = 0
x = -2
Fifth and Sixth roots
x^2 + 1 = 0
x^2 = -1
x = +/-
x = +/- i
The equation of the line passing through and
is
. Here
is the slope of the line.
Substituting numerical values, the equation of the line is
The equation of the line is
The maximum value of the objective function is 330
<h3>How to maximize the objective function?</h3>The given parameters are:
Max w = 5y₁ + 3y₂
Subject to
y₁ + y₂ ≤ 50
2y₁ + 3y₂ ≤ 60
y₁ , y₂ ≥ 0
Start by plotting the graph of the constraints (see attachment)
From the attached graph, we have:
(y₁ , y₂) = (90, -40)
Substitute (y₁ , y₂) = (90, -40) in w = 5y₁ + 3y₂
w = 5 * 90 - 3 * 40
Evaluate
w = 330
Hence, the maximum value of the function is 330
Read more about objective functions at:
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