IJ=x-4
JK=5x+12
IK=9x-19
IJ+JK=IK
(x-4)+(5x+12)=(9x-19)
x-4+5x+12=9x-19
6x+8=9x-19
6x+8-6x+19=9x-19-6x+19
27=3x
3x=27
3x/3=27/3
x=9
IJ=x-4=9-4→IJ=5
JK=5x+12=5(9)+12=45+12→JK=57
IK=9x-19=9(9)-19=81-19→IK=62
Answer: Option B. IJ=5, JK=57, IK=62
Answer:
We'll use 3.14 as an approximation of pi. So, now we plug the values into the equation. A = 3.14 * 8^2/2.
Solving for the polynomial function of least degree with
integral coefficients whose zeros are -5, 3i
We have:
x = -5
Then x + 5 = 0
Therefore one of the factors of the polynomial function is
(x + 5)
Also, we have:
x = 3i
Which can be rewritten as:
x = Sqrt(-9)
Square both sides of the equation:
x^2 = -9
x^2 + 9 = 0
Therefore one of the factors of the polynomial function is (x^2
+ 9)
The polynomial function has factors: (x + 5)(x^2 + 9)
= x(x^2 + 9) + 5(x^2 + 9)
= x^3 + 9x + 5x^2 = 45
Therefore, x^3 + 5x^2 + 9x – 45 = 0
f(x) = x^3 + 5x^2 + 9x – 45
The polynomial function of least degree with integral coefficients
that has the given zeros, -5, 3i is f(x) = x^3 + 5x^2 + 9x – 45
Answer:
Yes
Step-by-step explanation:
True, you are correct.
