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:
see explanation
Step-by-step explanation:
The equation of a circle in standard form is
(x - h)² + (y - k)² = r²
where (h, k) are the coordinates of the centre and r is the radius
(x - 3)² + (y + 4)² = 36 ← is in standard form
with (h, k) = (3, - 4 ) = centre and r = 
= 6
Susan interpreted the values of h and k incorrectly by using the values - 3 and 4 instead of taking the negative of them.
She also divided 36 by 2 , thus omitting the root of 36
Answer:
12345678910
Step-by-step explanation:
CHARRRRRRRRR joke lang bestie
Answer:
Step-by-step explanation:
ΔSVX ≅ ΔUTX Given
SV // TU Given
SV ≅ TU Corresponding part of congruent triangle
VUTS parallelogram Definition of parallelogram.
Hint: 1) If any one pair of opposite side is congruent and parallel, then the quadrilateral is parallelogram.
2) If two pair of opposite sides are parallel then it is a parallelogram.
Step-by-step explanation:
The initial image of the photo is 2 in by 4 in. The mat is 4 in by 6 in.
The new image is dilated by a scale of 2. So we double the dimensions. The new photo is 4 in by 8 in. The new mat is 8 in by 12 in.
