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3 years ago

What is The polynomial for the question above

Mathematics

1 answer:

frosja888 [35]3 years ago

4 0

A polynomial of degree n with roots $r_1,r_2,...,r_n$ can be represented as: $f(x)=A(x-r_1)(x-r_2)\cdot...\cdot(x-r_n)$ , where A is a constant (The same "a" that appears outside the parentheses in the field where you will type the polynomial). In this question, the number of roots is 4, because the degree of f(x) is 4. If a polynomial has only real coefficients, the complex roots appear in pairs of conjugates. So, since $-3+4i$ is a zero, $-3-4i$ is a zero too. Then, we have the 4 zeros. Hence, $f(x)$ will be in the form:

$f(x)=A(x-5)^2(x-(-3+4i))(x-(-3-4i))\\\\f(x)=A(x-5)^2(x^2-(-3+4i)x-(-3-4i)x+(-3+4i)(-3-4i))\\\\f(x)=A(x^2-10x+25)(x^2+6x+25)\\\\f(x)=A(x^4-4x^3-10x^2-100x+625)$

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Find and simplify the difference quotient f(x + 1) – f(x) for the following function.

ra1l [238]

In your situation you said that

$f(x) = mx+b$
$m=0$

Putting those together, you'd have $f(x) = b$ because $0\cdot x+b=b$ .

To evaluate the difference quotient, first find each piece on it's own:

$f(x+1) = b$ because $f(x) = b$ no matter what your x-value is.
$f(x) = b$

So putting those together:

$f(x+1) - f(x) = b-b = 0$

Remember that the difference quotient is basically finding the slope of something. Since you were given that the slope is 0, the difference quotient should work out to match that.

3 0

3 years ago

Read 2 more answers

The surface area of a right cone is 400.2 m2. The radius of the cone is 6.0 m. Determine the height of the cone to the nearest m

svetoff [14.1K]

S = πr(r + √(h² + r²))
  400.2 = 3.14(6)(6 + √(h² + 6²))
  400.2 = 18.84(6 + √(h² + 36))
  18.84   18.84
  21¹⁰⁹/₄₇₁ = 6 + √(h² + 36))
  - 6 - 6
  15¹⁰⁹/₄₇₁ = √(h² + 36)
231²²¹⁰⁰⁵/₂₂₁₈₄₁ = h² + 36
- 36   - 36
195²²¹⁰⁰⁵/₂₂₁₈₄₁ = h²
  14 ≈ h

8 0

2 years ago

6 times the sum of a number and 5 is 16

Ulleksa [173]

The equation would be 6 times x+5=16. This would be kinda Impossible. But a really close answer is 1.833

8 0

3 years ago

Novay_Z [31]

Answer:

The sequence of transformations that maps ΔABC to ΔA'B'C' is the reflection across the line y = x and a translation 10 units right and 4 units up, equivalent to T₍₁₀, ₄₎

Step-by-step explanation:

For a reflection across the line y = -x, we have, (x, y) → (y, x)

Therefore, the point of the preimage A(-6, 2) before the reflection, becomes the point A''(2, -6) after the reflection across the line y = -x

The translation from the point A''(2, -6) to the point A'(12, -2) is T(10, 4)

Given that rotation and translation transformations are rigid transformations, the transformations that maps point A to A' will also map points B and C to points B' and C'

Therefore, a sequence of transformation maps ΔABC to ΔA'B'C'. The sequence of transformations that maps ΔABC to ΔA'B'C' is the reflection across the line y = x and a translation 10 units right and 4 units up, which is T₍₁₀, ₄₎

5 0

2 years ago

hat is the approximate distance between Portland, Oregon (45° north, 122° west), and San Francisco, California (37° north, 122°

juin [17]

Since both places are on the same longitude (122° west) and different lattitudes 45° north and 37° north, the distance between the two place lies on a great circle (along a line of longitude). Difference in latitudes = 45 - 37 = 8° (subtract the lattitudes because both sides are on the same side in the latitude i.e. both are north) The distance between two points along the line of longitude is given by theta / 360 x 2 x pi x R: where theta = 8° and R is the radius of the earth = 3,960 miles. d = 8 / 360 x 2 x pi x 3960 = 552.9 miles

8 0

3 years ago