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2 years ago

Find an equation of a line parallel to the given line and contains the given point. Write the equation in slope-intercept form.

Mathematics

2 answers:

nikdorinn [45]2 years ago

8 0

Answer: y = 3x-17 or y = 3x^2 - 4x - 37

Step-by-step explanation:

Murljashka [212]2 years ago

5 0

Answer:

NO HABLO TAKA TAKA

Step-by-step explanation:

OZHY

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1. Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial i

Katena32 [7]

Polynomial with real coefficients always has even number of complex roots. We know that one of them is 2 + 5i so the second one will be 2 - 5i and:

$f(x)=\big(x-4\big)\big(x-(-8)\big)\big(x-(2+5i)\big)\big(x-(2-5i)\big)=\\\\=(x-4)(x+8)(x-2-5i)(x-2+5i)=\\\\=(x^2-4x+8x-32)(x^2-2x+5ix-2x+4-10i-5ix+10i-25i^2)\\\\=\big(x^2+4x-32\big)\big(x^2-4x+4-25\cdot(-1)\big)=\\\\=(x^2+4x-32)(x^2-4x+29)=\\\\=x^4-4x^3+29x^2+4x^3-16x^2+116x-32x^2+128x-928=\\\\=\boxed{x^4-19x^2+244x-928}$

Answer B.

8 0

3 years ago

Read 2 more answers

What is the equation of the following graph?

nignag [31]

Answer:

${y}^{2} = 24x$

Step-by-step explanation:

This is horizontal parabola.

It has a directrix at x=-6.

The vertex is (0,0).

The focus is (6,0)

The equation of this parabola is of the form

${y}^{2} = 4px$

Where (p,0) is the focus

By comparison, p=6

Therefore the equation of the parabola is

${y}^{2} = 4 \times 6x$

This implies that

${y}^{2} =2 4 x$

6 0

3 years ago

If we inscribe a circle such that it is touching all six corners of a regular hexagon of side 10 inches, what is the area of the

Brrunno [24]

Answer:

$\left(100\pi - 150\sqrt{3}\right)$ square inches.

Step-by-step explanation:

<h3>Area of the Inscribed Hexagon</h3>

Refer to the first diagram attached. This inscribed regular hexagon can be split into six equilateral triangles. The length of each side of these triangle will be $10$ inches (same as the length of each side of the regular hexagon.)

Refer to the second attachment for one of these equilateral triangles.

Let segment $\sf CH$ be a height on side $\sf AB$ . Since this triangle is equilateral, the size of each internal angle will be $\sf 60^\circ$ . The length of segment

$\displaystyle 10\, \sin\left(60^\circ\right) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ .

The area (in square inches) of this equilateral triangle will be:

$\begin{aligned}&\frac{1}{2} \times \text{Base} \times\text{Height} \\ &= \frac{1}{2} \times 10 \times 5\sqrt{3}= 25\sqrt{3} \end{aligned}$ .

Note that the inscribed hexagon in this question is made up of six equilateral triangles like this one. Therefore, the area (in square inches) of this hexagon will be:

$\displaystyle 6 \times 25\sqrt{3} = 150\sqrt{3}$ .

<h3>Area of of the circle that is not covered</h3>

Refer to the first diagram. The length of each side of these equilateral triangles is the same as the radius of the circle. Since the length of one such side is $10$ inches, the radius of this circle will also be $10$ inches.

The area (in square inches) of a circle of radius $10$ inches is:

$\pi \times (\text{radius})^2 = \pi \times 10^2 = 100\pi$ .