1. If a quadratic expression has x-intercepts of

Determine the center and radius of the following circle equation: x2 + y2 – 10x + 8y + 40 = 0

antiseptic1488 [7]

Answer:

The center of this circle is (5, -4) and the radius is 1.

Step-by-step explanation:

First regroup these terms according to x and y :

x^2 - 10x + y^2 + 8y = -40

Next, complete the square for x^2 - 10x: x^2 - 10x + 5^2 - 5^2.

and the same for y^2 + 8y: y^2 + 8y + 16 - 16

Substituting these results into x^2 - 10x + y^2 + 8y = -40, we get:

x^2 - 10x + 5^2 - 5^2 y^2 + 8y + 16 - 16 = -40.

Next, rewrite x^2 - 10x + 25 and y^2 + 8y + 16 as squares of binomials:

Then x^2 - 10x + 5^2 - 5^2 y^2 + 8y + 16 - 16 = -40 becomes:

(x - 5)^2 + (y + 4)^2 - 25 - 16 = -40, or:

(x - 5)^2 + (y + 4)^2 = 1

This equation has the form (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius. Matching like terms, we get h = 5, k = -4 and r = 1.

The center of this circle is (5, -4) and the radius is 1.

6 0

3 years ago

Read 2 more answers

Algebra 1 CR-8

antiseptic1488 [7]

$\mathrm{Domain\:of\:}\:x^2+2x-15\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:$

$\mathrm{Range\:of\:}x^2+2x-15:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:-16\:\\ \:\mathrm{Interval\:Notation:}&\:[-16,\:\infty \:)\end{bmatrix}$

$\mathrm{Axis\:interception\:points\:of}\:x^2+2x-15:\quad \mathrm{X\:Intercepts}:\:\left(3,\:0\right),\:\left(-5,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:-15\right)$

$\mathrm{Vertex\:of}\:x^2+2x-15:\quad \mathrm{Minimum}\space\left(-1,\:-16\right)$

4 0

2 years ago

The sum of a number times 6 and 5 equals 4 translate the sentence into an equation

yan [13]

If x is the number we get:-

6x + 5 = 4

3 0

3 years ago

PLEASE SOMEONE HELP ILL GIVE BRAINLIEST!

grin007 [14]

Answer:

Part A

The rotation of the point (x, y) 180° about the origin gives the point (-x, -y)

Therefore, we have the points, ΔLMN, L(1, 1), and M(2, 2), and N(3, 3)

The coordinates of the vertices of the image, ΔL'M'N' are L'(-1, -1), and M'(-2, -2), and N'(-3, -3)

Please find attached the graph of triangle ΔLMN and ΔL'M'N'

Part B

The lines drawn through L and L' and through M and M' are colinear

Part C

A line is defined by two points, such as L and M. Rotation of a line through 180° about the origin will give an image location on the same path as the extension of the original line.

The third point, N, however, defines a plane, and the rotation of a plane by 180° will give an image which is turned upside down with regards to the preimage

Therefore, a trough in the preimage becomes a peak in the image and the lines drawn through N and N' crosses the colinear lines drawn throgh M and M' and L and L'

Step-by-step explanation:

5 0

3 years ago

Which point is a reflection of Z(5 1/2,3) across the y axis

RSB [31]

Answer: C?

Step-by-step explanation:

Because reflecting means like think of a mirror it’s reflecting you so the opposite side would be the bottom right quadrant please tell me if I’m right?

5 0

3 years ago