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3 years ago

Using the graph below, find the point(s) such that the sum of their coordinates is equal to 7.

Mathematics

1 answer:

gregori [183]3 years ago

7 0

Answer:

Only A, B and C are the points whose coordinates are equal to 7.

Step-by-step explanation:

Considering the point A

As the location of point A is (0, 7)

The sum of their coordinates is equal to 0 + 7 = 7

Considering the point B

As the location of point B is (2, 5)

The sum of their coordinates is equal to 2 + 5 = 7

Considering the point C

As the location of point C is (6, 1)

The sum of their coordinates is equal to 6 + 1 = 7

Considering the point D

As the location of point D is (4, -3)

The sum of their coordinates is equal to 4 + ( -3 ) = 1

Considering the point E

As the location of point E is (1, -6)

The sum of their coordinates is equal to 1 + ( -6 ) = -5

Considering the point F

As the location of point F is (-2, -4)

The sum of their coordinates is equal to -2 + ( -4 ) = -2 - 4 = -6

Considering the point G

As the location of point F is (-5, -2)

The sum of their coordinates is equal to -5 + ( -2 ) = -5 - 2 = -7

Considering the point H

As the location of point H is (-4, 3)

The sum of their coordinates is equal to -4 + ( 3 ) = -4 + 3 = -1

From the above discussion, only A, B and C are the points whose coordinates are equal to 7.

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Find the equation of the sphere if one of its diameters has endpoints (4, 2, -9) and (6, 6, -3) which has been normalized so tha

Pavel [41]

Answer:

$(x - 5)^2 + (y - 4)^2 + (z - 6)^2 = 14$ .

(Expand to obtain an equivalent expression for the sphere: $x^2 - 10\,x + y^2 - 8\, y + z^2 - 12\, z + 63 = 0$ )

Step-by-step explanation:

Apply the Pythagorean Theorem to find the distance between these two endpoints:

$\begin{aligned}&\text{Distance}\cr &= \sqrt{\left(x_2 - x_1\right)^2 + \left(y_2 - y_1\right)^2 + \left(z_2 - z_1\right)^2} \cr &= \sqrt{(6 - 4)^2 + (6 - 2)^2 + ((-3) - (-9))^2 \cr &= \sqrt{56}}\end{aligned}$ .

Since the two endpoints form a diameter of the sphere, the distance between them would be equal to the diameter of the sphere. The radius of a sphere is one-half of its diameter. In this case, that would be equal to:

$\begin{aligned} r &= \frac{1}{2} \, \sqrt{56} \cr &= \sqrt{\left(\frac{1}{2}\right)^2 \times 56} \cr &= \sqrt{\frac{1}{4} \times 56} \cr &= \sqrt{14} \end{aligned}$ .

In a sphere, the midpoint of every diameter would be the center of the sphere. Each component of the midpoint of a segment (such as the diameter in this question) is equal to the arithmetic mean of that component of the two endpoints. In other words, the midpoint of a segment between $\left(x_1, \, y_1, \, z_1\right)$ and $\left(x_2, \, y_2, \, z_2\right)$ would be:

$\displaystyle \left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right)$ .

In this case, the midpoint of the diameter, which is the same as the center of the sphere, would be at:

$\begin{aligned}&\left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right) \cr &= \left(\frac{4 + 6}{2},\, \frac{2 + 6}{2}, \, \frac{(-9) + (-3)}{2}\right) \cr &= (5,\, 4\, -6)\end{aligned}$ .