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2 years ago

Can someone pls help me/?

Mathematics

2 answers:

zubka84 [21]2 years ago

4 0

Answer:

Yes I can help you but I am not sure

It is 17 radius

Explanation:

navik [9.2K]2 years ago

3 0

Answer:

since the answer is 17.59 and if you round it to the nearest tenth it would be 17.60 but if if rounding to the nearest whole numbers it would be 18

hope it help

Step-by-step explanation:

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Thank you and have a good day

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Help? Thanks also kinda easy

frosja888 [35]

Answer: 292

Step-by-step explanation:

5 0

3 years ago

4.4x10^-5 in standard form

QveST [7]

The answer is

0.000044

4 0

3 years ago

Read 2 more answers

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:

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

The equation for a sphere of radius $r$ and center $\left(x_0,\, y_0,\, z_0\right)$ would be:

$\left(x - x_0\right)^2 + \left(y - y_0\right)^2 + \left(z - z_0\right)^2 = r^2$ .

In this case, the equation would be:

$\left(x - 5\right)^2 + \left(y - 4\right)^2 + \left(z - (-6)\right)^2 = \left(\sqrt{56}\right)^2$ .

Simplify to obtain:

$\left(x - 5\right)^2 + \left(y - 4\right)^2 + \left(z + 6\right)^2 = 56$ .

Expand the squares and simplify to obtain:

$x^2 - 10\,x + y^2 - 8\, y + z^2 - 12\, z + 63 = 0$ .

8 0

3 years ago

Conplete the square to solve the quadratic equation. x<br><img src="https://tex.z-dn.net/?f=1%20%7Bx%7D%5E%7B2%7D%20%20-%206x%20

o-na [289]

Answer:

$\large\boxed{x=4\pm2\sqrt6}$

Step-by-step explanation:

$x^2-6x+12=2x+20\qquad\text{subtract}\ 2x\ \text{and}\ 12\ \text{from both sides}\\\\x^2-8x=8\\\\x^2-2(x)(4)=8\qquad\text{add}\ 4^2=16\ \text{to both sides}\\\\x^2-2(x)(4)+4^2=8+4^2\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\(x-4)^2=24\iff x-4=\pm\sqrt{24}\\\\x-4=\pm\sqrt{4\cdot6}\\\\x-4=\pm\sqrt4\cdot\sqrt6\\\\x-4=\pm2\sqrt6\qquad\text{add 4 to both sides}\\\\x=4\pm2\sqrt6$

6 0

3 years ago

Write an equation for the line passing through the points (2,4) and (2,7).

Nataly_w [17]

Answer:

x=2

Step-by-step explanation:

Since the points (2,4) and (2,7) are on the line whose equation is x = 2,

The equation of the line passing through them is hence x = 2

4 0

3 years ago