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3 years ago

How is the point (2, 9) in the scatterplot described?

Mathematics

1 answer:

AnnZ [28]3 years ago

7 0

Im not sure but u can search up the name and see online i would help if i could i hope the advise i gave helps just seach up if no one can help im so sorry i wanted to help but idk this stuff yet sorry

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HELP i need it i will give brainly

iogann1982 [59]

Answer:

2/9^8

Step-by-step explanation:

3^0 = 1 so it turns into 9^-8/ 2^-1

Then put 9^-8 on the bottom to make it positive

Now its 1/9^8 & 2^-1

Then put 2 on top to make it positive and 2^1 is just 2

So now you have 2/9^8

5 0

2 years ago

Read 2 more answers

Please I need it now

soldi70 [24.7K]

Answer:

it's C. 2 and 5...

Step-by-step explanation:

If the denominator of rational number contains no prime factors other then 2 or 5 or both, then this rational number can be expressed as terminating decimal. So the correct answer is option is C

5 0

3 years ago

Plz help I'm halfway through the lesson

Solnce55 [7]

3.5555555556 / the third one

5 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:

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