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2 years ago

I need help please ASAP !

Mathematics

2 answers:

pav-90 [236]2 years ago

6 0

Answer:

lll only

Step-by-step explanation:

I did a quiz similar to that I hope you make a good grade!

Natasha2012 [34]2 years ago

3 0

Answer:

III only

Step-by-step explanation:

lkjhgfdsaasdfghjkjhgf

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Factor x^2-x+3=3 please :D

Bess [88]

Answer:

x(x - 1 ) = 0

Step-by-step explanation:

x^2 - x + 3 = 3

subtract 3 from each side:

x^2 - x = 0

x(x - 1 ) = 0

4 0

2 years ago

Are there always the same number of prime numbers between 2 consecutive multiples of 10? Why?

galina1969 [7]

Answer:

Yes it is 5

Step-by-step explanation:

It is 5 because you divide 2 and 10 and it is 5

4 0

3 years ago

PLEASE HELP!!!!<br><br> Find the value of x.<br> 130 degrees<br> Xdegrees

strojnjashka [21]

Answer:

Step-by-step explanation:

130+25+25=180

..........

.....

3 0

2 years ago

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3 liters of water are shared equally by 5 people how much water does each person get? division equation to represent the situati

kirill [66]

600 mililiters each

Step-by-step explanation:

1 liter is equal to 1000 mililiter and is multiplied by 3 and the answer is devided into 5.

5 0

2 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:

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