Simplify quantity 18 x minus 6 over 9 x to the fifth power all over quantity 15 x plus 5 over 21 x squared

2 answers:

6 0

[(18x - 6) / 9x^5] / [(15x + 5) / 21x^2]

The initial equation formulated would be:

[6(3x - 1) / 9x^5][21x^2 / 5(3x + 1)] 

By simplifying:

[126x^2(3x - 1) / [45x^5(3x + 1)]

14(3x - 1) / 5x^3(3x + 1)

Therefore, the simplest form would be:

(42x - 14) / (15x^4 + 5x^3)

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

Vladimir79 [104]3 years ago

3 0

It is important to read every word of this question carefully. Then, it will become easy to solve the question in hand.

[(18x - 6) / 9x^5] / [(15x + 5) / 21x^2]

[6(3x - 1) / 9x^5][21x^2 / 5(3x + 1)] 

[126x^2(3x - 1) / [45x^5(3x + 1)] 

14(3x - 1) / 5x^3(3x + 1) 

(42x - 14) / (15x^4 + 5x^3)

I hope that this is the answer that you were looking for and the answer has actually come to your desired help. I also hope that the procedure is clear enough for you to understand.

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in a first aid kit the ratio of large bandages to small bandages is 5 to 2. Based on this ratio, how many large bandages are in

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3 years ago

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:

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