$\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=2\\ h=5 \end{cases}\implies V=\cfrac{\pi (2)^2(5)}{3} \\\\\\ V=\cfrac{20\pi }{3}\implies V\approx 20.94$

6 0

3 years ago

If x+1/x= 3, then prove that m^5+1/m^5= 123

alina1380 [7]

9514 1404 393

Explanation:

We can start with the relations ...

$\displaystyle\left(x+\frac{1}{x}\right)^3=\left(x^3+\frac{1}{x^3}\right)+3\left(x+\frac{1}{x}\right)\\\\\left(x+\frac{1}{x}\right)^5=\left(x^5+\frac{1}{x^5}\right)+5\left(x^3+\frac{1}{x^3}\right)+10\left(x+\frac{1}{x}\right)\\\\\textsf{From these, we can derive ...}\\\\x^5+\frac{1}{x^5}=\left(x+\frac{1}{x}\right)^5-5\left(\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)\right)-10\left(x+\frac{1}{x}\right)$

$\displaystyle x^5+\frac{1}{x^5}=\left(x+\frac{1}{x}\right)^5-5\left(x+\frac{1}{x}\right)^3+5\left(x+\frac{1}{x}\right)\right)\\\\x^5+\frac{1}{x^5}=3^5 -5(3^3)+5(3)\\\\=((3^2-5)3^2+5)\cdot3=(4\cdot9+5)\cdot3=(41)(3)\\\\=\boxed{123}$

7 0

3 years ago

A (0, 2) and B (6,6) are points on the straight line ABCD.

elena-14-01-66 [18.8K]

Answer:

(18, 14)

Step-by-step explanation:

We know that C and D lie on the line AB and BC = CD = AB. Then we need to use the distance formula and equation of the line AB to find the other two coordinates.

The distance formula states that the distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ , the distance is denoted by: $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ . Let's find the distance between A and B:

d = $\sqrt{(6-0)^2+(6-2)^2}=\sqrt{6^2+4^2} =\sqrt{36+16} =\sqrt{52} =2\sqrt{13}$

Now say the coordinates of D are (a, b). Then the distance between D and B will be twice of 2√13, which is 4√13:

4√13 = $\sqrt{(6-a)^2+(6-b)^2}$

Square both sides:

208 = (6 - a)² + (6 - b)²

Let's also find the equation of the line AB. The y-intercept we know is 2, so in y = mx + b, b = 2. The slope is (6 - 2) / (6 - 0) = 4/6 = 2/3. So the equation of the line is: y = (2/3)x + 2. Since (a, b) lines on this line, we can put in a for x and b for y: b = (2/3)a + 2. Substitute this expression in for b in the previous equation:

208 = (6 - a)² + (6 - b)²

208 = (6 - a)² + (6 - (2/3a + 2))² = (6 - a)² + (-2/3a + 4)²

208 = a² - 12a + 36 + 4/9a² - 16/3a + 16 = 13/9a² - 52/3a + 52

0 = 13/9a² - 52/3a - 156

13a² - 156a - 1404 = 0

a² - 12a - 108 = 0

(a + 6)(a - 18) = 0

a = -6 or a = 18

We know a can't be negative so a = 18. Plug this back in to find b:

b = 2/3a + 2 = (2/3) * 18 + 2 = 12 + 2 = 14

So point D has coordinates (18, 14).

8 0

3 years ago

Read 2 more answers

Write an equation (-2, 10), (10, -14)

Brilliant_brown [7]

$\bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{10})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{-14}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-14}-\stackrel{y1}{10}}}{\underset{run} {\underset{x_2}{10}-\underset{x_1}{(-2)}}}\implies \cfrac{-24}{10+2}\implies \cfrac{-24}{12}\implies -2$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{10}=\stackrel{m}{-2}[x-\stackrel{x_1}{(-2)}]\implies y-10=-2(x+2) \\\\\\ y-10=-2x-4\implies y=-2x+6$

5 0

3 years ago