The answer is 452.16 cm³.
Formula to calculate the volume of cone = V = πr²x h/3
Now r = 6 cm and h = 12cm, and value of π = 3.14
By putting the values,
V = (3.14) (6)² (12/3) = (3.14)(36)(4)
V = 452.16 cm³

3 0

3 years ago

Simplify -12x^4/x^4+8x^5

alexgriva [62]

Answer:

OPTION A: $\frac{12}{1 + 8x}$ , where $x \ne - \frac{1}{8}$ .

Step-by-step explanation:

Given: $\frac{- 12x^4}{x^4 + 8x^5}$

Taking $x^4$ common outside in the denominator, we get:

$= \frac{-12x^4}{(x^4)(1 + 8x)}$

$x^4$ will get cancelled on the numerator and denominator, we get:

$= \frac{-12}{1 + 8x}$

we know that the denominator can not be zero.

That means, 1 + 8x $\ne$ 0.

$\implies 8x \ne -1$

$\implies x \ne \frac{-1}{8}$

So, the answer is: $\frac{-12}{1 + 8x}$ , where $x \ne \frac{-1}{8}$ .

6 0

3 years ago

Describe the steps required to determine the equation of a quadratic function given its zeros and a point.

faltersainse [42]

Answer:

Procedure:

1) Form a system of 3 linear equations based on the two zeroes and a point.

2) Solve the resulting system by analytical methods.

3) Substitute all coefficients.

Step-by-step explanation:

A quadratic function is a polynomial of the form:

$y = a\cdot x^{2}+b\cdot x + c$ (1)

Where:

$x$ - Independent variable.

$y$ - Dependent variable.

$a$ , $b$ , $c$ - Coefficients.

A value of $x$ is a zero of the quadratic function if and only if $y = 0$ . By Fundamental Theorem of Algebra, quadratic functions with real coefficients may have two real solutions. We know the following three points: $A(x,y) = (r_{1}, 0)$ , $B(x,y) = (r_{2},0)$ and $C(x,y) = (x,y)$

Based on such information, we form the following system of linear equations:

$a\cdot r_{1}^{2}+b\cdot r_{1} + c = 0$ (2)

$a\cdot r_{2}^{2}+b\cdot r_{2} + c = 0$ (3)

$a\cdot x^{2} + b\cdot x + c = y$ (4)

There are several forms of solving the system of equations. We decide to solve for all coefficients by determinants:

$a = \frac{\left|\begin{array}{ccc}0&r_{1}&1\\0&r_{2}&1\\y&x&1\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }$

$a = \frac{y\cdot r_{1}-y\cdot r_{2}}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x+x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}$

$a = \frac{y\cdot (r_{1}-r_{2})}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x +x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}$

$b = \frac{\left|\begin{array}{ccc}r_{1}^{2}&0&1\\r_{2}^{2}&0&1\\x^{2}&y&1\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }$

$b = \frac{(r_{2}^{2}-r_{1}^{2})\cdot y}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x +x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}$

$c = \frac{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&0\\r_{2}^{2}&r_{2}&0\\x^{2}&x&y\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }$

$c = \frac{(r_{1}^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1})\cdot y}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x + x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}$

And finally we obtain the equation of the quadratic function given two zeroes and a point.

6 0

3 years ago