All categories

2 years ago

How to calculate roots when the equation is non quadratic?

Mathematics

1 answer:

likoan [24]2 years ago

5 0

Solve like a linear algebra equation
$y = mx + b$
use for slope
$m = y1 - y2 \div x1 - x2$
the b is intersept

You might be interested in

Question number 4 please help

Korvikt [17]

$\frac{x^2+x-30}{x^2-3x+2}\\\\x^2-3x+2=0\\\\x^2-x-2x+2=0\\\\x(x-1)-2(x-1)=0\\\\(x-1)(x-2)=0\iff x-1=0\ \vee\ x-2=0\\\\x=1\ \vee\ x=2\\\\Answer:For\ x=1\ or\ x=2.$

8 0

2 years ago

10 POINTS! PLEASE HELP!! The illustration shows a compact disc tray that holds five CDs. The radius of each compact disc is 9 cm

MariettaO [177]

We can see that there are 5 CDs, each of radius 9 cm

<u>Area occupied by 1 disc:</u>

Area of a circle = πr²

Area of disc = π(9)²

Area of disc = 3.14 * 81 = 254 cm²

<u>Area occupied by 5 discs:</u>

Area occupied by 5 discs = Area occupied by 1 disc * 5

Area occupied by 5 discs = 254 * 5

Area occupied by 5 discs = 1270 cm²

3 0

2 years ago

Find the volume of the solid.

dmitriy555 [2]

In Cartesian coordinates, the region (call it $R$ ) is the set

$R = \left\{(x,y,z) ~:~ x\ge0 \text{ and } y\ge0 \text{ and } 2 \le z \le 4-x^2-y^2\right\}$

In the plane $z=2$ , we have

$2 = 4 - x^2 - y^2 \implies x^2 + y^2 = 2 = \left(\sqrt2\right)^2$

which is a circle with radius $\sqrt2$ . Then we can better describe the solid by

$R = \left\{(x,y,z) ~:~ 0 \le x \le \sqrt2 \text{ and } 0 \le y \le \sqrt{2 - x^2} \text{ and } 2 \le z \le 4 - x^2 - y^2 \right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\sqrt2} \int_0^{\sqrt{2-x^2}} \int_2^{4-x^2-y^2} dz \, dy \, dx$

While doable, it's easier to compute the volume in cylindrical coordinates.

$\begin{cases} x = r \cos(\theta) \\ y = r\sin(\theta) \\ z = \zeta \end{cases} \implies \begin{cases}x^2 + y^2 = r^2 \\ dV = r\,dr\,d\theta\,d\zeta\end{cases}$

Then we can describe $R$ in cylindrical coordinates by

$R = \left\{(r,\theta,\zeta) ~:~ 0 \le r \le \sqrt2 \text{ and } 0 \le \theta \le\dfrac\pi2 \text{ and } 2 \le \zeta \le 4 - r^2\right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\pi/2} \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta \, dr \, d\theta \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta\,dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} r((4 - r^2) - 2) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} (2r-r^3) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \left(\left(\sqrt2\right)^2 - \frac{\left(\sqrt2\right)^4}4\right) = \boxed{\frac\pi2}$

3 0

1 year ago

Which of the following is a factor of 32x + 4? (1 point)

Neporo4naja [7]

Answer:

2x-1

Step-by-step explanation:

hope you get help from me

3 0

3 years ago

1. What will be the letter in the following sequence?

svetlana [45]

Answer:

1. G

2. Not sure what you are ask

5 0

3 years ago