All categories

3 years ago

If x is a positive number greater than 10, what is always true about the solution to the problem to -3 + x ?

Mathematics

1 answer:

alexandr402 [8]3 years ago

3 0

Answer:

The equation answer will always be positive

Step-by-step explanation:

You might be interested in

Neeed helppp pleasseeee

OleMash [197]

-1/24 is the answer

7 0

3 years ago

Find the general solution to the homogeneous differential equation. d2ydx2−1dydx−30y=0 d2ydx2−1dydx−30y=0 Use c1c1 and c2c2 in y

lukranit [14]

Answer:

The general solution of second order homogeneous differential equation is $y(x)=c_1e^{6x}+c_2e^{-5x}$

Step-by-step explanation:

To find the general solution of this second order homogeneous differential equation $\frac{d^2y}{dx^2}-\frac{dy}{dx}-30y=0$ we are going to use this Theorem:

Given the differential equation $a \ddot{y}+b\dot{y}+cy =0, a\neq 0$ , consider the quadratic polynomial $ax^2+bx+c$ , called the characteristic polynomial. Using the quadratic formula, this polynomial always has one or two roots, call them $r$ and $s$ . The general solution of the differential equation is:

(a) $\ds y=Ae^{rt}+Be^{st}$ if the roots $r$ and $s$ are real numbers and $r\not=s$ .

(b) $\ds y=Ae^{rt}+Bte^{rt}$ , if $r=s$ is real.

(c) $\ds y=A\cos(\beta t)e^{\alpha t}+B\sin(\beta t)e^{\alpha t}$ , if the roots $r$ and $s$ are complex numbers $\alpha+\beta i$ and $\alpha-\beta i$

Applying the above Theorem we have:

$\mathrm{Substitute\quad }\frac{d^2y}{dx^2},\:\frac{dy}{dx}\mathrm{\:with\:}\ddot{y},\dot{y}$

$\ddot{y}-\dot{y}-30y=0$

The characteristic polynomial is $x^2-x-30$ and we find the roots as follows:

$\mathrm{Break\:the\:expression\:into\:groups}$

$\left(x^2+5x\right)+\left(-6x-30\right)$

$\mathrm{Factor\:out\:}x\mathrm{\:from\:}x^2+5x\mathrm{:\quad }x\left(x+5\right)$

$\mathrm{Factor\:out\:}-6\mathrm{\:from\:}-6x-30\mathrm{:\quad }-6\left(x+5\right)$

$\mathrm{Factor\:out\:common\:term\:}x+5$

$\left(x+5\right)\left(x-6\right)$

The roots of characteristic polynomial are $r=-5$ and $s=6$

Therefore the general solution of second order homogeneous differential equation is $y(x)=c_1e^{6x}+c_2e^{-5x}$

5 0

3 years ago

Use the quadratic formula to solve the equation 4x^2+6x-4=0

Vesnalui [34]

Answer:

$\large\boxed{x=-2\ or\ x=\dfrac{1}{2}}$

Step-by-step explanation:

$\text{The quadratic formula of an equation}\ ax^2+bx+c=0\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\text{We have}\ 4x^2+6x-4=0\to a=4,\ b=6,\ c=-4.\\\\\text{Substitute:}\\\\x=\dfrac{-6\pm\sqrt{6^2-4(4)(-4)}}{2(4)}=\dfrac{-6\pm\sqrt{36+64}}{8}=\dfrac{-6\pm\sqrt{100}}{8}\\\\x=\dfrac{-6-10}{8}=\dfrac{-16}{8}=-2\ or\ x=\dfrac{-6+10}{8}=\dfrac{4}{8}=\dfrac{1}{2}$