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2 years ago

I need help asapp...

Mathematics

1 answer:

prohojiy [21]2 years ago

3 0

The value of b is -16 and the value of ac is 60 after comparing with the standard equation.

<h3>What is a quadratic equation?</h3>

Any equation of the form $\rm ax^2+bx+c=0$ where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.

As we know, the formula for the roots of the quadratic equation is given by:

$\rm x = \dfrac{-b \pm\sqrt{b^2-4ac}}{2a}$

We have a quadratic function:

= 3x² - 16x + 20

On comparing with standard function:

b = -16

a = 3

c = 20

ac = 3(20) = 60

Thus, the value of b is -16 and the value of ac is 60 after comparing with the standard equation.

Learn more about quadratic equations here:

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Answer:

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Step-by-step explanation:

-6 - x = 15

15 + 6 = 21 / -1 = -21

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3 years ago

Multiply. 2.4 x 0.002 =

aleksandrvk [35]

2.4 ×2 ×10*-3
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3 years ago

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andrezito [222]

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3 years ago

Solve the separable differential equation:dx/dt= x^2+ (1/9) and find the particular solution satisfying the initial condition: x

sveticcg [70]

Answer:

The particular solution satisfying the initial condition, $x(0)=6$ , of the differential equation $\frac{dx}{dt}=x^2+\frac{1}{9}$ is $x=\frac{\tan \left(\frac{t+3\arctan \left(18\right)}{3}\right)}{3}$ .

Step-by-step explanation:

A separable differential equation is any differential equation that we can write in the following form.

$N(y)\frac{dy}{dx}=M(x)$

We may find the solutions to certain separable differential equations by separating variables, integrating with respect to x, and ultimately solving the resulting algebraic equation for y.

To find the solution of the differential equation $\frac{dx}{dt}=x^2+\frac{1}{9}$ you must:

Separate the differential equation and integrate both sides.

$dx=(x^2+\frac{1}{9})\cdot dt\\ \\\frac{dx}{x^2+\frac{1}{9}} =dt\\\\\int {\frac{dx}{x^2+\frac{1}{9}}} =\int dt$

Solving $\int \frac{dx}{x^2+\frac{1}{9}}$

$\mathrm{Apply\:Integral\:Substitution:}\:x=\frac{1}{3}u\\\\\int \frac{3}{u^2+1}du\\\\3\cdot \int \frac{1}{u^2+1}du\\\\\mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u^2+1}du=\arctan \left(u\right)\\\\3\arctan \left(u\right)\\\\\mathrm{Substitute\:back}\:u=\frac{x}{\frac{1}{3}}\\\\3\arctan \left(3x\right)\\\\\int \frac{1}{x^2+\frac{1}{9}}dx=3\arctan \left(3x\right)+C$

Therefore,

$\int {\frac{dx}{x^2+\frac{1}{9}}} =\int dt\\\\3\arctan \left(3x\right)+C=t+D\\\\3\arctan \left(3x\right)=t+D-C\\\\3\arctan \left(3x\right)=t+E$

Now, we use the initial condition $x(0)=6$ to find the value of the constant E.

$3\arctan \left(3(6)\right)=0+E\\E=3\arctan \left(18\right)$

Thus,

$3\arctan \left(3x\right)=t+3\arctan \left(18\right)$

and we solve for x,

$\frac{3\arctan \left(3x\right)}{3}=\frac{t}{3}+\frac{3\arctan \left(18\right)}{3}\\\\\arctan \left(3x\right)=\frac{t+3\arctan \left(18\right)}{3}\\\\\arctan \left(x\right)=a\quad \Rightarrow \quad \:x=\tan \left(a\right)\\\\3x=\tan \left(\frac{t+3\arctan \left(18\right)}{3}\right)\\\\x=\frac{\tan \left(\frac{t+3\arctan \left(18\right)}{3}\right)}{3}$

8 0

3 years ago

How many solutions does the following equation have?<br> 20z – 5 – 12z = 10z + 8

tensa zangetsu [6.8K]

The equation $20z-5-12z = 10z + 8$ have one solution i.e $z=-\frac{13}{2}$

Step-by-step explanation:

Finding the solutions of the equation:

$20z-5-12z = 10z + 8$

Solving:

$20z-5-12z = 10z + 8\\8z-5=10z+8\\Adding\,\,5\,\,on\,\,both\,\,sides\\8z=10z+8+5\\8z=10z+13\\Adding\,\,-10z\,\,on\,\,both\,\,sides\\8z-10z=13\\-2z=13\\z=\frac{13}{-2} \,\,or\\z=-\frac{13}{2}$

The equation $20z-5-12z = 10z + 8$ have one solution i.e $z=-\frac{13}{2}$

Keywords: Solving the equations:

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6 0

3 years ago

Read 2 more answers