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3 years ago

I need to show how to find these solutions 1/4 and 3/4 by completing the square!

Mathematics

2 answers:

Arlecino [84]3 years ago

8 0

Answer:

See explanation, and ask for more details if unclear!

Step-by-step explanation:

The perfect square of this equation is $x^2-x+\frac{1}{4}$ , since the square would be $(x-\frac{1}{2})^2$ . 1/4=4/16, meaning that you can set up the equation in the following way:

$(x^2-x+\frac{1}{4})-\frac{1}{16}=0$

$(x-\frac{1}{2})^2=\frac{1}{16}$

Take the square root of both sides:

$x-\frac{1}{2}=\pm \frac{1}{4}$

Add 1/2 to both sides:

$x=\frac{1}{2}\pm \frac{1}{4}=\frac{3}{4}, \frac{1}{4}$ . Hope this helps!

frosja888 [35]3 years ago

7 0

Answer:

see below

Step-by-step explanation:

x^2 - x + 3/16 = 0

Subtract 3/16 from each side

x^2 - x + 3/16 - 3/16 = -3/16

x^2 -x = -3/16

Take the coefficient of x

-1

Divide by 2

-1/2 call this a

Square it

(-1/2)^2 = 1/4

Add it to each side

x^2 - x +1/4 =- 3/16+1/4

Changing to the "square"

( x +a)^2 = - 3/16+1/4

getting a common denominator

( x -1/2) ^2 = -3/16+ 4/16

( x -1/2) ^2 =1/16

Take the square root of each side

sqrt( ( x -1/2) ^2) = ± sqrt(1/16)

x -1/2 = ± 1/4

Add 1/2 to each side

x -1/2 + 1/2 = 1/2 ± 1/4

x = 1/2 ± 1/4

Separate into 2 equations

x = 1/2 + 1/4 x = 1/2 - 1/4

2/4 +1/4 2/4 -1/4

3/4 1/4

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0.253

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leonid [27]

Answer with explanation:

The given differential equation is

y" -y'+y=2 sin 3x------(1)

Let, y'=z

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$\frac{dy}{dx}=z\\\\d y=zdx\\\\y=z x$

Substituting the value of , y, y' and y" in equation (1)

z'-z+zx=2 sin 3 x

z'+z(x-1)=2 sin 3 x-----------(1)

This is a type of linear differential equation.

Integrating factor

$=e^{\int (x-1) dx}\\\\=e^{\frac{x^2}{2}-x}$

Multiplying both sides of equation (1) by integrating factor and integrating we get

$\rightarrow z\times e^{\frac{x^2}{2}-x}=\int 2 sin 3 x \times e^{\frac{x^2}{2}-x} dx=I$

$I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx -\int \frac{2\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx-\frac{2I}{3}\\\\\frac{5I}{3}=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{5}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{5} dx$

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3 years ago

An instructor who taught two sections of Math 161A, the first with 20 students and the second with 30 students, gave a midterm e

uranmaximum [27]

Answer:

The answers are for option (a) 0.2070 (b)0.3798 (c) 0.3938

Step-by-step explanation:

Given:

Here Section 1 students = 20

Section 2 students = 30

Here there are 15 graded exam papers.

(a )Here Pr(10 are from second section) = ²⁰C₅ * ³⁰C₁₀/⁵⁰C₁₅= 0.2070

(b) Here if x is the number of students copies of section 2 out of 15 exam papers.

here the distribution is hyper-geometric one, where N = 50, K = 30 ; n = 15

Then,

Pr( x ≥ 10 ; 15; 30 ; 50) = 0.3798

(c) Here we have to find that at least 10 are from the same section that means if x ≥ 10 (at least 10 from section B) or x ≤ 5 (at least 10 from section 1)

so,

Pr(at least 10 of these are from the same section) = Pr(x ≤ 5 or x ≥ 10 ; 15 ; 30 ; 50) = Pr(x ≤ 5 ; 15 ; 30 ; 50) + Pr(x ≥ 10 ; 15 ; 30 ; 50) = 0.0140 + 0.3798 = 0.3938

Note : Here the given distribution is Hyper-geometric distribution

where f(x) = kCₓ)(N-K)C(n-x)/ NCK in that way all these above values can be calculated.

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3 years ago

Which represents the reflection of f(x)=sqrt x over the y axis

zloy xaker [14]

reflection of $f(x)=\sqrt{x}$ over the y axis we get $g(x)=\sqrt{-x}$

Step-by-step explanation:

We need to find the reflection of f(x)=sqrt x over the y axis

So, reflection over y- axis

If the function f(x) is transformed over y-axis then the new function g(x) will be $g(x)= f(-x)$

So, finding reflection of $f(x)=\sqrt{x}$ over the y axis we get:

$g(x)=\sqrt{-x}$

So, reflection of $f(x)=\sqrt{x}$ over the y axis we get $g(x)=\sqrt{-x}$

Keywords: Transformations

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#learnwithBrainly

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3 years ago

(10 points) Consider the initial value problem y′+3y=9t,y(0)=7. Take the Laplace transform of both sides of the given differenti

Rashid [163]

Answer:

The solution

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3 t}$

Step-by-step explanation:

Explanation:-

Consider the initial value problem y′+3 y=9 t,y(0)=7

Step(i):-

Given differential problem

y′+3 y=9 t

Take the Laplace transform of both sides of the differential equation

L( y′+3 y) = L(9 t)

Using Formula Transform of derivatives

L(y¹(t)) = s y⁻(s)-y(0)

By using Laplace transform formula

$L(t) = \frac{1}{S^{2} }$

Step(ii):-

Given

L( y′(t)) + 3 L (y(t)) = 9 L( t)

$s y^{-} (s) - y(0) + 3y^{-}(s) = \frac{9}{s^{2} }$

$s y^{-} (s) - 7 + 3y^{-}(s) = \frac{9}{s^{2} }$

Taking common y⁻(s) and simplification, we get

$( s + 3)y^{-}(s) = \frac{9}{s^{2} }+7$

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

Step(iii):-

By using partial fractions , we get

$\frac{9}{s^{2} (s+3} = \frac{A}{s} + \frac{B}{s^{2} } + \frac{C}{s+3}$

$\frac{9}{s^{2} (s+3} = \frac{As(s+3)+B(s+3)+Cs^{2} }{s^{2} (s+3)}$

On simplification we get

9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Put s =0 in equation(i)

9 = B(0+3)

B = 9/3 = 3

Put s = -3 in equation(i)

9 = C(-3)²

C = 1

Given Equation 9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Comparing 'S²' coefficient on both sides, we get

9 = A s²+3 A s +B(s)+3 B +C(s²)

0 = A + C

put C=1 , becomes A = -1

$\frac{9}{s^{2} (s+3} = \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}$

Step(iv):-

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

$y^{-}(s) =9( \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}) + \frac{7}{s+3}$

Applying inverse Laplace transform on both sides

$L^{-1} (y^{-}(s) ) =L^{-1} (9( \frac{-1}{s}) + L^{-1} (\frac{3}{s^{2} }) + L^{-1} (\frac{1}{s+3}) )+ L^{-1} (\frac{7}{s+3})$

By using inverse Laplace transform

$L^{-1} (\frac{1}{s} ) =1$

$L^{-1} (\frac{1}{s^{2} } ) = \frac{t}{1!}$

$L^{-1} (\frac{1}{s+a} ) =e^{-at}$

Final answer:-

Now the solution , we get

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3t}$

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3 years ago