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3 years ago

Four ninths plus one ninth?

Mathematics

2 answers:

Semenov [28]3 years ago

5 0

Answer would be 5/9 bc keep same denominator add numerator

Kazeer [188]3 years ago

3 0

Answer:

5/9

Step-by-step explanation:

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Use Definition 7.1.1, DEFINITION 7.1.1 Laplace Transform Let f be a function defined for t ≥ 0. Then the integral ℒ{f(t)} = ∞ e−

poizon [28]

$f(t)=\begin{cases}\cos t&\text{for }0\le t$

The Laplace transform is then

$\mathcal L_s\{f(t)\}=\displaystyle\int_0^\infty f(t)e^{-st}\,\mathrm dt=\int_0^\pi e^{-st}\cos t\,\mathrm dt$

Let $I$ denote the integral we want to compute. Integrating by parts, setting

$u=e^{-st}\implies\mathrm du=-se^{-st}\,\mathrm dt$

$\mathrm dv=\cos t\,\mathrm dt\implies v=\sin t$

gives

$\displaystyle I=e^{-st}\sin t\bigg|_{t=0}^{t=\pi}+s\int_0^\pi e^{-st}\sin t\,\mathrm dt$

Integrate by parts again, setting

$u=e^{-st}\implies\mathrm du=-se^{-st}\,\mathrm dt$

$\mathrm dv=\sin t\,\mathrm dt\implies v=-\cos t$

Then

$\displaystyle I=e^{-st}\sin t\bigg|_{t=0}^{t=\pi}+s\left(-e^{-st}\cos t\bigg|_{t=0}^{t=\pi}-s\int_0^\pi e^{-st}\cos t\,\mathrm dt\right)$

$I=e^{-st}(\sin t-s\cos t)\bigg|_{t=0}^{t=\pi}-s^2I$

$(s^2+1)I=s(e^{-\pi s}+1)$

$I=\dfrac s{s^2+1}(e^{-\pi s}+1)$

7 0

4 years ago

Write any two quadratic equations.

Thepotemich [5.8K]

To find:

Any two quadratic equations.

Solution:

The general form of a quadratic equation is:

$ax^2+bx+c=0$

Where, a,b and c are real number and a is non zero.

We know that a,b,c can take any values but a cannot be 0.

For a=1, b=1,c=1, we get

$1x^2+1x+1=0$

$x^2+x+1=0$

One quadratic equation is $x^2+x+1=0$ .

For a=3,b=-1,c=5, we get

$3x^2+(-1)x+5=0$

$3x^2-x+5=0$

Therefore, the two quadratic equations are $x^2+x+1=0$ and $3x^2-x+5=0$ .

3 0

3 years ago

Pls help due in 10 mins

BabaBlast [244]

Answer:

Try inserting the Problem

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Evaluate the perimeter of a rectangle.

likoan [24]

Answer= 22

Explanation
————————

2l+2w=p

2(8)+2(3)= 22

4 0

4 years ago

How do I solve for the minimum and maximum of the function y=-1/2x^2 -5x+2

Sever21 [200]

Try this solution:

There are several ways to find the max or min of the given function:

1. to use derivative of the function. For more details see the attachment (3 basic steps); the coordinates of max-point are marked with green (-5; 14.5)

2. to use formulas. The given function is the standart function with common equation y=ax²+bx+c, it means the correspond formulas are (where a<0, the vertex of this function is its maximum):

$X_0=-\frac{b}{2a} ; \ X_0=-\frac{-5}{2*(-\frac{1}{2})} =-5.$

$Y_0=-\frac{D}{4a}; \ Y_0=-\frac{25+4*2*0.5}{4*(-\frac{1}{2})} =14.5$

Finally: point (-5;14.5) - maximum of the given function.

3. to draw a graph.

7 0

3 years ago