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3 years ago

7

Solve (9xy+5) (8xy+7)

1 answer:

Maurinko [17]3 years ago

7 0

It’s gonna look like that

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Plz Help me for this question

kicyunya [14]

Answer:

what do u want me to do

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

A bag of marbles contains 15 blue marbles and 25 green marbles. Which of the

d1i1m1o1n [39]

Answer:

5/8

Step-by-step explanation:

6 0

2 years ago

What is the perimeter of a square which has the same area as a circle with circumfrence of 4π

Kruka [31]

Answer:

Perimeter square = 8 sqrt(pi)

Step-by-step explanation:

The perimeter of a square is 4*s

The area of a circle is Area = pi * r^2

The circumference of a circle is C = 2*pi * r

C = 4 pi

4pi = 2*pi * r

r = 2

So the area of the circle is pi * r^2 = pi * 2^2 = 4pi

The square has the same area

Area = 4*pi

Square = 4*pi

s^2 = 4*pi

s = sqrt(4*pi)

s = 2*sqrt(pi)

The perimeter = 4 * 2 * sqrt(pi)

The perimeter = 8 * sqrt(pi)

8 0

2 years ago

the height y of a ball thrown by a child is given by y = -1/12x^2 + 2x + 4. how high is the ball when it is at its maximum heigh

zhuklara [117]

Answer:

The height of ball when it is at maximum height is 12 unit

Step-by-step explanation:

Given as :

A ball is thrown by child to height y

And y as a function x is y = $-\frac{1}{12}x^{2}+2x+4$

So, for maximum value of height ,

$\frac{\partial y}{\partial x}$ = 0

Or , $\frac{\partial (-\frac{1}{12}x^{2}+2x+4)}{\partial x}$ = 0

Or, $-\frac{1}{12}(2x)+2$ = 0

Or, $\frac{1}{12}(2x)$ = 2

or, x = 12

Hence The height of ball when it is at maximum height is 12 unit Answer

3 0

3 years ago

Consider the initial value problem y′+5y=⎧⎩⎨⎪⎪0110 if 0≤t<3 if 3≤t<5 if 5≤t<[infinity],y(0)=4. y′+5y={0 if 0≤t<311 i

rosijanka [135]

It looks like the ODE is

$y'+5y=\begin{cases}0&\text{for }0\le t$

with the initial condition of $y(0)=4$ .

Rewrite the right side in terms of the unit step function,

$u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t$

In this case, we have

$\begin{cases}0&\text{for }0\le t$

The Laplace transform of the step function is easy to compute:

$\displaystyle\int_0^\infty u(t-c)e^{-st}\,\mathrm dt=\int_c^\infty e^{-st}\,\mathrm dt=\frac{e^{-cs}}s$

So, taking the Laplace transform of both sides of the ODE, we get

$sY(s)-y(0)+5Y(s)=\dfrac{e^{-3s}-e^{-5s}}s$

Solve for $Y(s)$ :

$(s+5)Y(s)-4=\dfrac{e^{-3s}-e^{-5s}}s\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}{s(s+5)}+\dfrac4{s+5}$

We can split the first term into partial fractions:

$\dfrac1{s(s+5)}=\dfrac as+\dfrac b{s+5}\implies1=a(s+5)+bs$

If $s=0$ , then $1=5a\implies a=\frac15$ .

If $s=-5$ , then $1=-5b\implies b=-\frac15$ .

$\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}5\left(\frac1s-\frac1{s+5}\right)+\dfrac4{s+5}$

$\implies Y(s)=\dfrac15\left(\dfrac{e^{-3s}}s-\dfrac{e^{-3s}}{s+5}-\dfrac{e^{-5s}}s+\dfrac{e^{-5s}}{s+5}\right)+\dfrac4{s+5}$

Take the inverse transform of both sides, recalling that

$Y(s)=e^{-cs}F(s)\implies y(t)=u(t-c)f(t-c)$

where $F(s)$ is the Laplace transform of the function $f(t)$ . We have

$F(s)=\dfrac1s\implies f(t)=1$

$F(s)=\dfrac1{s+5}\implies f(t)=e^{-5t}$

We then end up with

$y(t)=\dfrac{u(t-3)(1-e^{-5t})-u(t-5)(1-e^{-5t})}5+5e^{-5t}$

3 0

3 years ago