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2 years ago

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Please help me with this I attached the pictures!

1 answer:

german2 years ago

8 0

For number one. Go to may then up to 55. Crawl before you walk. Just plot the points on the graph and make sure the line goes through 0,0

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a grocery store has 5 pounds of granola. One customer buys 2/3 pounds of granola and another buys 5/6 pounds. after these purcha

Pavlova-9 [17]

There would be 3 and 1/2 pounds left.

2/3 x 2 = 4/6
5/6
5/6 + 4/6 = 6/6 (1) and 3/6 (1/2)

3 0

2 years ago

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Studentka2010 [4]

Answer:

What the english?

Step-by-step explanation:

bdhznrbzjdnxhenaFjskrnx

4 0

2 years ago

A figure is made up of three triangles. Each triangle has a base of 5 feet and a height of 4.5 feet. What is the total area of t

aliina [53]

Answer:

Step-by-step explanation:

3 * 1/2 * 5 * 4.5

= 33.75 fT^2.

6 0

1 year 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

What is a property to write an equivalent expression for 12 X (100/5)

emmainna [20.7K]

Answer:

240

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers