Ask question

Ask question

All categories

3 years ago

5

HELP ME PLS....

2 answers:

elena55 [62]3 years ago

7 0

Answer:

Step-by-step explanation:

a) 1/4

b)9/4

c)9/4

d)He ate more berries on tuesday as he at 1/4 on monday and 3/4 on tuesday.

3/4 is 3 times more than 1/4

Galina-37 [17]3 years ago

7 0

To confirm, everything that BlueOrnage · Ace said was correct! I saw the explanation(s), and solved myself. I'd retype the answers, but that would be a waste. I hope I helped you! Great job, BlueOrnage! You are so kind!

You might be interested in

Mrs. Buchanan is planning marching routines for the school band. The band has 36 members. Band

olga nikolaevna [1]

This is another answer for brainly.

5 0

3 years ago

Read 2 more answers

John buys a computer with a 20% discount. If he paid $760 for the computer, how much was the discount?

Mumz [18]

If there was 20% taken off you add 20% back on and that is how you will get your answer. So take 760 plus the 20% which gives you 912 dollars

8 0

3 years ago

Read 2 more answers

A table with a round top is cut in half so that it can be used against a wall.after it is cut,the edge along the wall is 6 feet

MrRissso [65]

Answer:

The area of the table after it is cut is 12 or 36 because

Step-by-step explanation:

6 times 2= 12 and 6 times 6= 36 so year

4 0

3 years ago

Would appreciate some help first to answer gets brainliest

elixir [45]

The correct answer is B

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