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3 years ago

SOMEONE HELP ME I ONLY HAVE 5 MIN TO TURN THIS IN!!!!!!

Mathematics

1 answer:

SVETLANKA909090 [29]3 years ago

4 0

Answer:

i wanna say 8 bc 12 would make all the other numbers to big and 8 would prob make the the numbers the highest you can go without going over 80

Step-by-step explanation:

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I don’t understand need help

kodGreya [7K]

Answer:

6y^2-12y-4

Step-by-step explanation:

(8y^2-5y+7) - (2y^2+7y+11)

(8y^2-5y+7)-2y^2-7y-11

Combine like terms:

6y^2-12y-4

5 0

4 years ago

Read 2 more answers

The sum of two numbers is 43, and their product is 450. What is their difference?

kow [346]

Answer:

List the factors of 450

1,450

3,150

5,90

6,75

9,50

10,45

15,30

18,25

The factor with the sum of 43 is 18+25

25-18=7 so the difference is 7

hope this helps

Step-by-step explanation:

8 0

4 years ago

Read 2 more answers

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

4 years ago

the number of pizzas eaten at the class party is equal to one-fourth and number of students plus two more pizzas for the adults

maria [59]

If s represents the number of students, then the number of pizzas is
s/4 + 2 = 9
s/4 = 7 . . . . . . subtract 2
s = 28 . . . . . . multiply by 4

There are 28 students.

7 0

3 years ago

Need help someome help me plz

elena55 [62]

A mapping diagram shows relationships

3 0

4 years ago

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