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krek1111 [17]
2 years ago
6

A total of 729 students participate in a school pep rally event. If every 30 minutes, 2/3 of the students leave, after 2 hours,

how many students will be still staying at the pep rally?
Mathematics
1 answer:
trapecia [35]2 years ago
5 0

There will be 9 students left at the prep rally after 2 hours.

After 30 minutes

  • Total Number of students = 729
  • 2/3 of 729 = 486
  • Number of students left = 729 - 486 = 243

After 1 hour:

  • 2/3 of 243 = 162
  • Number of students left = 243 - 162 = 81

After 1 hour 30 minutes :

  • 2/3 of 81 = 54
  • Number of students left = 81 - 54 = 27

After 2 hours :

  • 2/3 of 27 = 18
  • Number of students left = 27 - 18 = 9

Therefore, the Number of students left after 2 hours will be 9.

Learn more : brainly.com/question/18112348

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You paint four walls. Each wall is a rectangle with a length of 20 feet and a height of 10 feet. One gallon of paint covers abou
Sphinxa [80]

Answer:

Step-by-step explanation:

20 x 10= 200 ft

1 gallon is 330 ft

paint 4 walls

200 x 4

800 ft/ 330

You would need 2.434 gallons of paint.

6 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
Determine the discriminant for the quadratic equation -3 = x^2+4x+1 . Based on the discriminant value, how many real number solu
djverab [1.8K]

Answer:

1 real

Step-by-step explanation:

The discriminant is:

b^2-4ac

which comes from the quadratic formula.

The rules are:

If the discriminant is > 0, there are 2 real solutions

If the discriminant is = 0, there is 1 real solution

If the discriminant is < 0, there are 2 imaginary solutions

(there are 2 other situations involving square roots, but these are the most basic ones in Algebra 2 books when you study quadratics)

First, we have to get that quadratic in standard form, which means getting everything on one side of the equals sign and setting it equal to 0:

x^2+4x+4 = 0

where a = 1, b = 4, c = 4

Filling in the discriminant:

4^2-4(1)(4)

gives us

16 - 16 = 0

Therefore, our quadratic has one real solution.  Factor it and see if you'd like, to prove that it is true.

7 0
3 years ago
Blake burns 41/2 calories in 2/3 of a minute riding his bike. what is blakes unit rate
MA_775_DIABLO [31]
(9/2) (3)
--------- X -----
(2/3) (3)

(27/2)
---------
(2)

(27)
------
(4)

6 and (3/4) calories:1 minute
3 0
3 years ago
NEED HELP PLZ!!!!!!George and John travel 392 miles in 7 hours. If they travel at the same rate, how long will it take them to t
Paraphin [41]

Answer:

Step-by-step explanation:

a. Find the unit rate, or how many miles they traveled in one hour, by dividing:

\frac{392}{7}=56 mph

b. Now that we know how fast they were going, we can use it in either a proportion or in the d = rt equation. For the proportion set up, put miles on the top and hours on the bottom and solve for the unknown:

\frac{mi.}{hr}:\frac{56}{1}=\frac{728}{x} and cross multiply to solve for x:

56x = 728 so

13 hours.

Now use the d = rt equation:

728 = 56t so

t = 13 hours

5 0
2 years ago
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