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2 years ago

PLEASE HELP ASAP, WILL GIVE BRAINLIEST IF CORRECT

Mathematics

1 answer:

dimaraw [331]2 years ago

8 0

Answer:

Line R

Step-by-step explanation:

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Scenario: Two people in a major city with a population of 450,000 get infected with a rapidly spreading virus. If each day the i

zmey [24]

1) See attachment.

2) See attachment

3) $y=2^{x+1}$

4) 17.8 days

Step-by-step explanation:

The table is in attachment.

In this problem, we are told that the initial number of people infected ad day zero is two, so the first row is (0,2).

Then, we are told that each day, an infected person infects 2 additional people. Therefore, at day 1, the number of infected people will be 2*2=4.

Then, each of the 4 persons infect 2 additional persons, so the number of infected people at day 2 will be 4*2=8.

Continuing the sequence, the following days the number of infected people will be:

8*2 = 16

16*2 = 32

32*2 = 64

The graph representing the situation is shown in attachment.

On the x-axis, we have represented the day, from zero to 5.

On the y-axis, we have represented the number of infected people.

We see that the points on the graph are:

0, 2

1, 4

2, 8

3, 16

4, 32

5, 64

Here we have to create a mathematics model (so, an equation) representing this scenario.

First of all, we notice that the number of infected people at day 0 is 2:

$p(0)=2$

To write an equation, we call $x$ the number of the day; this means that at x = 0, the value of y (number of infected people) is 2:

$y=2$

Then, at day 1 (x=1), the number of infected people is doubled:

$y(1)=2y(0)=2\cdot 2 = 4$

And so on. This means that for each increase of x of 1 unit, the value of y doubles: so, we can represents the model as

$y=2\cdot 2^x$

$y=2^{x+1}$

Here we are told that the entire city has a population of

p = 450,000

people.

In order for the virus to infect the whole population, it means that the value of y must be equal to the total population:

y = 450,000

Substituting into the equation of the model, this means that

$450,000 = 2^{x+1}$

And solving for x, we find the number of days after which this will happen:

$log_2(450,000)=x+1\\x=log_2(450,000)-1=17.8 d$

So, after 17.8 days.

8 0

3 years ago

Increase 12 1/2 cm by 25 mm

nadezda [96]

Change the 12 and 1/2 cm to mm which is 125mm + 25mm = 150mm

3 0

3 years ago

What k-1/3=5/6 equal ?

JulijaS [17]

The value of "K"

$k -\frac{1}{3} = \frac{5}{6}$

MMC (3,6)
3,6 | 3
1,2 | 2
1,1 |__ 3*2 = 6

Solving:
$k -\frac{1}{3} = \frac{5}{6}$
$\frac{6k}{6} - \frac{2}{6} = \frac{5}{6}$
cancel denominators (6)
$6k - 2 = 5$
$6k = 5+2$
$6k = 7$
$\boxed{k = \frac{7}{6} }$

7 0

3 years ago

For the function defined by f(t)=2-t, 0≤t<1, sketch 3 periods and find:

Oksi-84 [34.3K]

The half-range sine series is the expansion for $f(t)$ with the assumption that $f(t)$ is considered to be an odd function over its full range, $-1$ . So for (a), you're essentially finding the full range expansion of the function

$f(t)=\begin{cases}2-t&\text{for }0\le t$

with period 2 so that $f(t)=f(t+2n)$ for $|t|$ and integers $n$ .

Now, since $f(t)$ is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with

$f(t)=\displaystyle\sum_{n\ge1}b_n\sin\frac{n\pi t}L$

where

$b_n=\displaystyle\frac2L\int_0^Lf(t)\sin\frac{n\pi t}L\,\mathrm dt$

In this case, $L=1$ , so

$b_n=\displaystyle2\int_0^1(2-t)\sin n\pi t\,\mathrm dt$
$b_n=\dfrac4{n\pi}-\dfrac{2\cos n\pi}{n\pi}-\dfrac{2\sin n\pi}{n^2\pi^2}$
$b_n=\dfrac{4-2(-1)^n}{n\pi}$

The half-range sine series expansion for $f(t)$ is then

$f(t)\sim\displaystyle\sum_{n\ge1}\frac{4-2(-1)^n}{n\pi}\sin n\pi t$

which can be further simplified by considering the even/odd cases of $n$ , but there's no need for that here.

The half-range cosine series is computed similarly, this time assuming $f(t)$ is even/symmetric across its full range. In other words, you are finding the full range series expansion for

$f(t)=\begin{cases}2-t&\text{for }0\le t$

Now the sine series expansion vanishes, leaving you with

$f(t)\sim\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi t}L$

where

$a_n=\displaystyle\frac2L\int_0^Lf(t)\cos\frac{n\pi t}L\,\mathrm dt$

for $n\ge0$ . Again, $L=1$ . You should find that

$a_0=\displaystyle2\int_0^1(2-t)\,\mathrm dt=3$

$a_n=\displaystyle2\int_0^1(2-t)\cos n\pi t\,\mathrm dt$
$a_n=\dfrac2{n^2\pi^2}-\dfrac{2\cos n\pi}{n^2\pi^2}+\dfrac{2\sin n\pi}{n\pi}$
$a_n=\dfrac{2-2(-1)^n}{n^2\pi^2}$

Here, splitting into even/odd cases actually reduces this further. Notice that when $n$ is even, the expression above simplifies to

$a_{n=2k}=\dfrac{2-2(-1)^{2k}}{(2k)^2\pi^2}=0$

while for odd $n$ , you have

$a_{n=2k-1}=\dfrac{2-2(-1)^{2k-1}}{(2k-1)^2\pi^2}=\dfrac4{(2k-1)^2\pi^2}$

So the half-range cosine series expansion would be

$f(t)\sim\dfrac32+\displaystyle\sum_{n\ge1}a_n\cos n\pi t$
$f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}a_{2k-1}\cos(2k-1)\pi t$
$f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}\frac4{(2k-1)^2\pi^2}\cos(2k-1)\pi t$

Attached are plots of the first few terms of each series overlaid onto plots of $f(t)$ . In the half-range sine series (right), I use $n=10$ terms, and in the half-range cosine series (left), I use $k=2$ or $n=2(2)-1=3$ terms. (It's a bit more difficult to distinguish $f(t)$ from the latter because the cosine series converges so much faster.)

5 0

3 years ago

Question 8<br> What is (4x - 5)(6x + 5) expressed as a trinomial?

erica [24]

Answer:

24x^2 -10x - 25

Step-by-step explanation:

24x^2 + 20x - 30x - 25

24^2 -10x - 25

8 0

3 years ago

Read 2 more answers