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Dahasolnce [82]

4 years ago

6

Please help me on this one

1 answer:

lara31 [8.8K]4 years ago

8 0

Answer:8600

Step-by-step explanation:

477

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Consider the first order differential equation

AURORKA [14]

Answer:

For y(-7) =6.4

The largest interval is between

$-\infty \to -5$

For y(-2.5) = -0.5.

The largest interval is between

$-5 \to 5$

For y(0) = 0

The largest interval is between

$-5 \to 5$

For y(4.5) = -2.1.

The largest interval is between

$-5 \to 5$

For y(14)= 1.7.

The largest interval is between

$9 \to \infty$

Step-by-step explanation:

From m the question we are told that

The first order differential equation is $\frac{y' - t}{ t^2 -25} = \frac{e^t}{t-9}$

Now the first step is to obtain the domain of the differential equation

Now to do that let consider the denominators

Now generally

$t^2 - 25 \ne 0$ side calculation

=> $t\ne \pm5$ $t^2 - 25 = 0$

$t = \pm 5$

Also $t-9\ne 0$ $t -9 = 0$

=> $t\ne 9$ $t= 9$

This means that this first order differential equation is discontinuous at

$t = -5 , \ \ t = 5 \ \ t = 9$

$This \ is \ illustrated \ below \ \\ ------------------\\\. \ \ \ | \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \ \ \ \ \ \ \ \ \ \ \ \ \ |\\. \ -5 \ \ \ \ \ \ \ \ \ \ \ \ 5 \ \ \ \ \ \ \ \ \ \ \ \ \ 9$

So

For y(-7) =6.4

The largest interval is between

$-\infty \to -5$

For y(-2.5) = -0.5.

The largest interval is between

$-5 \to 5$

For y(0) = 0

The largest interval is between

$-5 \to 5$

For y(4.5) = -2.1.

The largest interval is between

$-5 \to 5$

For y(14)= 1.7.

The largest interval is between

$9 \to \infty$

4 0

3 years ago

In 2005, 120 students at Lincoln High School

Mrrafil [7]

Answer:

The average increase in the number of smart

phones per year is 45

Step-by-step explanation:

Given that :

Number of students who had phones in 2005 = 120

Number of students who had phones in 2010 = 345

(345 - 120) / 5

225 / 5

= 45

Hence, average increase in number of smartphones per year = 45

3 0

3 years ago

The value of an investment A (in dollars) after t years is given by the function A(t) = A0ekt. If it takes 10 years for an inves

Ludmilka [50]

Answer:

20 years.

Step-by-step explanation:

We have been given a formula $A(t)=A_0\cdot e^{kt}$ , which represents the value of an investment A (in dollars) after t years.

Substitute the given values:

$\$3,000=\$1,000\cdot e^{k*10}$

Let us solve for k.

$\frac{\$3,000}{\$1,000}=\frac{\$1,000\cdot e^{k*10}}{\$1,000}$

$3=e^{k*10}$

Take natural log of both sides:

$\text{ln}(3)=\text{ln}(e^{k*10})$

Using property $\text{ln}(a^b)=b\cdot \text{ln}(a)$ , we will get:

$\text{ln}(3)=10k\cdot\text{ln}(e)$

We know that $\text{ln}(e)=1$ , so

$\text{ln}(3)=10k\cdot 1$

$\text{ln}(3)=10k$

$\frac{\text{ln}(3)}{10}=\frac{10k}{10}$

$\frac{\text{ln}(3)}{10}=k$

$\$9,000=\$1,000\cdot e^{\frac{\text{ln}(3)}{10}*t}$

Dividing both sides by 1000, we will get:

$9=e^{\frac{\text{ln}(3)}{10}*t}$

Take natural log of both sides:

$\text{ln}(9)=\text{ln}(e^{\frac{\text{ln}(3)}{10}*t)$

$\text{ln}(9)=\frac{\text{ln}(3)}{10}*t\cdot\text{ln}(e)$

$\text{ln}(9)=\frac{\text{ln}(3)}{10}*t\cdot1$

$10*\text{ln}(9)=10*\frac{\text{ln}(3)}{10}*t$

$10\text{ln}(9)=\text{ln}(3)*t$

$10\text{ln}(3^2)=\text{ln}(3)*t$

$2\cdot 10\text{ln}(3)=\text{ln}(3)*t$

$20\text{ln}(3)=\text{ln}(3)*t$

Divide both sides by $\text{ln}(3)$ :

$\frac{20\text{ln}(3)}{\text{ln}(3)}=\frac{\text{ln}(3)*t}{\text{ln}(3)}$

$20=t$

Therefore, it will take 20 years for the investment to be $9,000.

5 0

3 years ago

Calculate the area of triangle ABC with altitude CD, given A(-7,-1), B(-1,5), C(0,0), and D(-3,3)

Zigmanuir [339]

Answer:

d

Step-by-step explanation:

6 0

3 years ago

Analyzing Student Work

Romashka-Z-Leto [24]

Answer: B

Step-by-step explanation:

Edge2020

4 0

4 years ago