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3 years ago

12

What is 10^{2} /19^{4}

1 answer:

MatroZZZ [7]3 years ago

4 0

Answer:

$\frac{100}{130321}$

Step-by-step explanation:

10² ÷ $19^{4}$

100 ÷ $19^{4}$

100 ÷ 130321

$\frac{100}{130321}$

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Find the 25th term of an arithmetic sequence whose first term is 12 and whose common difference is ‒6.

blondinia [14]

C -132 is the answer

8 0

3 years ago

Determine if the sequence is arithmetic, geometric, or neither: 15, 18, 21, 24,

Vikki [24]

Answer:

arithmitic

Step-by-step explanation:

7 0

3 years ago

Let y(t) be the solution to y˙=3te−y satisfying y(0)=3 . (a) Use Euler's Method with time step h=0.2 to approximate y(0.2),y(0.4

OLEGan [10]

Answer:

$y(0.2)=3$ , $y(0.4)=3.005974448$ , $y(0.6)=3.017852169$ , $y(0.8)=3.035458382$ , and $y(1.0)=3.058523645$

The general solution is $y=\ln \left(\frac{3t^2}{2}+e^3\right)$

The error in the approximations to y(0.2), y(0.6), and y(1):

$|y(0.2)-y_{1}|=0.002982771$

$|y(0.6)-y_{3}|=0.008677796$

$|y(1)-y_{5}|=0.013499859$

Step-by-step explanation:

<em>Point a:</em>

The Euler's method states that:

$y_{n+1}=y_n+h \cdot f \left(t_n, y_n \right)$ where $t_{n+1}=t_n + h$

We have that $h=0.2$ , $t_{0}=0$ , $y_{0} =3$ , $f(t,y)=3te^{-y}$

We need to find $y(0.2)$ for $y'=3te^{-y}$ , when $y(0)=3$ , $h=0.2$ using the Euler's method.

So you need to:

$t_{1}=t_{0}+h=0+0.2=0.2$

$y\left(t_{1}\right)=y\left(0.2)=y_{1}=y_{0}+h \cdot f \left(t_{0}, y_{0} \right)=3+h \cdot f \left(0, 3 \right)=$

$=3 + 0.2 \cdot \left(0 \right)= 3$

$y(0.2)=3$

We need to find $y(0.4)$ for $y'=3te^{-y}$ , when $y(0)=3$ , $h=0.2$ using the Euler's method.

So you need to:

$t_{2}=t_{1}+h=0.2+0.2=0.4$

$y\left(t_{2}\right)=y\left(0.4)=y_{2}=y_{1}+h \cdot f \left(t_{1}, y_{1} \right)=3+h \cdot f \left(0.2, 3 \right)=$

$=3 + 0.2 \cdot \left(0.02987224102)= 3.005974448$

$y(0.4)=3.005974448$

The Euler's Method is detailed in the following table.

<em>Point b:</em>

To find the general solution of $y'=3te^{-y}$ you need to:

Rewrite in the form of a first order separable ODE:

$e^yy'\:=3t\\e^y\cdot \frac{dy}{dt} =3t\\e^y \:dy\:=3t\:dt$

Integrate each side:

$\int \:e^ydy=e^y+C$

$\int \:3t\:dt=\frac{3t^2}{2}+C$

$e^y+C=\frac{3t^2}{2}+C\\e^y=\frac{3t^2}{2}+C_{1}$

We know the initial condition y(0) = 3, we are going to use it to find the value of $C_{1}$

$e^3=\frac{3\left(0\right)^2}{2}+C_1\\C_1=e^3$

So we have:

$e^y=\frac{3t^2}{2}+e^3$

Solving for <em>y</em> we get:

$\ln \left(e^y\right)=\ln \left(\frac{3t^2}{2}+e^3\right)\\y\ln \left(e\right)=\ln \left(\frac{3t^2}{2}+e^3\right)\\y=\ln \left(\frac{3t^2}{2}+e^3\right)$

<em>Point c:</em>

To compute the error in the approximations y(0.2), y(0.6), and y(1) you need to:

Find the values y(0.2), y(0.6), and y(1) using $y=\ln \left(\frac{3t^2}{2}+e^3\right)$

$y(0.2)=\ln \left(\frac{3(0.2)^2}{2}+e^3\right)=3.002982771$

$y(0.6)=\ln \left(\frac{3(0.6)^2}{2}+e^3\right)=3.026529965$

$y(1)=\ln \left(\frac{3(1)^2}{2}+e^3\right)=3.072023504$

Next, where $y_{1}, y_{3}, \:and \:y_{5}$ are from the table.

$|y(0.2)-y_{1}|=|3.002982771-3|=0.002982771$

$|y(0.6)-y_{3}|=|3.026529965-3.017852169|=0.008677796$

$|y(1)-y_{5}|=|3.072023504-3.058523645|=0.013499859$

3 0

4 years ago

Need an answer thanks in advance!

slamgirl [31]

Answer:

1806 seats.

Step-by-step explanation:

From the question given above, the following data were obtained:

Row 1 = 24 seats

Row 2 = 27 seats

Row 3 = 30 seats

Total roll = 28

Total number of seat =?

From the above data, we can liken the roll to be in arithmetic progress.

Also, we are asked to determine the total number of seats in the theater.

Thus the sum of the sequence can be written as:

Roll 1 + Roll 2 + Roll 3 +... + Roll 28 i.e

24 + 27 + 30 +...

Thus, we can obtain obtained the total number of seats in the theater by applying the sum of arithmetic progress formula. This can be obtained as follow:

First term (a) = 24

Common difference (d) = 2nd term – 1st term

Common difference (d) = 27 – 24 = 3

Number of term (n) = 28

Sum of the 28th term (S₂₈) =?

Sₙ = n/2 [2a + (n –1)d]

S₂₈ = 28/2 [2×24 + (28 –1)3]

S₂₈ = 14 [48 + 27×3]

S₂₈ = 14 [48 + 81]

S₂₈ = 14 [129]

S₂₈ = 1806

Thus, the number of seats in the theater is 1806.

6 0

3 years ago

The Jurassic Zoo charges $14 for each adult admission and $9 for each child. The total bill for the 214 people from a school t

Butoxors [25]

a=adult

c=child

a+c=214

c=214-a

9c+14a=2081

9(214-a)+14a=2081

1926-9a+14a=2081

5a=155

a=155/5=31

31 adults

183 children

check

31*14 = 434

183*9=1647

1647+434=2081

4 0

4 years ago