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2 years ago

Please Answer These questions (on the picture) 20 Points

Mathematics

1 answer:

telo118 [61]2 years ago

8 0

#13 is 17
#14 is 42
#15 is 15
#16 is 7
#17 is 3
#18 is 4

only giving you those fam good luck homie

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Can someone help me with this one

HACTEHA [7]

X = 21 + 34
Answer: x = 55°

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:

Point a:

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.

Point b:

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 y we get:

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

Point c:

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

3 years ago

NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. One hundred tickets

fomenos

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

In Jeremiah class

Irina18 [472]

Answer:

2/5 = 4/10

4/10 = 40%

Explanation:

2/5 is a simplified fraction, and one of the bigger fractions it’s made by 4/10. Do you can use 4/10, but can also use 8/20, 20/50, 40/100, so on. Since we know that any fraction that of ten can be turned into a percentage easily (2/10=20%, 5/10=50%,9/10=%90), then 4/10 is equal to 40%. So 40% of the people in Jeremiah’s class are boys.

6 0

1 year ago

How many liters of pure water should be mixed with a 18-L solution of 80% acid to produce a mixture that is 90% water?

Sphinxa [80]

126 liters of pure water should be added

Solution:

Let "x" be the liters of pure water

Then, 18 + x is the liters of final solution

18 Liter solution of 80% acid, which means 100 - 80 = 20 % of water is used

Pure water is 100%

Then according to question, we can say,

"x" liters of pure water 100 % is mixed with a 18 liter solution of 20 % water to produce a mixture (18 + x) liter that is 90 % water

Thus we frame a equation as:

Equation:

water + water = water

$x \times 100\% + 18 \times 20 \% = (18+x) \times 90\%$

$x \times \frac{100}{100} + 18 \times \frac{20}{100} = (18+x) \times \frac{90}{100}\\\\Simplify\ the\ above\ equation\\\\x + 0.2 \times 18 = (18+x) \times 0.9\\\\x + 3.6 = 16.2 + 0.9x\\\\0.1x = 12.6\\\\Divide\ both\ sides\ of\ equation\ by\ 0.1\\\\x = 126$

Thus 126 liters of pure water should be added

8 0

3 years ago