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3 years ago

7

There are n gallons of oil. After m gallons have been used then, in terms of m and n, what percent of oil remains? What percent

was used? PLZ HELP

1 answer:

ExtremeBDS [4]3 years ago

8 0

Answer:

remains is (n-m/n)*100 used is (m/n)*100

Step-by-step explanation:

You might be interested in

If F(x)=x^3-2x^2, which expression is equivalent to f(I) -2+i,-2-i,2+i,2-i

Romashka-Z-Leto [24]

Hello from MrBillDoesMath!

Answer:

2 - i

Discussion:

Evaluate F(x) = x^3 - 2 x^2 when x = i.

i^2 = -1 for i^3 = i ( i^2) = i (-1) = -i

So F(i) = (i)^3 - 2 (i)^2

= -i - 2(-1)

= -i + 2

= 2 - i

which is the last answer shown

Thank you,

MrB

6 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

3 years ago

Tìm n: 133^5 + 110^5 + 84^5 + 27^5 = n^5

slavikrds [6]

Answer: n=144. . . . .

5 0

3 years ago

Jimmy Andrew and Richard collect comic books. Jimmy has five more comic books than Andrew, and Richard has four times as many co

adell [148]

Answer: Andrew = 20

Jimmy = 25

Richard = 100

Step-by-step explanation:

Let the number of Andrew's comic books be y.

Jimmy has five more comic books than Andrew. This means Jimmy has: y+5

Richard has four times as many comic book as jimmy. This means Richard has: 4(y+5) = 4y + 20

Since there are 145 comic books together. This can be rewritten as:

y + (y+5) + (4y+20) = 145

6y+25 = 145

6y = 145 - 25

y = 120/6

y = 20

Andrew has 20 comic books

Jimmy has five more comic books than Andrew. Jimmy has: 20+5 = 25

Richard has: 4y + 20 = 4(20) + 20 = 100

5 0

3 years ago

Lesson 1 Homework

Nonamiya [84]

128

This question requires us to solve an infinite GP

As we can see the first unit is 16 wide, then 8, then 4 and so on; this can be written as

$64(1,0.5,0.25,0.125,....$ $\infinity$ ∞ $)$ (64 used as 16*4 = 64 is the area of the first rectangle)

We have $S$ ∞ $=a/(1-r)$ where a is the first term and r is the common ratio of the GP for summing an infinite GP.

here we have

$a=64 \\ r=0.5$

Therefore

$S$ ∞ $=\frac{64}{1-0.5} \\\\=\frac{64}{0.5} \\\\=128$

Learn more about GP here

brainly.com/question/16954106

#SPJ9

5 0

2 years ago