Ask question

Ask question

All categories

3 years ago

12

What is another way to write (3x1000)+(3x10)+(3x1/100)+(3x1/1000

1 answer:

Paha777 [63]3 years ago

3 0

That'd be 3000 + 30 + 0.03 + 0.003, or 3030.033

hope this helped :)
alisa202

plzz tell me if I was wrong

You might be interested in

MATH HELP PLEASE DUE SOON!

julsineya [31]

Answer:

a

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

The function Q(x)=2x^2-kx+18. For what value of k does Q(x) have one distinct real solution? (NEEDS TO BE DONE WITHIN 2 HOURS!)

Aliun [14]

Answer:

k = 12

Step-by-step explanation:

Given:

The equation $Q(x)=2x^2-kx+18$

To find:

Value of $k = ?$ for which the given equation has one distinct real solution.

Solution:

The given equation is a quadratic equation.

There are always two solutions of a quadratic equation.

For the equation: $ax^{2} +bx+c=0$ to have one distinct solution:

$b^2 - 4ac = 0$

Here,

a = 2,

b = -k and

c = 18

Putting the values, we get:

$(-k)^2 - 4\times 2\times 18 = 0\\\Rightarrow k^2 = 18\times 8\\\Rightarrow k^2 =144\\\Rightarrow k = 12$

The equation becomes:

$Q(x)=2x^2-12x+18$

And the one root is:

$2(x^2-6x+9 ) = 0\\\Rightarrow 2(x-3)^2=0\\\Rightarrow x = 3$

4 0

3 years ago

Rational expression 8/x+5 + x/x+5

mina [271]

Https://www.symbolab.com/solver/function-range-calculator
this should help

6 0

3 years ago

Write the equations for the graphs below (worth 5 points)

serg [7]

Answers:

10) y= 1/2x - 2
11) y= 2x + 3
12) y= 2/3x - 4

I found this by using y=mx+ b

3 0

2 years ago

Read 2 more answers

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