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3 years ago

HELP!!!!

Mathematics

2 answers:

salantis [7]3 years ago

6 0

Answer:

Step-by-step explanation:

egoroff_w [7]3 years ago

5 0

Answer:

x axis

Step-by-step explanation:

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Which is the greater than or less than sign I cannot remember it is confusing sorry if you help thank you!

iVinArrow [24]

< greater than
> less than

4 0

4 years ago

Read 2 more answers

PLAEASE HELP ME If the equation of a line is 2y+3x=3, what is the y intercept of the line?

Alja [10]

Answer:

3/2

Step-by-step explanation:

2y+3x=3

This represents a linear equation and the format for a linear equation is

y = mx+b

m = slope

b= y-intercept

we have to subtract 3x from both sides to make this the y=mx+b form

2y=-3x+3

and divide both sides by 2

y = (-3x+3)/2

3/2 or 1.5 is the y-intercept

the constant of a linear equation (or 3) is the y-intercept, if there is no constant then the y-intercept is 0

6 0

3 years ago

Read 2 more answers

Help me out please. i need to find the function rule,

Ksivusya [100]

Replace x in the equations and see which one gets the matching y :

-5(0) = 0 +1 = 1

-5(1) = -5 + 1 = -4

-5(-1) = 5 +6 = 6

The first equation works.

7 0

3 years ago

Let f(x) = 1/x . Find the number b such that the average rate of change of f on the interval [2, b] is − 1/8

vekshin1

Answer:

b=4

Step-by-step explanation:

So, we have the function $f(x)=1/x$ . We need to find b such that the average rate of change or the slope is -1/8 between the intervel [2, b]. First, let's find f(2).

f(2) = 1/(2) = 1/2

So, we have the point (2, 1/2)

At point b, f(b) = 1/b.

Let's plug this into the slope formula:

$\frac{y_2-y_1}{x_2-x_1}=\frac{.5-\frac{1}{b} }{2-b} =-1/8$

Now, we just need to solve for b. First, let's multiply both the numerator and denominator by b (to get rid of the annoying fraction in the numerator).

$\frac{.5b-1}{2b-b^2} =\frac{-1}{8}$

Now, cross multiply.

$4b-8=b^2-2b$

$b^2-6b+8=0$

Solve for b. Factor using the numbers -4 and -2.

$=(b-4)(b-2)=0$

Thus, b=4 or b=2.

However, b=2 is not a possible solution since the interval [2,2] means nothing. Thus, b=4.

4 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:

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