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2 years ago

WILL GIVE 32 POINTS FOR WHOEVER GIVES CORRECT ANSWER

Mathematics

1 answer:

Brilliant_brown [7]2 years ago

8 0

Graph the line that goes through the point (3,-1) and has a y-intercept of -4:

You can simply graph two points to create the line for this problem. The points are (3, -1) and (0, -4). The slope for this line is 1/1 too, which means that if you need more points, you would start at either of the two given points and go up one unit and over one unit to the right.

Select three points that the line would go through (y = x - 3):

In the equation, the y-intercept is given, which is one point - (0,-3). Then, we can go up one unit and right one unit to find a second point - (1,-2). And also down one unit and left one unit to find a third point - (-1,4).

Which equation represents the same relationship shown in the graph:

First, let's find the y-intercept, which is (0,6). Then, we can find other points where the line crosses and count how many units up and to the right that would be, which is 1/1. Therefore, the correct equation is B - y = x + 6.

Which of the following graphs represents y = x - 2:

The y-intercept is (0,-2) and the slope is positive and 1/1. We are looking for a graph that crosses through the point (0,-2), points up and to the right, and increases by 1 unit up and 1 unit right each time. Therefore, the correct graph is B.

Hope this helps!! :)

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If a=2, b-3 and c=5. Thr value of 4a-2b+5c are

inn [45]

(4*2)-(2*-3)+(5*5)= 8-6+25= 27

4 0

2 years ago

Please help ASAP Simplify <img src="https://tex.z-dn.net/?f=%5Csqrt%7B.64x%5E2%7D" id="TexFormula1" title="\sqrt{.64x

Brilliant_brown [7]

Answer: 4/5x

Explanation: simply the expression, then convert it and calculate your answer.

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2 years ago

Which step is included in the graph of the function f(x)=[x-1]? (the brackets are ceiling functions symbols)

KiRa [710]

we are given

f(x)=[x=1]

where bracket means ceiling functions

we know that

Ceiling function returns the least value of the integer that is greater than or equal to the specified number

so, we can check each options

option-A:

$-4\leq x$

At x=-4:

f(x)=[-4-1] =-5

For x<-3:

Let's assume

x=-3.1

f(x)=[-3.1-1] =[-4.1]=-5

so, this interval is TRUE

option-B:

$-2\leq x$

At x=-2:

f(x)=[-2-1] =-3

For x<-1:

Let's assume

x=-1.1

f(x)=[-1.1-1] =[-2.1]=-3

so, this is FALSE

7 0

2 years ago

Read 2 more answers

A paint store mixed 4 3/4 cans of paint and 7 2/3 cans of white paint did this mixture make?

stira [4]

Answer:

12 5/12 cans of white make

Step-by-step explanation:

6 0

2 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:

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