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3 years ago

PLEASE HELP OH LORD JESUS PLEASE HELP ME!! I NEED HELP!! PLEASE HELP!!!

Mathematics

1 answer:

ddd [48]3 years ago

4 0

Answer:

-1 <---------- slope

Step-by-step explanation:

Since the formula is y2-y1=x2-x1

since the points are (-2,6) and (0,4)

Enter the points in

4-6/0-(-2) ----> (2 negatives is a +)

-2/2= -1

The slope is -1

If the slope is -1, the equation will be y=-x+b

then, enter one of the given points, (-2,6)

6=2+b

b=4

So the slope int form is y=-x+b

to find the point slope form,

y-y1=m(x1-x) <--- formula

So,

take 1 point (-2,6) and enter it in

y-6=-1(x+2) <--- point slope form

to graph y=-3x+2

make x=0 and enter into the equation

y=0+2

y=2

so if x=0, y=2 so thats one point (0,2)

and lets say if x=-6 (enter into equation)

y=-3(-6)+2

y=18+2

y=20

so if x=-6, y=20 (another point) (-6,20)

you only need 2 points to graph a line

so enter (0,2) and (-6,20) and draw a line

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Solve the following differential equation: (2x+5y)dx+(5x−4y)dy=0 *Hint: they are exact<br><br> C=.

Tpy6a [65]

Answer with Step-by-step explanation:

The given differential equation is

$(2x+5y)dx+(5x-4y)dy=0$

Now the above differential equation can be re-written as

$P(x,y)dx+Q(x,y)dy=0$

Checking for exactness we should have

$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$

$\frac{\partial P}{\partial y}=\frac{\partial (2x+5y)}{\partial y}=5$

$\frac{\partial Q}{\partial x}=\frac{\partial (5x-4y)}{\partial x}=5$

As we see that the 2 values are equal thus we conclude that the given differential equation is exact

The solution of exact differential equation is given by

$u(x,y)=\int P(x,y)dx+\phi(y)\\\\u(x,y)=\int (2x+5y)dx+\phi (y)\\\\u(x,y)=x^2+5xy+\phi (y)$

The value of $\phi (y)$ can be obtained by differentiating u(x,y) partially with respect to 'y' and equating the result with P(x,y)

$\frac{\partial u}{\partial y}=\frac{\partial (x^2+5xy+\phi (y)))}{\partial y}=Q(x,y))\\\\5y+\phi '(y)=(5x-4y)\\\\\phi '(y)=5x-9y\\\\\int\phi '(y)\partial y=\int (5x-9y)\partial y\\\\\phi (y)=5xy-\frac{9y^2}{2}\\\\\therefore u(x,y)=x^2+10xy-\frac{9y^2}{2}+c$

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3 years ago

If a₁ = 4 and an = 5an-1 then find the value of a5.

igor_vitrenko [27]

The value of $a_{5}$ is 2500, when $a_{1}=4$ and $5a_{n-1}$ .

Given that, $a_{1}=4$ and $5a_{n-1}$ .

We need to find the value of $a_{5}$ .

<h3>What is an arithmetic sequence?</h3>

An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.

Now, to find the value of $a_{5}$ :

$a_{2} =5a_{2-1}=5a_{1}=5 \times4=20$