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1 year ago

Write an equation using (-1,0) and (4,-5) in slope intercept form help pls :')

1 answer:

5 0

ANSWER:
y=-x-1

EXPLANATION:
First you need to find the slope of the points (-1,0) and (4,-5). To find the slope you use the equation m=y2-y1/x2-x1
-5=y2
0=y1
4=x2
-1=x1
So, plugging the numbers into the equation, you should get m=-5-0/4-(-1)
Then you get -5/5 which simplifies to -1
Since slope intercept form is y=Mx+B, m=slope, you substitute -1 into the m spot.
To find the y-intercept, you need to first pick one of the points to use. (I will use (4,-5))
The 4 is in the x position in (x,y) so you plug that into the equation. y=-1(4)+b
The -5 is in the y position in (x,y) so you have to plug that into the equation as well. -5=-1(4)+b
Simplified you get -5=-4+b
You then have to add 4 to both side of the equation, which ends you up with -1=b. b is the y intercept so when you plug every thing into the equation you get y=-x-1
Hope that helps!

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Consider the differential equation y'' − y' − 20y = 0. Verify that the functions e−4x and e5x form a fundamental set of solution

KIM [24]

Answer:

Therefore $e^{-4x}$ and $e^{5x}$ are fundamental solution of the given differential equation.

Therefore $e^{-4x}$ and $e^{5x}$ are linearly independent, since $W(e^{-4x},e^{5x})=9e^x\neq 0$

The general solution of the differential equation is

$y=c_1e^{-4x}+c_2e^{5x}$

Step-by-step explanation:

Given differential equation is

y''-y'-20y =0

Here P(x)= -1, Q(x)= -20 and R(x)=0

Let trial solution be $y=e^{mx}$

Then, $y'=me^{mx}$ and $y''=m^2e^{mx}$

$\therefore m^2e^{mx}-m e^{mx}-20e^{mx}=0$

$\Rightarrow m^2-m-20=0$

$\Rightarrow m^2-5m+4m-20=0$

$\Rightarrow m(m-5)+4(m-5)=0$

$\Rightarrow (m-5)(m+4)=0$

$\Rightarrow m=-4,5$

Therefore the complementary function is = $c_1e^{-4x}+c_2e^{5x}$

Therefore $e^{-4x}$ and $e^{5x}$ are fundamental solution of the given differential equation.

If $y_1$ and $y_2$ are the fundamental solution of differential equation, then

$W(y_1,y_2)=\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|\neq 0$

Then $y_1$ and $y_2$ are linearly independent.

$W(e^{-4x},e^{5x})=\left|\begin{array}{cc}e^{-4x}&e^{5x}\\-4e^{-4x}&5e^{5x}\end{array}\right|$

$=e^{-4x}.5e^{5x}-e^{5x}.(-4e^{-4x})$

$=5e^x+4e^x$

$=9e^x\neq 0$

Therefore $e^{-4x}$ and $e^{5x}$ are linearly independent, since $W(e^{-4x},e^{5x})=9e^x\neq 0$

Let the the particular solution of the differential equation is

$y_p=v_1e^{-4x}+v_2e^{5x}$

$\therefore v_1=\int \frac{-y_2R(x)}{W(y_1,y_2)} dx$

and

$\therefore v_2=\int \frac{y_1R(x)}{W(y_1,y_2)} dx$

Here $y_1= e^{-4x}$ , $y_2=e^{5x}$ , $W(e^{-4x},e^{5x})=9e^x$ ,and $R(x)=0$

$\therefore v_1=\int \frac{-e^{5x}.0}{9e^x}dx$

and

$\therefore v_2=\int \frac{e^{5x}.0}{9e^x}dx$

The the P.I = 0

The general solution of the differential equation is

$y=c_1e^{-4x}+c_2e^{5x}$

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2 years ago

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julsineya [31]

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Step-by-step explanation:

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3 years ago

Loren solved the equation 10 = StartFraction 19 Over 9 EndFraction (149) + b for b as part of her work to find the equation of a

Elden [556K]

The error that Loren made is C. She should have solved 149 = StartFraction 19 Over 9 EndFraction (10) + b for b.

<h3>What is an equation?</h3>

It should be noted that an equation is a statement that shows that two mathematical expressions are equal.

In this case, for Loren to find the slope, this will be:

m = y2 - y1/x2 - x1

= (149 - 130)/(10 - 1)

= 19/9

In this case, the error that Loren made is that she should have solved 149 = StartFraction 19 Over 9 EndFraction (10) + b for b.

Learn more about equations on:

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1 year ago

How much much interest is on earned on an account if... Principal= $1,250 Rate= 5.7% Time= 4 year

Free_Kalibri [48]

Alright, lets gets started.

I suppose , we are asked to find simple interest for above question.

Principal P = $1,250

Rate r = 5.7 %

Time t = 4 years

The formula for simple interest = $\frac{P*r*t}{100}$

Hence simple interest $= \frac{1250*5.7*4}{100}$

Simple interest $= \frac{28500}{100}$

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andre [41]

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3 years ago