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3 years ago

Y = -5x+1 y = 3x – 2 Is (3.8) a solution of the system?

Mathematics

1 answer:

Bezzdna [24]3 years ago

3 0

Answer: No, 3.8 is not a solution.

Step-by-step explanation:

Set -5x+1 equal to 3x-2

-5x+1=3x-2 subtract 3x making it -8x+1=-2 subtract one from both sides and get -8x=-3 divide by -8 and get -3/-8 which equals in decimal form 0.375

Check it

plug in 0.375 for X

-5(0.375)+1= -0.875

3(0.375)-2 = -0.875

It checks out ^_^

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Find the coordinates of P so that P partitions AB in the ratio 3:4 with A(-9,-9) and B(5,-2).

ivolga24 [154]

The coordinates of point P are (-3 , -6)

Step-by-step explanation:

If point (x , y) partitions line AB where A = $(x_{1},y_{1})$ and B = $(x_{2},y_{2})$

at a ratio $m_{1}:m_{2}$ from A then:

1. $x=\frac{x_{1}m_{2}+x_{2}m_{1}}{m_{1}+m_{2}}$

2. $y=\frac{y_{1}m_{2}+y_{2}m_{1}}{m_{1}+m_{2}}$

∵ P partitions AB in the ratio 3 : 4

∴ $m_{1}:m_{2}$ = 3 : 4

∵ A = (-9 , -9) and B = (5 , -2)

- Substitute the coordinates of points A and B in the rules above

∵ $x=\frac{(-9)(4)+(5)(3)}{3+4}$

∴ $x=\frac{-36+15}{7}$

∴ $x=\frac{-21}{7}$

∴ x = -3

The x-coordinate of P is -3

∵ $y=\frac{(-9)(4)+(-2)(3)}{3+4}$

∴ $y=\frac{-36+(-6)}{7}$

∴ $y=\frac{-42}{7}$

∴ y = -6

The y-coordinate of P is -6

The coordinates of point P are (-3 , -6)

Learn more:

You can learn more about the mid-point of a segment in brainly.com/question/5223123

#LearnwithBrainly

5 0

4 years ago

Rewrite the equation to describe 4 as a function of 8:

masya89 [10]

Answer:

The steps below show how I transformed the equation in the question into standard form, where the slope and y-intercept is clearest. The slope is -1/2 and the y-intercept is 32.

$4x + 8y = 512 \\ x + 2y = 64 \\ 2y = - x + 64 \\ y = - \frac{1}{2} x + 32$

6 0

3 years ago

find the equation of parabola when point of intersection of directrix and axis is at (0,4) and focus at (6,4)

Allushta [10]

Answer:

Find the coordinates of the point of intersection of the axis and the directrix of the parabola whose focus is (3,3) and directrix is 3x−4y=2. Find also the length of the latus-rectum.

Step-by-step explanation:

6 0

3 years ago

You are given the following information about an investment account:

Flura [38]

Answer:

The calculated dollar-weighted rate of return, Y = -25%.

Step-by-step explanation:

The time-weighted rate of return is 0%

Therefore, (12/10)*( X/(12 + X)) = 1

12X = 120 + 10X -> X = 60

The dollar-weighted rate of return,Y is calculated below as:

Y = (X-(10 + X))/(1*10+X/2)

Y = (60-(10+60))/(1*10+60/2)

Y = (60 - 70)/(10+30)

Y= -10/40

Y = -25%

Therefore, the calculated dollar-weighted rate of return, Y = -25%.

7 0

4 years ago

Read 2 more answers

Solve the initial value problem 2ty" + 10ty' + 8y = 0, for t > 0, y(1) = 1, y'(1) = 0.

Eva8 [605]

I think you meant to write

$2t^2y''+10ty'+8y=0$

which is an ODE of Cauchy-Euler type. Let $y=t^m$ . Then

$y'=mt^{m-1}$

$y''=m(m-1)t^{m-2}$

Substituting $y$ and its derivatives into the ODE gives

$2m(m-1)t^m+10mt^m+8t^m=0$

Divide through by $t^m$ , which we can do because $t\neq0$ :

$2m(m-1)+10m+8=2m^2+8m+8=2(m+2)^2=0\implies m=-2$

Since this root has multiplicity 2, we get the characteristic solution

$y_c=C_1t^{-2}+C_2t^{-2}\ln t$

If you're not sure where the logarithm comes from, scroll to the bottom for a bit more in-depth explanation.

With the given initial values, we find

$y(1)=1\implies1=C_1$

$y'(1)=0\implies0=-2C_1+C_2\implies C_2=2$

so that the particular solution is

$\boxed{y(t)=t^{-2}+2t^{-2}\ln t}$

# # #

Under the hood, we're actually substituting $t=e^u$ , so that $u=\ln t$ . When we do this, we need to account for the derivative of $y$ wrt the new variable $u$ . By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{\mathrm dy}{\mathrm du}\dfrac{\mathrm du}{\mathrm dt}=\dfrac1t\dfrac{\mathrm dy}{\mathrm du}$

Since $\frac{\mathrm dy}{\mathrm dt}$ is a function of $t$ , we can treat $\frac{\mathrm dy}{\mathrm du}$ in the same way, so denote this by $f(t)$ . By the quotient rule,

$\dfrac{\mathrm d^2y}{\mathrm dt^2}=\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac ft\right]=\dfrac{t\frac{\mathrm df}{\mathrm dt}-f}{t^2}$

and by the chain rule,

$\dfrac{\mathrm df}{\mathrm dt}=\dfrac{\mathrm df}{\mathrm du}\dfrac{\mathrm du}{\mathrm dt}=\dfrac1t\dfrac{\mathrm df}{\mathrm du}$

where

$\dfrac{\mathrm df}{\mathrm du}=\dfrac{\mathrm d}{\mathrm du}\left[\dfrac{\mathrm dy}{\mathrm du}\right]=\dfrac{\mathrm d^2y}{\mathrm du^2}$

so that

$\dfrac{\mathrm d^2y}{\mathrm dt^2}=\dfrac{\frac{\mathrm d^2y}{\mathrm du^2}-\frac{\mathrm dy}{\mathrm du}}{t^2}=\dfrac1{t^2}\left(\dfrac{\mathrm d^2y}{\mathrm du^2}-\dfrac{\mathrm dy}{\mathrm du}\right)$

Plug all this into the original ODE to get a new one that is linear in $u$ with constant coefficients:

$2t^2\left(\dfrac{\frac{\mathrm d^2y}{\mathrm du^2}-\frac{\mathrm d y}{\mathrm du}}{t^2}\right)+10t\left(\dfrac{\frac{\mathrm dy}{\mathrm du}}t\right)+8y=0$

$2y''+8y'+8y=0$

which has characteristic equation

$2r^2+8r+8=2(r+2)^2=0$

and admits the characteristic solution

$y_c(u)=C_1e^{-2u}+C_2ue^{-2u}$

Finally replace $u=\ln t$ to get the solution we found earlier,

$y_c(t)=C_1t^{-2}+C_2t^{-2}\ln t$

4 0

4 years ago