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3 years ago

I'll give brainliest :)

Mathematics

1 answer:

nevsk [136]3 years ago

4 0

Answer:

Step-by-step explanation:

In the equation y = -x - 4, the gradient is -1.

While in the second equation,

5x + 5y = 20

y = -x + 4

So the gradient is -1 too

Both sides are not perpendicular to each other because if you apply the formula, m1m2 = -1, and if substitute both gradient, (-1)(-1) = 1 ≠ -1

Therefore, no they are not perpendicular but parallel instead.

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Answer:

y-axis

Step-by-step explanation:

The shape is reflected across the y-axis which is the vertical axis.

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3 years ago

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How do you solve this?Please explain.

Readme [11.4K]

Answer:

first you take the square root or 0.9, 2.7, 1.7, and 0.4 which will equal=

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4 years ago

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If f(x) = x3 + 2x2 − 5, then what is f(−3)?

Effectus [21]

Answer:

our answer is 8

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4 years ago

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If it takes 10gallons of gas to travel 130 miles. Plot a chart showing the distance covered by 25 gallons, 30 gallons and 32 gal

matrenka [14]

Answer:

325, 390, 416

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2 years ago

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