All categories

3 years ago

Slope -5,y-intercept (0,3) What is f(x)

Mathematics

1 answer:

MA_775_DIABLO [31]3 years ago

8 0

Answer:

$f(x)=-5\ x+3$

Step-by-step explanation:

Given:

Slope of line =-5

y-intercept = $(0,3)$

The equation of line is given by :

$f(x)=m\ x+b$

where $m$ represents slope of line and $b$ represents y-intercept.

So from the given data, we have $m=-5$ and $b=3$

∴ The equation of line is given by:

$f(x)=-5\ x+3$

You might be interested in

Diego measured the length of a pen to be 22 cm.

solmaris [256]

Answer:

Step-by-step explanation:

From the question, we are informed that Diego measured the length of a pen to be 22 cm while the actual length of the pen is 23 cm.

The percentage decrease:

= (23 - 22)/22 × 100

= 1/22 × 100

= <em>4.5%</em>

good luck :)

4 0

3 years ago

Read 2 more answers

A movie theater buys popcorn kernals for $5.88 per pound. (A) Which equation shows the cost of popcorn kernals? (B) what is the

inessss [21]

Answer:

Given that the manager of Theatre A says that they usually go through about 15 cups of popcorn kernels and about 5 cups of oil each weeknight.

Then, the ratio value of oil to popcorn kernels for theatre A is 5 / 15 = 1 / 5.

Given that the manager of Theatre B says that they order 18 cups of oil and 72 cups of popcorn kernels each week.

Then, the ratio value of oil to popcorn kernels for theatre B is 18 / 72 = 1 / 4.

Given that the manager of Theatre C says that their concessions use 6 cups of oil and 32 cups of popcorn kernels on a busy Saturday.

Then, the ratio value of oil to popcorn kernels for theatre C is 6 / 32 = 3 / 16.

Step-by-step explanation:

4 0

3 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:

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