All categories

3 years ago

Which direction does the graph of the equation shown below open? Y^2-4x+4y-4=0

Mathematics

2 answers:

zavuch27 [327]3 years ago

8 0

We have that
Y²-4x+4y-4=0
you can rewrite the equation as
4x=y²+4y-4
x=(y²/4)+y-1
x=0.25*(y²+4y)-1
x=0.25*(y²+4y+4-4)-1
x=0.25*(y²+4y+4)-5
x=0.25*(y+2)²-5

the equation of the parabola is of the form
x=a(y-k)²+h
is a horizontal parabola
where
(h,k)-----> is the vertex-----> (-5,-2)
a=0.25
a> 0--------> open to the right

the answer is
the option D. Right

see the attached figure

Taya2010 [7]3 years ago

3 0

The answer is D, right because you rewrite the equation to a x= equation. Just by doing that, you know the equation has to be left or right. When rewriting the equation, the slope is positive, therefore it is facing right

You might be interested in

What is the value of the expression below when x = 6 and y -4 ?

tester [92]

Answer:

-2

Step-by-step explanation:

Plug in the respected values for x and y.

Divide and get -2

8 0

2 years ago

A pharmaceutical company sells bottles of 500 calcium tablets in two dosages: 250 milligram and 500 milligram. Last month, the c

Svetach [21]

X = 250 mg tabs
y = 500 mg tabs

2200x + 1800y = 39200....multiply by -1
2200x + 2200y = 44000
--------------------------------
-2200x - 1800y = - 39200 (result of multiplying by -1)
2200x + 2200y = 44000
--------------------------------add
400y = 4800
y = 4800/400
y = 12...... the 500 mg tabs (y) costs $ 12 per bottle

2200x + 2200y = 44000
2200x + 2200(12) = 44000
2200x + 26400 = 44000
2200x = 44000 - 26400
2200x = 17600
x = 17600/2200
x = 8 <== the 250 mg tabs (x) costs $ 8 per bottle

3 0

2 years ago

Read 2 more answers

Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f

drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

$y''+3y'=0$

The characteristic equation is

$r^2+3r=r(r+3)=0$

with roots $r=0$ and $r=-3$ , giving the two solutions $C_1$ and $C_2e^{-3t}$ .

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

$y''+3y'=2t^4$

Assume the ansatz solution,

${y_p}=at^5+bt^4+ct^3+dt^2+et$

$\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e$

$\implies {y_p}''=20at^3+12bt^2+6ct+2d$

(You could include a constant term f here, but it would get absorbed by the first solution $C_1$ anyway.)

Substitute these into the ODE:

$(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4$

$15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4$

$\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}$

$y''+3y'=t^2e^{-3t}$

$e^{-3t}$ is already accounted for, so assume an ansatz of the form

$y_p=(at^3+bt^2+ct)e^{-3t}$

$\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}$

$\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}$

Substitute into the ODE:

$(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}$

$9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2$

$-9at^2+(6a-6b)t+2b-3c=t^2$

$\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}$

$y''+3y'=\sin(3t)$

Assume an ansatz solution

$y_p=a\sin(3t)+b\cos(3t)$

$\implies {y_p}'=3a\cos(3t)-3b\sin(3t)$

$\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)$

Substitute into the ODE:

$(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)$

$(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)$

$\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}$

So, the general solution of the original ODE is

$y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}$

3 0

3 years ago

A store bought a tent for $230 and

MatroZZZ [7]

Answer:

409.86

Step-by-step explanation:

230×.65=149.50

230+149.50=379.50×.08=30.36

379.50+30.36=409.86

8 0

3 years ago

Please help, super easy problem

katen-ka-za [31]

Let the volume be x.

Therefore, answer of the given question = x cm³

6 0

3 years ago