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11111nata11111 [884]

3 years ago

14

G(x) = -0.5x^2 + 4x – 2

1 answer:

-BARSIC- [3]3 years ago

5 0

Answer:

Step-by-step explanation:

What is the question?

g(x) = -0.5x² + 4x - 2 is a down-opening parabola. Do you need to know how to put it in vertex form?

vertex (4,6)

focus (4,5.5)

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What represents the inverse of the function f(x) = 4x

Katarina [22]

Answer:

$f { - 1}^{ } = \frac{x}{4}$

Step-by-step explanation:

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8 0

3 years ago

ABDE is a rectangle on coordinate axes, the sides of the rectangle are parallel to the axes,

Hitman42 [59]

The coordinates of B and D are the location of the vertices B and D in the rectangle

The coordinates of B and D are (55, 30) and (55, 14)

<h3>How to determine the coordinates of B and D?</h3>

From the figure of the rectangle, we have:

A = (25,30)

C = (40, 22)

Point B has the same y-coordinate as point A

So, we have:

B = (x, 30)

The x-coordinate is then calculated as:

Bx = 2Cx - Ax

This gives

Bx = 2*40 - 25

Bx = 55

So, the coordinates of B are:

B = (55, 30)

Point D has the same x-coordinate as point B

So, we have:

D = (55, y)

The y-coordinate is then calculated as:

Dy = Cy - (By - Cy)

This gives

Dy = 22 - (30 - 22)

Dy = 14

So, the coordinates of B are:

D = (55, 14)

Hence, the coordinates of B and D are (55, 30) and (55, 14)

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8 0

2 years ago

Apply the method of undetermined coefficients to find a particular solution to the following system.wing system.

jarptica [38.1K]

$y''-y'+y=\sin x$

The corresponding homogeneous ODE has characteristic equation $r^2-r+1=0$ with roots at $r=\dfrac{1\pm\sqrt3}2$ , thus admitting the characteristic solution

$y_c=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x$

For the particular solution, assume one of the form

$y_p=a\sin x+b\cos x$

${y_p}'=a\cos x-b\sin x$

${y_p}''=-a\sin x-b\cos x$

Substituting into the ODE gives

$(-a\sin x-b\cos x)-(a\cos x-b\sin x)+(a\sin x+b\cos x)=\sin x$

$-b\cos x+a\sin x=\sin x$

$\implies a=1,b=0$

Then the general solution to this ODE is

$\boxed{y(x)=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x+\sin x}$

$y''-3y'+2y=e^x\sin x$

$\implies r^2-3r+2=(r-1)(r-2)=0\implies r=1,r=2$

$\implies y_c=C_1e^x+C_2e^{2x}$

Assume a solution of the form

$y_p=e^x(a\sin x+b\cos x)$

${y_p}'=e^x((a+b)\cos x+(a-b)\sin x)$

${y_p}''=2e^x(a\cos x-b\sin x)$

Substituting into the ODE gives

$2e^x(a\cos x-b\sin x)-3e^x((a+b)\cos x+(a-b)\sin x)+2e^x(a\sin x+b\cos x)=e^x\sin x$

$-e^x((a+b)\cos x+(a-b)\sin x)=e^x\sin x$

$\implies\begin{cases}-a-b=0\\-a+b=1\end{cases}\implies a=-\dfrac12,b=\dfrac12$

so the solution is

$\boxed{y(x)=C_1e^x+C_2e^{2x}-\dfrac{e^x}2(\sin x-\cos x)}$

$y''+y=x\cos(2x)$

$r^2+1=0\implies r=\pm i$

$\implies y_c=C_1\cos x+C_2\sin x$

Assume a solution of the form

$y_p=(ax+b)\cos(2x)+(cx+d)\sin(2x)$

${y_p}''=-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x)$

Substituting into the ODE gives

$(-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x))+((ax+b)\cos(2x)+(cx+d)\sin(2x))=x\cos(2x)$

$-(3ax+3b-4c)\cos(2x)-(3cx+3d+4a)\sin(2x)=x\cos(2x)$

$\implies\begin{cases}-3a=1\\-3b+4c=0\\-3c=0\\-4a-3d=0\end{cases}\implies a=-\dfrac13,b=c=0,d=\dfrac49$

so the solution is

$\boxed{y(x)=C_1\cos x+C_2\sin x-\dfrac13x\cos(2x)+\dfrac49\sin(2x)}$

7 0

3 years ago

Solve for x. Show your work.<br><br> 34x-5 = (1/27) 2x+10

Ilia_Sergeevich [38]

Answer:The answer in a fraction is 405/916 and the decimal form is 0.44213973

…

Step-by-step explanation:

6 0

3 years ago

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The area of the Everglades is about 1,900 square kilometers. What is the approximate size of the Everglades in acres?

Vera_Pavlovna [14]

America's Everglades<span> once covered almost 11,000 </span>square miles<span> of south Florida. ... half of the </span>1900<span>, the </span>Everglades<span> today is half the </span>size<span> it was a century ago.
This is what I think of it

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8 0

3 years ago