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3 years ago

10

Find the standard equation of a circle that satisfies the conditions. Radius 5, center (1,negative 3)

1 answer:

professor190 [17]3 years ago

6 0

Answer: $(x-1)^{2}$ + $(y+3)^{2}$ = 25

Step-by-step explanation:

You might be interested in

QUESTION 5 of 10: You are considering selling your home. You consult with an agent who gives you the asking prices of five homes

san4es73 [151]

Answer: $169,500

Step-by-step explanation:

The mean simply means the average price of a set of a numbers that are given. For one t calculate the mean, one has to add the numbers that are given and then divide them by the amount of numbers.

I'm the case above, the numbers given are $165,000, $188,000, $175,000, and $150,000.

Mean = ($165,000 ,+ $188,000 +$175,000 + $150,000) ÷ 4

= $678,000/4

= $169,500

3 0

3 years ago

Please help me ASAP with this math problem

ser-zykov [4K]

Answer:

Root( c square -9)

Step-by-step explanation:

The answer u have is the answer for a^2

8 0

2 years ago

Read 2 more answers

What’s the equation for g(x)

taurus [48]

Answer:

g(4)=10

Step-by-step explanation:

4 0

3 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

Consider the quadratic function f(x) = –2x2 5x – 4. the leading coefficient of the function is?

evablogger [386]

Answer:

$\text{-2 is the leading coefficient of the quadratic equation }f(x)=-2x^2+5x-4$

Step-by-step explanation:

Given the quadratic function

$f(x)=-2x^2+5x-4$

we have to find the leading coefficient of given quadratic equation.

$\text{The general form of quadratic equation is }ax^2+bx+c=0$

The coefficient of $x^2$ i.e a is known as the leading coefficient.

Comparing given equation with the standard equation, we get

a=-2

$\text{Hence, -2 is the leading coefficient of the quadratic equation }f(x)=-2x^2+5x-4$

3 0

3 years ago

Read 2 more answers