All categories

3 years ago

how can you use the definition of a circle and the distance formula to derive the standard equation of a circle?

1 answer:

5 0

Circle definition : a round plane figure whose boundary (the circumference) consists of points equidistant (=radius r) from a fixed point (the center (h,k)).

let (x,y) is a point on the circle.
its distance (r) to center is
(x-h)^2 + (y-k)^2 = r^2
... the circle equation

You might be interested in

keira has 83 spelling words to study in 8 weeks. she already knows 3 words. she will study the same number of words each week. h

Lemur [1.5K]

10 because she already knows 3 so then you divide 80 (number of words she hasn’t learned) by 8 (the number of weeks) and you get 10.

5 0

3 years ago

Read 2 more answers

. A triangle has vertices on a coordinate grid at points J(-1, 5), K(4, 5), and L(4, -2). What is the length, in units, of ?

Neko [114]

Answer:

20.6

Step-by-step explanation:

Given data

J(-1, 5)

K(4, 5), and

L(4, -2)

Required

The perimeter of the traingle

Let us find the distance between the vertices

J(-1, 5) amd

K(4, 5)

The expression for the distance between two coordinates is given as

d=√((x_2-x_1)²+(y_2-y_1)²)

substitute

d=√((4+1)²+(5-5)²)

d=√5²

d= √25

d= 5

Let us find the distance between the vertices

K(4, 5), and

L(4, -2)

The expression for the distance between two coordinates is given as

d=√((x_2-x_1)²+(y_2-y_1)²)

substitute

d=√((4-4)²+(-2-5)²)

d=√-7²

d= √49

d= 7

Let us find the distance between the vertices

L(4, -2) and

J(-1, 5)

The expression for the distance between two coordinates is given as

d=√((x_2-x_1)²+(y_2-y_1)²)

substitute

d=√((-1-4)²+(5+2)²)

d=√-5²+7²

d= √25+49

d= √74

d=8.6

Hence the total length of the triangle is

=5+7+8.6

=20.6

4 0

3 years ago

Can someone help me? I’ll reward brainalist + points

iren [92.7K]

Answer:

B, -25

Step-by-step explanation:

3-4 = -1

-1(3 + 3^2 + 13) = -1(25), the answer is -25

8 0

3 years ago

Read 2 more answers

Pleeease open the image and hellllp me

Verdich [7]

1. Rewrite the expression in terms of logarithms:

$y=x^x=e^{\ln x^x}=e^{x\ln x}$

Then differentiate with the chain rule (I'll use prime notation to save space; that is, the derivative of y is denoted y' )

$y'=e^{x\ln x}(x\ln x)'=x^x(x\ln x)'$

$y'=x^x(x'\ln x+x(\ln x)')$

$y'=x^x\left(\ln x+\dfrac xx\right)$

$y'=x^x(\ln x+1)$

2. Chain rule:

$y=\ln(\csc(3x))$

$y'=\dfrac1{\csc(3x)}(\csc(3x))'$

$y'=\sin(3x)\left(-\cot^2(3x)(3x)'\right)$

$y'=-3\sin(3x)\cot^2(3x)$

Since $\cot x=\frac{\cos x}{\sin x}$ , we can cancel one factor of sine:

$y'=-3\dfrac{\cos^2(3x)}{\sin(3x)}=-3\cos(3x)\cot(3x)$

3. Chain rule:

$y=e^{e^{\sin x}}$

$y'=e^{e^{\sin x}}\left(e^{\sin x}\right)'$

$y'=e^{e^{\sin x}}e^{\sin x}(\sin x)'$

$y'=e^{e^{\sin x}+\sin x}\cos x$

4. If you're like me and don't remember the rule for differentiating logarithms of bases not equal to e, you can use the change-of-base formula first:

$\log_2x=\dfrac{\ln x}{\ln2}$

Then

$(\log_2x)'=\left(\dfrac{\ln x}{\ln 2}\right)'=\dfrac1{\ln 2}$

So we have

$y=\cos^2(\log_2x)$

$y'=2\cos(\log_2x)\left(\cos(\log_2x)\right)'$

$y'=2\cos(\log_2x)(-\sin(\log_2x))(\log_2x)'$

$y'=-\dfrac2{\ln2}\cos(\log_2x)\sin(\log_2x)$

and we can use the double angle identity and logarithm properties to condense this result:

$y'=-\dfrac1{\ln2}\sin(2\log_2x)=-\dfrac1{\ln2}\sin(\log_2x^2)$

5. Differentiate both sides:

$\left(x^2-y^2+\sin x\,e^y+\ln y\,x\right)'=0'$

$2x-2yy'+\cos x\,e^y+\sin x\,e^yy'+\dfrac{xy'}y+\ln y=0$

$-\left(2y-\sin x\,e^y-\dfrac xy\right)y'=-\left(2x+\cos x\,e^y+\ln y\right)$

$y'=\dfrac{2x+\cos x\,e^y\ln y}{2y-\sin x\,e^y-\frac xy}$

$y'=\dfrac{2xy+\cos x\,ye^y\ln y}{2y^2-\sin x\,ye^y-x}$

6. Same as with (5):

$\left(\sin(x^2+\tan y)+e^{x^3\sec y}+2x-y+2\right)'=0'$

$\cos(x^2+\tan y)(x^2+\tan y)'+e^{x^3\sec y}(x^3\sec y)'+2-y'=0$

$\cos(x^2+\tan y)(2x+\sec^2y y')+e^{x^3\sec y}(3x^2\sec y+x^3\sec y\tan y\,y')+2-y'=0$

$\left(\cos(x^2+\tan y)\sec^2y+x^3\sec y\tan y\,e^{x^3\sec y}-1\right)y'=-\left(2x\cos(x^2+\tan y)+3x^2\sec y\,e^{x^3\sec y}+2\right)$

$y'=-\dfrac{2x\cos(x^2+\tan y)+3x^2\sec y\,e^{x^3\sec y}+2}{\cos(x^2+\tan y)\sec^2y+x^3\sec y\tan y\,e^{x^3\sec y}-1}$

7. Looks like

$y=x^2-e^{2x}$

Compute the second derivative:

$y'=2x-2e^{2x}$

$y''=2-4e^{2x}$

Set this equal to 0 and solve for x :

$2-4e^{2x}=0$

$4e^{2x}=2$

$e^{2x}=\dfrac12$

$2x=\ln\dfrac12=-\ln2$

$x=-\dfrac{\ln2}2$

7 0

3 years ago

What are the values of a, b, and c in the quadratic equation 0 = 1/2 x2 – 3x – 2?

MA_775_DIABLO [31]

In a quadratic equation, the standard equation is represented as follows:

Ax^2 + Bx + C = 0

where A, B and C are constants and A should not have a zero value.

The values of A, B, and C would be as follows:

A = 1/2
B = -3
C = -2

Hope this answers the question. Have a nice day.

8 0

3 years ago