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2 years ago

50 points!! help asap pls

Mathematics

2 answers:

Roman55 [17]2 years ago

8 0

Answer:

Option D

Step-by-step explanation:

Given:

Radius = 10 units
Center point = (h, k) = (-4, -9)

∴ x coordinate of center point: (h, k) ⇒ (-4, -9)

∴ y coordinate of center point: (h, k) ⇒ (-4, -9)

Equation of circle formula:

⇒ (x - h)² + (y - k)² = r²

Substitute the radius in the formula;

⇒ (x - h)² + (y - k)² = r²

⇒ (x - h)² + (y - k)² = (10)²

Substitute the x and y coordinate in the formula;

⇒ [x - (-4)]² + [y - (-9)]² = (10)² [x coordinate (h): -4; y coordinate (k): -9]

⇒ [x + 4]² + [y + 9]² = 100 (Option D)

Therefore, Option D is correct.

Korvikt [17]2 years ago

6 0

Answer:

(x+4)² + (y+9)² = 100

Step-by-step explanation:

We want to find the equation of a circle with a radius of length 10 and a center at (-4, -9).

general equation of circle : (x-h)² + (y-k)² = r²

where (h,k) is at center and r = radius

the center is at (-4,-9) so h = -4 and k = -9 and the circle has a radius of 10 so r = 10

so we have (x-h)² + (y-k)² = r²

h = -4 , k = -9 and r = 10

(x-(-4))² + (y-(-9))² = 10²

==> apply double negative sign rule and remove parenthesis for -(-4) and -(-9)

(x+4)² + (y+9)² = 10²

==> simplify exponent

(x+4)² + (y+9)² = 100

and we are done!

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Find an equation of the tangent to the curve x =5+lnt, y=t2+5 at the point (5,6) by both eliminating the parameter and without e

svet-max [94.6K]

ANSWER

$y = 2x -4$

EXPLANATION

Part a)

Eliminating the parameter:

The parametric equation is

$x = 5 + ln(t)$

$y = {t}^{2} + 5$

From the first equation we make t the subject to get;

$x - 5 = ln(t)$

$t = {e}^{x - 5}$

We put it into the second equation.

$y = { ({e}^{x - 5}) }^{2} + 5$

$y = { ({e}^{2(x - 5)}) } + 5$

We differentiate to get;

$\frac{dy}{dx} = 2 {e}^{2(x - 5)}$

At x=5,

$\frac{dy}{dx} = 2 {e}^{2(5 - 5)}$

$\frac{dy}{dx} = 2 {e}^{0} = 2$

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

$y-y_1=m(x-x_1)$

$y - 6 = 2(x - 5)$

$y = 2x - 10 + 6$

$y = 2x -4$

Without eliminating the parameter,

$\frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} }$

$\frac{dy}{dx} = \frac{ 2t}{ \frac{1}{t} }$

$\frac{dy}{dx} = 2 {t}^{2}$

At x=5,

$5 = 5 + ln(t)$

$ln(t) = 0$

$t = {e}^{0} = 1$

This implies that,

$\frac{dy}{dx} = 2 {(1)}^{2} = 2$

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

$y-y_1=m(x-x_1)$

$y - 6 = 2(x - 5) =$

$y = 2x -4$

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