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preciadogabriel40

preciadogabriel40

21.12.2019 •

Mathematics

What is the center of the circle given by the equation x^2+y^2-14y-15=0

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Ответ:

haleylayne74

13.01.2022

(0, 7)

Step-by-step explanation:

The equation of a circle in the standard form:

(a-h)^2+(y-k)^2=r^2

(h, k) - center

r - radius

We have the equation:

x^2+y^2-12y-15=0

Convert into a standard form using

$(a-b)^2=a^2-2ab+b^2\qquad(*)$

$x^2+\underbrace{y^2-2(y)(7)+7^2}_{(*)}-7^2-15=0\\\\(x-0)^2+(y-7)^2-49-15=0$

(x-0)^2+(y-7)^2-64=0 add 64 to both sides

(x-0)^2+(y-7)^2=64

The center (0, 7)

The radius: r = √64 = 8

0,0(0 оценок)

Ответ:

timithythaxton

16.10.2020

$A = (\frac{80}{3},9)$

B = (5,2.5)

The slope is 0.3

Step-by-step explanation:

Given

y = 0.3x + 1

A = (-,9)

B = (5,-)

Required:

Fill in the gap and calculate the slope

In A = (-,9)

y = 9 and x is ??

Substitute 9 for y in y = 0.3x + 1

9 = 0.3x + 1

Collect Like Terms

0.3x = 9 - 1

0.3x = 8

$x = \frac{8}{0.3}$

$x = \frac{80}{3}$

So:

A = (-,9)

$A = (\frac{80}{3},9)$

In B = (5,-)

x = 5 and y is ??

Substitute 5 for x in y = 0.3x + 1

y = 0.3 * 5 + 1

y = 1.5 + 1

y = 2.5

So:

B = (5,-)

B = (5,2.5)

Calculating the slope (m)

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{2.5-9}{5 - \frac{80}{3}}$

$m = \frac{-6.5}{\frac{15- 80}{3}}$

$m = \frac{-6.5}{\frac{-65}{3}}$

$m = \frac{-6.5*3}{-65}$

$m = \frac{-19.5}{-65}$

m = 0.3

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