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2 years ago

Find the center and radius of the circle represented by the equation below.

Mathematics

1 answer:

pantera1 [17]2 years ago

8 0

Answer:

Center= (-6,-8)

Radius= 6

Step-by-step explanation:

We use the formula (x-h)^2+(y-k)^2 then plug in our numbers:

$(x - h)^2 + (y - k)^2 = r^2\\(x+6)^2 +(y-k)^2=36$

In the equation -h is 6 so our h (x-coordinate) is -6

In the equation -k is 8 so our k (y-coordinate) is -8

In the equation r^2 is 36 so we take the square root of 36 to get r which is 6.

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In the diagram, the measures of 21, 24, and 25 are 120". The measure of

topjm [15]

Answer:

C) Yes, because <1 and <5 are congruent.

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2 years ago

3x - 2y = 10

natka813 [3]

3x - 2y = 10
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3 years ago

Item 5 Identify the terms, coefficients, and constants in the expression 6 + n x n + 1/2d

Alenkasestr [34]

Answer:

Terms (Variables) = x , d . Their corresponding coefficients = n^2 , 1/2 . Constant = 6

Step-by-step explanation:

6 + n x n + 1/2d

6 + n^2 x + 1/2d

Terms (Variables) = x , d . Their corresponding coefficients = n^2 , 1/2 . Constant = 6

7 0

3 years ago

Help find length of ac n explain please thanks

Debora [2.8K]

Answer:

AC = 12.5 cm

Step-by-step explanation:

use cosine law c^2 = a^2 + b^2 − 2ab cos(C)

c^2 = 6^2 + 10^2 - 2(6)(10)(cosin100)

c^2 = 156.84

c= √156.84

c= 12.5 cm

The cosine rule is used when we are given either three sides or two sides and the included angle.

6 0

2 years ago

Read 2 more answers

A^-1/[a^-1 - b^-1] + a^-1/[a^-1 + b^-1]

o-na [289]

$\dfrac{a^{-1}}{a^{-1}-b^{-1}}+\dfrac{a^{-1}}{a^{-1}+b^{-1}}=a^{-1}\left(\dfrac{a^{-1}+b^{-1}}{(a^{-1}-b^{-1})(a^{-1}+b^{-1})}+\dfrac{a^{-1}-b^{-1}}{(a^{-1}+b^{-1})(a^{-1}-b^{-1})}\right)$

$=a^{-1}\left(\dfrac{(a^{-1}+b^{-1})+(a^{-1}-b^{-1})}{a^{-2}-b^{-2}}\right)$

$=\dfrac{2a^{-2}}{a^{-2}-b^{-2}}$

$=\dfrac{2b^2}{b^2-a^2}$

5 0

3 years ago

Read 2 more answers