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2 years ago

Find an equation of the circle that has center (-1,2) and passes through (5,-4).

1 answer:

8 0

General equation for a circle
(x - a)² + (y - b)² = r²
with (a,b) represents the center, (x,y) represents one of the points lie on the circle, and r represents the radius

Determine r² by substituting the points into the general equation
(x - a)² + (y - b)² = r²
(5 - (-1))² + (-4 - 2)² = r²
(5 + 1)² + (-6)² = r²
6² + 36 = r²
36 + 36 = r²
72 = r²

Determine the equation of the circle
(x - a)² + (y - b)² = r²
(x - (-1))² + (y - 2)² = 72
(x + 1)² + (y - 2)² = 72 (This is the equation of the circle)

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Can u help me. if answer is correct, i will give u brainliest

den301095 [7]

Answer:

135 units²

Step-by-step explanation:

The area (A) of a parallelogram is calculated as

A = bh ( b is the base and h the perpendicular height )

To calculate h use Pythagoras' identity on the right triangle on the left

h² + 8² = 17²

h² + 64 = 289 ( subtract 64 from both sides )

h² = 225 ( take the square root of both sides )

h = $\sqrt{225}$ = 15 , thus

A = 9 × 15 = 135 units²

7 0

2 years ago

I’m not sure what to do, this was what I got. Someone please help and solve problem #14 in the picture. Thanks!

mars1129 [50]

Answer:

x = 8
AM = BM = 16

Step-by-step explanation:

Your problem setup is correct. The triangle is isosceles, so the marked segments are the same length.

3x - 8 = 2x

You can solve this by subtracting 2x (from both sides) ...

x - 8 = 0

Then adding 8 (to both sides):

x = 8

MB = MA = 2x = 16.

4 0

2 years ago

The quadratic function g(x) = a.ca + bx+c has the

Mumz [18]

<em>The value of b is 14 and the value of c is 65</em>

<h2>Explanation:</h2>

The quadratic function is a function of the form:

$f(x)=ax^2+bx+c$

Here we know that the leading coefficient $a=1$ so we reduce our equation to:

$g(x)=x^2+bx+c$

The roots are those values at which $g(x)=0$

So:

$x^2+bx+c=0 \\ \\ First \ root: \\ \\ (-7+4i)^2+b(-7+4i)+c=0 \\ \\ (-7)^2-2(7)(4i)+(4i)^2-7b+4bi+c=0 \\ \\ 49-56i+16i^2-7b+4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49-56i+16(-1)-7b+4bi+c=0 \\ \\ 49-56i-16-7b+4bi+c=0 \\ \\ 33-56i-7b+4bi+c=0 \\ \\ \\$

$Second \ root: \\ \\ (-7-4i)^2+b(-7-4i)+c=0 \\ \\ (-1)^2(7+4i)^2+b(-7-4i)+c=0 \\ \\ (7)^2+2(7)(4i)+(4i)^2-7b-4bi+c=0 \\ \\ 49+56i+16i^2-7b-4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49+56i+16(-1)-7b-4bi+c=0 \\ \\ 49+56i-16-7b-4bi+c=0 \\ \\ 33+56i-7b-4bi+c=0$

So we have:

$(1) \ 33-56i-7b+4bi+c=0 \\ \\ (2) \ 33+56i-7b-4bi+c=0 \\ \\ \\ Subtract \ 2 \ from: \\ \\ 33-56i-7b+4bi+c-(33+56i-7b-4bi+c)=0 \\ \\ 33-56i-7b+4bi+c-33-56i+7b+4bi-c=0 \\ \\ \\ Combine \ like \ terms: \\ \\ 33-33-56i-56i-7b+7b+4bi+4bi+c-c=0 \\ \\ -112i+8bi=0 \\ \\ Isolating \ b: \\ \\ b=\frac{112i}{8i} \\ \\ \boxed{b=14}$