What’s the solution set for x? Can someone please answer and explain how they got they’re answer for 40 points.

Mathematics

1 answer:

xz_007 [3.2K]2 years ago

3 0

Answer:

i think the answer is 2.I had a problem like this and i think u have to find out what both of them equal

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Find the first, fourth, and eight terms of the sequence: A(n)=-5•3n-1

lys-0071 [83]

A (1)= -5×3(1)-1

a(1)= -5×3-1

a (1)= -16

A(4)= -5×3(4)-1

a (4)= -5×12-1

a( 4) = -60-1

a(4)= -61

A(8)= -5×3(8)-1

a(8)= -5×24-1

a(8)= -120-1

a(8)= -121

Please vote my answer branliest! Thanks.

6 0

2 years ago

Write the equation of the circle whose endpoints of a diameter are (2,-5) and (8,-1)

Goryan [66]

if the endpoints are there, that means that segment with those endpoints is the diameter of the circle, and that also means that the midpoint of that diameter is the center of the circle.

it also means that the distance from the midpoint to either endpoint, is the radius of the circle.

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{2}~,~\stackrel{y_1}{-5})\qquad (\stackrel{x_2}{8}~,~\stackrel{y_2}{-1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{8+2}{2}~~,~~\cfrac{-1-5}{2} \right)\implies (5,-3)\impliedby center \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \stackrel{midpoint}{(\stackrel{x_1}{5}~,~\stackrel{y_1}{-3})}\qquad \stackrel{endpoint}{(\stackrel{x_2}{8}~,~\stackrel{y_2}{-1})}\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{radius}{r}=\sqrt{[8-5]^2+[-1-(-3)]^2}\implies r=\sqrt{(8-5)^2+(-1+3)^2} \\\\\\ r=\sqrt{3^2+2^2}\implies r=\sqrt{13} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{5}{ h},\stackrel{-3}{ k})\qquad \qquad radius=\stackrel{\sqrt{13}}{ r} \\[2em] [x-5]^2+[y-(-3)]^2=(\sqrt{13})^2\implies \boxed{(x-5)^2+(y+3)^2=13}$