(please actually answer this. i already gotten 3 answers that were unhelpful and i really need this answer.) Using the following equation, find the center and radius of the circle. You must show all work and calculations to receive credit. x2 − 4x + y2 + 8y = −4 (10 points)

option a is correct.

step-by-step explanation:

scale factor: it is used to find proportional measurements.

proportions means having the same ratio.

scale factor is the ratio of the model measurement to the actual measurement in a simplest form.

or we can say that the ratio of any two corresponding lengths in two similar geometric figures

if the scale factor is more than 1, then it is an enlargement,

if the scale factor less than 1, then it is a reduction.

given the two geometric figure:

by definition of proportions, we have

proofs involving isosceles triangles often require special consideration because an isosceles triangle has several distinct properties that do not apply to normal triangles.(more about triangle types) therefore, when you are trying to prove that two triangles are congruent, and one or both triangles, are isosceles you have a few theorems that you can use to make your life easier.

step-by-step explanation:

an isosceles triangle has two congruent sides and two congruent angles. the congruent angles are called the base angles and the other angle is known as the vertex angle. ∠bac and ∠bca are the base angles of the triangle picture on the left. the vertex angle is ∠abc

answer: solve 5\cdot 2^x=2405⋅2  

x

=2405, dot, 2, start superscript, x, end superscript, equals, 240.

solution

to solve for xxx, we must first isolate the exponential part. to do this, divide both sides by 555 as shown below. we do not multiply the 555 and the 222 as this goes against the order of operations!

\begin{aligned} 5\cdot 2^x& =240 2^x& =48 \\ \end{aligned}  

5⋅2  

x

2  

x

​    

=240

=48

​

step-by-step explanation:

тgє qυєѕтισи αѕкє∂: ιи тнє ѕєяιєѕ σя яєα¢тισиѕ тgαт σ¢¢υя ιи ρнσтσѕуитнєѕιѕ, ωнι¢н fσямѕ ωнєи ℓιgнт ѕтяιкєѕ α мσℓє¢υℓє σf ¢нℓσяσρнуℓℓ.

α. мσℓє¢υℓє σf ωαтєя

в.тнєє-¢αявσи ѕυgαя

¢. "єχ¢ιтє∂" єℓє¢тяσи

∂. иα∂ρн

му αиѕωєя: ιт ιѕ ¢. "єχ¢ιтє∂" єℓє¢тяσи

нσρє ι ¢συℓ∂ нєℓρ υ