Hi! ❄
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The eccentricity of the conic section below is <u>1, </u>since the conic section given is a parabola.
The eccentricity of a parabola is <u>1</u>.
Hope that made sense !!
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![\star\tiny\pmb{_{calligraphy}}\star](https://tex.z-dn.net/?f=%5Cstar%5Ctiny%5Cpmb%7B_%7Bcalligraphy%7D%7D%5Cstar)
Answer:
![\boxed{\sf \ \ \ -2(x+4)(x-3)^3 \ \ \ }](https://tex.z-dn.net/?f=%5Cboxed%7B%5Csf%20%5C%20%5C%20%5C%20-2%28x%2B4%29%28x-3%29%5E3%20%5C%20%5C%20%5C%20%7D)
Step-by-step explanation:
Hello,
let's note k a real, we can write the polynomial as
![k(x-(-4))^1(x-3)^3=k(x+4)(x-3)^3](https://tex.z-dn.net/?f=k%28x-%28-4%29%29%5E1%28x-3%29%5E3%3Dk%28x%2B4%29%28x-3%29%5E3)
and we know that f(0)=216 so
![216=k(0+4)(0-3)^3=k*4*(-1)^3*3^3=-27*4*k=-108k\\\\ k=-\dfrac{216}{108}=-2](https://tex.z-dn.net/?f=216%3Dk%280%2B4%29%280-3%29%5E3%3Dk%2A4%2A%28-1%29%5E3%2A3%5E3%3D-27%2A4%2Ak%3D-108k%5C%5C%5C%5C%3C%3D%3E%20k%3D-%5Cdfrac%7B216%7D%7B108%7D%3D-2)
So the solution is
![-2(x+4)(x-3)^3](https://tex.z-dn.net/?f=-2%28x%2B4%29%28x-3%29%5E3)
hope this helps
3.70000000000000000000000000000
so basically for this problem you're just going to plug in the point (3,-3) into every equation, keeping in mind that the 3 is the x value and that the -3 is the y value. once you plug it in you can just use order of operations (pemdas) to solve.
the equations that do work with the point (3,-3) are (2x+4y=-6)
(2x-3y=15)
(8x+3y=15)
(7x-2y=-24)
Step-by-step explanation:
so to work one of these problems you take (2x+4y=-6) and plug in your variables.
2(3)+4(-3)=-6
then you're going to multiply
6-12=-6
the subtract the negative twelve from the 6
-6=-6
so we get a true statement. if you do this and the two numbers don't match then the equation does not work with your point. I hope this helps!