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chubhunter [2.5K]
3 years ago
8

Do these 3 and show your work

Mathematics
1 answer:
Alinara [238K]3 years ago
6 0

Domain:\ x\neq-6\\\\\dfrac{5}{x+6}=10\to\dfrac{5}{x+6}=\dfrac{10}{1}\ \ \ \ |\text{cross multiply}\\\\10(x+6)=(5)(1)\ \ \ \ \ |\text{use distributive property}\\\\(10)(x)+(10)(6)=5\\\\10x+60=5\ \ \ \ |-60\\\\10x=-55\ \ \ \ |:10\\\\\boxed{x=-5.5}\in D

Domain:\\\dfrac{5}{x+6}\geq0\ \wedge\ x+6\neq0\to x+6 > 0\to x > -6\\\\\sqrt{\dfrac{5}{x+6}}=10\ \ \ \ \ |^2\\\\\left(\sqrt{\dfrac{5}{x+6}}\right)^2=10^2\\\\\dfrac{5}{x+6}=100\\\\\dfrac{5}{x+6}=\dfrac{100}{1}\ \ \ \ |\text{cross multiply}\\\\100(x+6)=5\ \ \ \ \ |:100\\\\x+6=0.05\ \ \ \ \ |-6\\\\\boxed{x=-5.95}\in D

Domain:x+3\geq0\to x\geq-3\\\\\sqrt{2x+3-x}=2\to\sqrt{x+3}=2\ \ \ \ |^2\\\\(\sqrt{x+3})^2=2^2\\\\x+3=4\ \ \ \ |-3\\\\\boxed{x=1}\in D

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Answer:

-\frac{1}{2}y^2-4y+1=x

Step-by-step explanation:

A parabola has a focus of F(8.5,−4) and a directrix of x=9.5.

General form of  horizontal parabola is

(y-k)^2=4p(x-h)

the distance between directrix and focus is the value of p

so p = 8.5 - 9.5 = -0.5

Focus is (h+p , k), given focus is (8.5, -4)

So k = -4 and h+p = 8.5

we know p = -0.5

h +p = 8.5

h - 0.5 = 8.5

so h= 9                   and k = -4

vertex is (h,k) that is (9, -4)

Now plug in the value in the general equation

(y-k)^2=4p(x-h), k= -4, h= 9 , p = -0.5

(y+4)^2=4(-0.5)(x-9)

(y+4)^2=-2(x-9)

y^2+8y+16=-2x+18

subtract 18 on both sides

y^2+8y-2=-2x

Divide whole equation by -2

-\frac{1}{2}y^2-4y+1=x




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