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vlabodo [156]
3 years ago
15

Determine the standard form of the equation for the circle with center (h,k) = (-5 ½) and radius r =1

Mathematics
1 answer:
wariber [46]3 years ago
8 0
Asuuing yo meant (h,k)=(-5,1/2)

equation is (x-h)^2+(y-k)^2=r^2
(h,k) is center and r=radious
given that (h,k)=(-5,1/2) and r=1
h=-5
k=1/2

(x-(-5))^2+(y- \frac{1}{2} )^2=1^2
(x+5)^2+(y- \frac{1}{2} )^2=1
if you wanted us to expand
x²+y^2+10x-y+25.25=1
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Will give brainliest answer
makvit [3.9K]

Answer:

28 . 26cm

Step-by-step explanation:

First find the radius

d = 2r \\ 9 = 2r \\  \frac{9}{2}  =  \frac{2r}{2}  \\ r = 4.5

Now find the CIRCUMFERENCE

c = 2\pi \: r \\  = 2 \times \pi \times 4.5 \\  = 9 \times \pi \\  = 28.26

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3 years ago
Jennie can make 7 quilts with 21 yards of material how many yards of material would be required to make 12 quilts ?
larisa [96]
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4 0
3 years ago
Does the following equation determine y to be a function of x? y square=x+3
Phantasy [73]

Answer:

∴ y² = x + 3 is not a function

Step-by-step explanation:

* Lets explain how to solve the problem

- The definition of the function is every input (x) has only one

  output (y)

- Ex:

# y = x + 1 where x ∈ R , is a function because every x has only

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# y² = x where x ∈ R , is not a function because y = ±√x, then one

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* Lets solve the problem

∵ y² = x + 3

- Find y by taking √ for both sides

∴ y = ± √(x + 3)

- That means y = √(x + 3)  and y = - √(x + 3)

∵ (x + 3) must be greater than or equal zero because there is no

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∴ x + 3 ≥ 0 ⇒ subtract 3 from both sides

∴ x ≥ -3

∴ x must be any number greater than or equal -3

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∴ y = √(0 + 3) = √3 and y = - √(0 + 3) = -√3

∴ x = 0 has two values of y ⇒ y = √3 and y = -√3

- Any value of x greater than or equal 3 will have two values of y

∴ y² = x + 3 is not a function

5 0
3 years ago
What is 300/2000 in the form of a fraction, decimal, and percent?
alex41 [277]
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7 0
3 years ago
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Answer:

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