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Doss [256]
3 years ago
8

Whats the radius for the circle? x^2+2x+y^2+4y-6=0

Mathematics
1 answer:
MAXImum [283]3 years ago
5 0

Answer:

r = \sqrt{11}

Step-by-step explanation:

So we need to complete the square for both parts of the equation

First though we can add the 6 to the other side so we have x² + 2x + y² + 4y = 6

So first we can complete the square for x² + 2x

To do so we need to use (\frac{b}{2} )^{2} to figure out the number we need to add to both sides

In this case our b is 2, so substituting this in we get (\frac{2}{2} )^{2} =(1)^{2} =1

Here we add 1 to both sides and now we have x² + 2x + 1 + y² + 4y = 6 + 1

Now we can follow the same steps to complete the square for y² + 4y

Here our b is 4, so substituting this in we get (\frac{4}{2} )^{2}=(2)^{2} =4

Now we add 4 to both sides and now we have x² + 2x + 1 + y² + 4y + 4 = 6 + 1 + 4

Now condensing everything we have (x + 1)² + (y + 2)² = 11

The formula for a circle is (x - h)² + (y - k)² = r²

In our equation we have r² = 11

To find the radius we need to take the square root of both sides \sqrt{r^{2}}  =\sqrt{11} to get r = \sqrt{11}

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43 and 56/12

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8 0
3 years ago
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Find the value of the discriminant for <img src="https://tex.z-dn.net/?f=7x%5E%7B2%7D%20%2B5x%2B1%3D0" id="TexFormula1" title="7
kondor19780726 [428]

Answer:

No real roots

Step-by-step explanation:

Given

7x² + 5x + 1 = 0 ← in standard form

with a = 7, b = 5, c = 1

To determine the nature of the roots use the discriminant

Δ = b² - 4ac

• If b² - 4ac > 0 then roots are real and distinct

• If b² - 4ac = 0 then roots are real and equal

• If b² - 4ac < 0 then the roots are not real

Here

b² - 4ac = 5² - (4 × 7 × 1) = 25 - 28 = - 3

Thus the 2 roots are not real

7 0
3 years ago
QUESTION 4
ANEK [815]

(2x-1)(x+4)=0

Step-by-step explanation:

A random zero property of multiplication is taken to find the solution

(2x-1)(x+4)=0

consider a=2x-1 and b=x-4                                      a.b=0

                                                                           either a or b or both must be 0

equating both the equations

2x-1=0 or x=4=0        x-4=0

2x-1=0                         x=4  

2x=1

x=1/2

substitute the values of x in the main equation

[2(1/2)-1][(1/2)+4]=0

3 0
3 years ago
Read 2 more answers
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