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murzikaleks [220]
3 years ago
12

HELP PLEASE I NEED IT ASAP I WILL MARK BRAINLIEST FOR CORRECT ANSWER

Mathematics
2 answers:
Liono4ka [1.6K]3 years ago
7 0

Answer:

wait whats "a" so confused

Step-by-step explanation:

Ymorist [56]3 years ago
5 0
6 that’s the answer
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200×39.37=7874
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3 years ago
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Find the radius of the circle with equation x²+y²+8x+8y+28=0
galina1969 [7]
<h2>Hello!</h2>

The answer is:

Center: (-4,-4)

Radius: 2 units.

<h2>Why?</h2>

To solve the problem, using the given formula of a circle, we need to find its standard equation form which is equal to:

(x-h)^{2}+(y-k)^{2}=r^{2}

Where:

"h" and "k" are the coordinates of the center of the circle and "r" is its radius.

So, we need to complete the square for both variable "x" and "y".

The given equation is:

x^{2}+y^{2}+8x+8y+28=0

So, solving we have:

x^{2}+y^{2}+8x+8y=-28

(x^2+8x+(\frac{8}{2})^{2})+(y^2+8y+(\frac{8}{2})^{2})=-28+((\frac{8}{2})^{2})++(\frac{8}{2})^{2})\\\\(x^2+8x+16 )+(y^2+8y+16)=-28+16+16\\\\(x^2+4)+(y^2+4)=4

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

Now, we have that:

h=-4\\k=-4\\r=\sqrt{4}=2

So,

Center: (-4,-4)

Radius: 2 units.

Have a nice day!

Note: I have attached a picture for better understanding.

3 0
3 years ago
Let f(x)=x+a/x+b such that f(f(1)=0 &amp; f(2)=-3 then a&amp;b resctivily are.
Yuliya22 [10]

Answer:

The value of a = - 1  ,   And b = \frac{ - 7 }{3}    

Step-by-step explanation:

Given as :

Function f(x) = \frac{(x + a)}{(x + b)}

And f(f(1)) = 0     And f(2) = - 3

Now For , x = 2 , y = - 3

I.e  f(2) = \frac{(2 + a)}{(2 + b)}

or,  - 3 = \frac{(2 + a)}{(2 + b)}

I.e 2 + a = - 6 - 3b

Or, a + 3b = - 8               ....... 1

Again  f(f(1)) = 0

So,  \frac{(1 + a)}{(1 + b)} = 0

Or,  1 + a = 0

∴    a = - 1

So , put htis value of a i n eq 1 , we get value of  b

So , - 1 + 3b = - 8

Or,   3b = - 7

∴       b = \frac{ - 7 }{3}

Hence The value of a = - 1  ,   And b = \frac{ - 7 }{3}    Answer

3 0
3 years ago
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Brilliant_brown [7]

Step-by-step explanation:

h(x)=-x-4

but h(x)=4

4=-x-4

x=-4-4

x=-8

5 0
3 years ago
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