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3 years ago

PLS USE the information given! DONT GUESS!

Mathematics

1 answer:

snow_tiger [21]3 years ago

4 0

Answer:

C. (x + 3)² + (y - 6)² = 4

Step-by-step explanation:

Get the center and radius from the graph.

Center is: (-3, 6)
Radius is: 2

Use the circle formula:

(x - h)² + (y - k)² = r²

We have:

h = -3, k = 6, r = 2

Substitute and get the equation:

(x - (-3))² + (y - 6)² = 2² ⇒
(x + 3)² + (y - 6)² = 4

Correct choice is C

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2 years ago

if a and b are the two odd positive integers such that a>b then one of the two numbers ( a+b) /2 and (a-b) /2is odd and even

Tamiku [17]

If $a$ and $b$ are odd positive integers, they are one more than a non-negative even number, i.e. there exists $m,n \in \mathbb{N}$ such that

$a = 2m+1,\quad b = 2n+1$

So, the first expression become

$\cfrac{a+b}{2} = \cfrac{2m+1+2n+1}{2} = \cfrac{2m+2n+2}{2} = m+n+1$

Similarly, we have

$\cfrac{a-b}{2} = \cfrac{2m+1-2n-1}{2} = \cfrac{2m-2n}{2} = m-n$

Now, the parity of these expressions depend on those of $m$ and $n$ . We have four cases:

If both m and n are even:

$m+n+1$ is odd, since $m+n$ is even, while $m-n$ is even

If one of the two is odd and the other is even:

$m+n+1$ is even, since $m+n$ is odd, while $m-n$ is odd

If both are odd:

$m+n+1$ is odd, since $m+n$ is even, while $m-n$ is even

So, in all cases, one between (a+b)/2 and (a-b)/2 is odd, and the other is even.

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3 years ago

What equation do you get when you solve z - m = z + bx for x?

Arisa [49]

the answer is z-m=z+bx so z-m-z=bx and x=-mb

So it's the first option

the solution is x= - mb

6 0

4 years ago

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4 years ago