What is the answer of this: 52.85-9.09 rounded to the nearest one?

A circle is translated 4 units to the right and then reflected over the x-axis. Complete the statement so that it will always be

irga5000 [103]

Answer:

The statement is now presented as:

$\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}$

In other words, this mathematical statement can be translated as:

There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r.

Step-by-step explanation:

Let $C = (h,k)$ the coordinates of the center of the circle, which must be transformed into $C'=(h', k')$ by operations of translation and reflection. From Analytical Geometry we understand that circles are represented by the following equation:

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

Where $r$ is the radius of the circle, which remains unchanged in every operation.

Now we proceed to describe the series of operations:

1) Center of the circle is translated 4 units to the right (+x direction):

$C''(x,y) = C(x, y) + U(x,y)$ (Eq. 1)

Where $U(x,y)$ is the translation vector, dimensionless.

If we know that $C(x, y) = (h,k)$ and $U(x,y) = (4, 0)$ , then:

$C''(x,y) = (h,k)+(4,0)$

$C''(x,y) =(h+4,k)$

2) Reflection over the x-axis:

$C'(x,y) = O(x,y) - [C''(x,y)-O(x,y)]$ (Eq. 2)

Where $O(x,y)$ is the reflection point, dimensionless.

If we know that $O(x,y) = (h+4,0)$ and $C''(x,y) =(h+4,k)$ , the new point is:

$C'(x,y) = (h+4,0)-[(h+4,k)-(h+4,0)]$

$C'(x,y) = (h+4, 0)-(0,k)$

$C'(x,y) = (h+4, -k)$

And thus, $h' = h+4$ and $k' = -k$ . The statement is now presented as:

$\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}$

In other words, this mathematical statement can be translated as:

There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r.

4 0

3 years ago

Prove the following statements: (a) For every integer x, if x is even, then for every integer y, xy is even. (b) For every integ

Murrr4er [49]

Answer and Step-by-step explanation:

(a) Given that x and y is even, we want to prove that xy is also even.

For x and y to be even, x and y have to be a multiple of 2. Let x = 2k and y = 2p where k and p are real numbers. xy = 2k x 2p = 4kp = 2(2kp). The product is a multiple of 2, this means the number is also even. Hence xy is even when x and y are even.

(b) in reality, if an odd number multiplies and odd number, the result is also an odd number. Therefore, the question is wrong. I assume they wanted to ask for the proof that the product is also odd. If that's the case, then this is the proof:

Given that x and y are odd, we want to prove that xy is odd. For x and y to be odd, they have to be multiples of 2 with 1 added to it. Therefore, suppose x = 2k + 1 and y = 2p + 1 then xy = (2k + 1)(2p + 1) = 4kp + 2k + 2p + 1 = 2(kp + k + p) + 1. Let kp + k + p = q, then we have 2q + 1 which is also odd.

(c) Given that x is odd we want to prove that 3x is also odd. Firstly, we've proven above that xy is odd if x and y are odd. 3 is an odd number and we are told that x is odd. Therefore it follows from the second proof that 3x is also odd.

3 0

3 years ago

I will give BRAINLIEST to whoever answers correctly first!!!

Tatiana [17]

$3.18

Explanation: I am a cashier hahaha

7 0

2 years ago

Read 2 more answers

Round to the nearest tenth. Asanji took a trip to Mexico. Upon leaving he decided to convert all of his Pesos back into dollars.

My name is Ann [436]

He would have around 15 dollars

4 0