4,523 ÷ 3 help me please

allochka39001 [22]

5x²y + 2xy² + x²y

Combining like terms would be
6x²y + 2xy²

The two terms are now unique and cannot be combined any further.

6 0

3 years ago

(5,9), (5-3) what is the slope?

monitta

Answer:

Hello!!! Princess Sakura here ^^

Step-by-step explanation:

When you want to find out what the slope is, you have to remember what the formula is first....

$m=\frac{y}{x} =\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

so in this case -3 would be the $y_{2}$ because it's the y in the ordered pair and it's the second one. Then one of the 5's would be $x_{2}$ and 9 would be $y_{1}$ and then the other 5 would be $x_{1}$ .

now if we plug it in

$\frac{-3-9}{5-5} =\frac{-12}{0}$

now we can tell that we can't divide by zero so the slope would be unidentified and that's it.

5 0

3 years ago

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

What is 7^5 equal to

Alona [7]

The answer to 7^5 = 16807

5 0

3 years ago

Manuel rented a truck for one day. There was a base fee of $15.99, and there was an additional charge of 92 cents for each mile

inysia [295]

Answer:

8

Step-by-step explanation:

5 0

2 years ago