I’m super confused on 1/3 times 60

Talja [164]

Answer:

B all square are regular polygons

8 0

3 years ago

W trójkącie ABC połączono wierzchołek C z punktem D który jest środkiem boku AB . Wyznaczono równierz środek odcinka CD punkt E

qwelly [4]

Translate into English please

8 0

3 years ago

For each equation, determine whether it represents a direct variation, an inverse variation, or neither. Find the constant of va

rewona [7]

Both inverse variation’s. K=constant=9,6

6 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

jen uses 8.2 of blue paint and white paint to paint her bedroom walls. 4/5ths of this amount is blue paint, and the rest is whit

lara31 [8.8K]

Answer:

The amount of white paint Jen used to paint the walls in her room is:

1.64 pints.

Step-by-step explanation:

To solve the exercise you only have to pay attention to the statement, in the section that says that 4/5 parts of the total painting is blue, therefore, if 1 is the total painting, you must do a subtraction:

1 - 4/5 = 1/5

Since the remaining white paint is 1/5 of the total, you have two ways to solve the exercise: multiply the total paint (8.2 pints) by 1/5 or divide the number by 5, as shown below:

Amount of white paint = 8.2 pints * (1/5) = 1.64 pints.
Amount of white paint = 8.2 pints / 5 = 1.64 pints.

As you can see, the two methods provide a value of 1.64 pints, which corresponds to 1/5 of the total paint and is the amount of white paint used.

5 0

3 years ago