Find the perimeter. Simplify your answer.

Which of the following expressions is equal to 1 divided by 16?

MrRissso [65]

$(\frac{1}{2})^{4}=\frac{1}{16}$

4 0

3 years ago

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Solve the following eqation: 4m - 12 = 0

OLEGan [10]

Answer:

m=3

Step-by-step explanation:

Move the 12 over to the other side so it will be 4m=12, then divide each side by 4 to get m=3

8 0

2 years ago

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The manager of a department store uses the given function to model the monthly opening inventory of down comforters at the store

tangare [24]

Answer: $d(x)=-750x^2-3750x-10500$

Step-by-step explanation:

Given: The manager of a department store uses the given function to model the monthly opening inventory of down comforters at the store, in dollars, where x represents the number of months from September.

$d(x)=-750(x^2-5x-14)$

To rewrite this hat reveals the store's opening inventory of down comforters in the month of September, multiply -750 on the bracket, we get

$d(x)=-750x^2-3750x-10500$

6 0

2 years ago

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What’s the quadrant

Vadim26 [7]

Hey there!!

Always remember this :

90-degree counterclockwise: (x, y) → (-y, x)

180-degree counterclockwise: (x, y) → (-x, -y)

270-degree counterclockwise: (x, y) → (y, -x)

1st quadrant : ( x , y )

second quadrant : ( -x , y )

Third quadrant : ( - x , - y )

Fourth quadrant : ( x , -y )

Now, the triangle abc lies in the 1st quadrant. So, all the vertexes will have positive values.

Let's take the vertex a as ( 1 , 2 )

Vertex b as ( 4 , 7 )

Vertex c as ( 5 , 9 )

If we rotate these values in 90 degress angle, these would become ( -y , x )

Vertex a : ( 1 , 2 ) ; After rotating : ( -2 , 1 )

Vertex b : ( 4 , 7 ) ; After rotating : ( -7 , 4 )

Vertex c : ( 5 , 9 ) ; After rotating : ( -9 , 5 )

Hope my answer helps!

3 0

2 years ago

A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an a

Scrat [10]

Answer:

The answer is (C) 8

Step-by-step explanation:

First, let's calculate the length of the side of the square.

$A_{square}=a^2$ , where $a$ is the length of the side. Now, let's try to build the square. First we need to find a point which distance from (0, 0) is 10. For this, we can use the distance formula in the plane:

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ which for $x_1=0$ and $y_1 = 0$ transforms as $d=\sqrt{(x_2)^2 + (y_2)^2}$ . The first point we are looking for is connected to the origin and therefore, its components will form a right triangle in which, the Pythagoras theorem holds, see the first attached figure. Then, $x_2$ , $y_2$ and 10 are a Pythagorean triple. From this, $x_2= 6$ or $x_2=8$ while $y_2= 6$ or $y_2=8$ . This leads us with the set of coordinates:

$(\pm 6, \pm 8)$ and $(\pm 8, \pm 6)$ . (A)

The next step is to find the coordinates of points that lie on lines which are perpendicular to the lines that joins the origin of the coordinate system with the set of points given in (A):

Let's do this for the point (6, 8).

The equation of the line that join the point (6, 8) with the origin (0, 0) has the equation $y = mx +n$ , however, we only need to find its slope in order to find a perpendicular line to it. Thus,

$m = \frac{y_2-y_1}{x_2-x_1} \\m = \frac{8-0}{6-0} \\m = 8/6$

Then, a perpendicular line has an slope $m_{\bot} = -\frac{1}{m} = -\frac{6}{8}$ (perpendicularity condition of two lines). With the equation of the slope of the perpendicular line and the given point (6, 8), together with the equation of the distance we can form a system of equations to find the coordinates of two points that lie on this perpendicular line.

$m_{\bot}=\frac{6}{8} = \frac{8-y}{6-x}\\ 6(6-x)+8(8-y)=0$ (1)

$d^2 = \sqrt{(y_o-y)^2+(x_o-x)^2} \\(10)^2=\sqrt{(8-y)^2+(6-x)^2}\\100 = \sqrt{(8-y)^2+(6-x)^2}$ (2)

This system has solutions in the coordinates (-2, 14) and (14, 2). Until here, we have three vertices of the square. Let's now find the fourth one in the same way we found the third one using the point (14,2). A line perpendicular to the line that joins the point (6, 8) and (14, 2) has an slope $m = 8/6$ based on the perpendicularity condition. Thus, we can form the system:

$\frac{8}{6} =\frac{2-y}{14-x} \\8(14-x) - 6(2-y) = 0$ (1)

$100 = \sqrt{(14-x)^2+(2-y)^2}$ (2)

with solution the coordinates (8, -6) and (20, 10). If you draw a line joining the coordinates (0, 0), (6, 8), (14, 2) and (8, -6) you will get one of the squares that fulfill the conditions of the problem. By repeating this process with the coordinates in (A), the following squares are found:

(0, 0), (6, 8), (14, 2), (8, -6)

(0, 0), (8, 6), (14, -2), (6, -8)
(0, 0), (-6, 8), (-14, 2), (-8, -6)
(0, 0), (-8, 6), (-14, -2), (-6, -8)

Now, notice that the equation of distance between the two points separated a distance of 10 has the trivial solution $(\pm10, 0)$ and $(0, \pm10)$ . By combining this points we get the following squares:

(0, 0), (10, 0), (10, 10), (0, 10)
(0, 0), (0, 10), (-10, 10), (-10, 0)
(0, 0), (-10, 0), (-10, -10), (0, -10)
(0, 0), (0, -10), (-10, -10), (10, 0)

See the attached second attached figure. Therefore, 8 squares can be drawn

8 0

2 years ago