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Sindrei [870]
2 years ago
5

The perimeter of the rectangle below is 94 units. Find the length of side BC.

Mathematics
1 answer:
Ilya [14]2 years ago
7 0

Answer:

Length BC = 26

Step-by-step explanation:

Perimetre = 94

; Perimetre = 2(2x + 3) + 2(3x - 1)

; 94 = 4x + 6 + 6x - 2

; 94 - 4 = 10x

; 90 = 10x

;Therefore the value of x = 9....Hence the length of BC

BC = 3(9) - 1

; BC = 26

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What are 2 equivalent expressions that are equal to 10m-40
yuradex [85]
It's simple, there is a ton
Multiply all the numbers by 2: 20m - 80
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An acute angle, θ, is in a right triangle such that cos θ = 12/13. What is the Value of csc θ? (100 Points!)
Hatshy [7]

Answer:

csc θ = 13/12

Step-by-step explanation:

cos θ = 12/13

Formula: cos θ = 1/cscθ

Its just the reciprocal.

solve:

1/csc θ = 12/13

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Joseph has 24 cousins he has 6 more females than males how many females does he have?
Brums [2.3K]
Given that Joseph has 24 cousins, and he has 6 more females than males, the number of females will be obtained as follows;
suppose he had x males, number of females will be x+6.
The total will be:
x+(x+6)=24
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6 0
3 years ago
You are going to create a circle graph to represent some data. How many degrees should a section representing 65 out of a hundre
hjlf

To figure out the degrees, you would create a ratio.

Since there are 360 degrees in your circle, that would be representative to 100 (because both are the total of what you want to find), and x would be representative to what degrees 65 is equal to.

your ratio should look like,

65/100=x/360

cross multiply and get

65(360)=100x

23400=100x

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It looks hard to read like this, but if you write it out on your paper it should make more sense!

8 0
3 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
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