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3 years ago

X - Зу = -2 X - Зу = 16

Mathematics

1 answer:

Stells [14]3 years ago

8 0

Answer: x - 3y = -2
x = -2 + 3y

x - 3y = 16
x = 16 + 3y

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I'm not sure if this is correct or not. I just learned how to plot these equations. Please tell me if i'm right or wrong and exp

solong [7]

Answer:

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8 0

3 years ago

The perimeter of a triangle is not greater than 12 and the lengths of sides are natural numbers.

weqwewe [10]

Answer:

3 triangles

Step-by-step explanation:

Perimeter of triangle = a + b + c

Given that :

P = 12

and a, b, c are natural numbers

Let :

Side A = a

Side B = b

Side C = 12 - (a + b)

Side A + side B > side C - - - (condition 1)

a + b > 12 - (a + b)

a + b > 12 - a - b

a + a + b + b > 12

2a + 2b > 12

2(a + b) > 12

a + b > 6

Side A - side B < side C

a - b < 12 - (a + b)

a - b + a + b < 12

2a < 12

a < 6

b < 6 (arbitrary point)

Going by the Constraint above :

The only three possibilities are :

(2, 5, 5)

(3, 4, 5)

(4, 4, 4)

Total number of triangle = 3

Equilateral triangle (all 3 sides equal) = (4, 4, 4) = 1

Isosceles triangle (only 2 sides equal) = (2, 5, 5) = 1

3 0

2 years ago

A square painting painting is surrounded by a frame. the outside edges of the frame are 3 inches in length, and the frame is 3 i

sdas [7]

Answer:

Area = 36 $inches ^ 2$

Step-by-step explanation:

There must be an error with the statement, because it is strange that the length of the frame is equal to its thickness.

Assuming that the data supplied is correct, we can do the following procedure.

Let's call L the length of the frame.

We know how much each side of the square measures.

L = 3 inches

We also know that the thickness (s) is:

s = 3 inches.

Then the area of the frame is equal to:

Area = Perimeter * thickness

Area = 4L*s

Area = 3*4*3

Area = 36 $inches ^ 2$

3 0

2 years ago

What is the equation of a line that has a slope of -1/2 and contains the point (8,1)?

Troyanec [42]

$(\stackrel{x_1}{8}~,~\stackrel{y_1}{1})\hspace{10em} \stackrel{slope}{m} ~=~ -\cfrac{1}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1}=\stackrel{m}{-\cfrac{1}{2}}(x-\stackrel{x_1}{8}) \\\\\\ y-1=-\cfrac{1}{2}x+4\implies y=-\cfrac{1}{2}x+5$