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zhannawk [14.2K]
4 years ago
11

Given the lengths of two sides of a triangle, find the range for the length of the third side. (Range means find between which t

wo numbers the length of the third side must fall.) Write an inequality. a 11.5 and 23.6
Mathematics
1 answer:
Dahasolnce [82]4 years ago
4 0

Answer:

12.1 < Y < 35.1

Step-by-step explanation:

Given

Sides: 11.5 and 23.6

Required

Determine the range of the third side

Let the third side with Y

<em>Two of the following conditions must be satisfied to calculate the range of the third side</em>

<em></em>11.5 + Y > 23.6

23.6+ Y > 11.5

11.5 + 23.6 > Y

We'll solve one after other

1.    11.5 + Y > 23.6

Y > 23.6 - 11.5

Y > 12.1

2.    23.6+ Y > 11.5

Y > 11.5- 23.6

Y > -12.1

3.    11.5 + 23.6 > Y

35.1 > Y

Y < 35.1

Inequalities with negative can't be used' So, we have

Y > 12.1 and Y < 35.1

Rewrite inequality

12.1 < Y   and   Y < 35.1

Combine inequality

12.1 < Y < 35.1

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Yes this equation is an identity.
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An irregular shape that can be split into familiar shapes such as rectangles, triangles, and circles is called a _____ figure.
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Answer:

A Composite figure

Step-by-step explanation:

An irregular shape that can be split into familiar shapes such as rectangle, triangles and circles is called a Composite figure.

A Composite figure may easily split into the familiar shapes such as triangles, rectangles and circles in order to find the area of these shapes.

Thus, area of such figures can be easily found out by a composite figure.

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Peter decided to buy a new car. He made a $2,160 down payment and then took a 48-month loan. The total amount for the car, plus
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16,560÷48=345
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8 0
3 years ago
Please help I will mark brainliest
Andreas93 [3]

is the same as picking from a points set hmmm, notice the table is an x,y table, so we can just pick any two points to get the slope and thereby the equation.


3)


let's use (100, 36.5) and (300, 106.5)


\bf (\stackrel{x_1}{100}~,~\stackrel{y_1}{36.5})\qquad  (\stackrel{x_2}{300}~,~\stackrel{y_2}{106.5}) \\\\\\ slope =  m\implies  \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{106.5-36.5}{300-100}\implies \cfrac{70}{200}\implies \cfrac{7}{20} \\\\\\ \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-36.5=\cfrac{7}{20}(x-100) \\\\\\ y-36.5=\cfrac{7}{20}x-35\implies y=\cfrac{7}{20}x+1.5


4)


and for this one let's pick say (35, 88) and (45, 82)


\bf (\stackrel{x_1}{35}~,~\stackrel{y_1}{88})\qquad  (\stackrel{x_2}{45}~,~\stackrel{y_2}{82}) \\\\\\ slope =  m\implies  \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{82-88}{45-35}\implies \cfrac{-6}{10}\implies -\cfrac{3}{5} \\\\\\ \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-88=-\cfrac{3}{5}(x-35) \\\\\\ y-88=-\cfrac{3}{5}x+21\implies y=-\cfrac{3}{8}x+109

5 0
3 years ago
Determine the values of a for which the following system of linear equations has no solutions, a unique solution, or infinitely
navik [9.2K]

Answer:

Never

Never

Never

Step-by-step explanation:

The equations given are

2x1−6x2−4x3 = 6 ....... (1)

−x1+ax2+4x3 = −1 ........(2)

2x1−5x2−2x3 = 9 ..........(3)

the values of a for which the system of linear equations has no solutions

Let first add equation 1 and 2. Also equation 2 and 3. This will result to

X1 + (a X2 - 6X2) - 0 = 5

And

X1 + (aX2-5X2) + 2X3 = 8

Since X2 and X3 can't be cancelled out, we conclude that the value of a is never.

a unique solution,

Let first add equation 1 and 2. Also equation 2 and 3. This will result to

X1 + (a X2 - 6X2) - 0 = 5

And

X1 + (aX2-5X2) + 2X3 = 8

The value of a = never

infinitely many solutions. 

Divide equation 1 by 2 we will get

X1 - 3X2 - 2X3 =3

Add the above equation with equation 3. This will result to

3X1 - 8X2 - 4X3 = 12

Everything ought to be the same. Since they're not.

Value of a = never.

4 0
3 years ago
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