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3 years ago

The sum of the interior angles of a quadrilateral equals. 180 270 360 540

2 answers:

7 0

The sum of the interior angles in any given quadrilateral is always going to be 360.

This means each angle inside the quadrilateral added up with always equal 360 degrees.

Hope this helps!
Brainliest is always appreciated if you feel its deserved! :)

bezimeni [28]3 years ago

3 0

Answer:

The sum of the interior angles of a quadrilateral equals _360° 

C is correct.

Step-by-step explanation:

The sum of the interior angles of a quadrilateral equals ______

The quadrilateral has four sides and four angles.

First we draw a diagonal of quadrilateral and it makes two triangle.

Sum of angles of triangle is 180° (Angle sum property of triangle)

A quadrilateral form by two triangle

Sum of interior angle of quadrilateral = 180° + 180°

= 360°

The sum of the interior angles of a quadrilateral equals _360° 

We have a general formula sum of interior angles of polygon.

Sum of angles of polygon having n sides = (n-2)180°

For quadrilateral, n = 4

Sum of angles of quadrilateral = (4-2)180°

= 360°

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Write the equation of the line in fully simplified slope-intercept form.

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x/3+y/-5=1

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3 years ago

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3 years ago

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emmasim [6.3K]

Answer:

7.5

Step-by-step explanation:

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3 years ago

Read 2 more answers

5m + 2[m+8] + 3 pls give answer m = 2

abruzzese [7]

So m=2, wherever m is put a 2

5x2+2(2+8)+3

Then do bidmas

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5x2+2x10+3

Multiplication-

10+20+3

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33

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7 0

4 years ago

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.

3 0

3 years ago