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lyudmila [28]
3 years ago
14

What the measure of a interior angle of a six sided figure

Mathematics
1 answer:
Sergio [31]3 years ago
7 0

Answer:

120 degrees.

Step-by-step explanation:

If it is a regular 6 sided figure then each interior angle is 180 - (360/6)

= 120 DEGREES.

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PLEASE HELP!
Degger [83]

Answer:

Perimeter = 21.58 units

Step-by-step explanation:

Perimeter of a polygon = Sum of measures of all sides of the polygon

Perimeter of the quadrilateral LEAP = PL + AE + EL + LP

                                                            = 4.47 + 4.47 + 6.32 + 6.32

                                                            = 21.58 units

[If the distances between two point are not given, use the formula to calculate the distance between two points (x_1, y_1) and (x_2,y_2)

Distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}]

7 0
3 years ago
PLEASE HELP WILL GIVE BRAINLIEST TO CORRECT ANSWER
nlexa [21]

Answer:

B. 6.000000 g

Step-by-step explanation:

The reason why its B is due to it having more 'sigs' than 'figs'

Hope this helped <3

6 0
3 years ago
Let T be the plane-2x-2y+z =-13. Find the shortest distance d from the point Po=(-5,-5,-3) to T, and the point Q in T that is cl
GaryK [48]

Answer:

d=10u

Q(5/3,5/3,-19/3)

Step-by-step explanation:

The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane n=(-2,-2,1), then r will have the next parametric equations:

x=-5-2\lambda\\y=-5-2\lambda\\z=-3+\lambda

To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T

-2x-2y+z =-13\\-2(-5-2\lambda)-2(-5-2\lambda)+(-3+\lambda) =-13\\10+4\lambda+10+4\lambda-3+\lambda=-13\\9\lambda+17=-13\\9\lambda=-13-17\\\lambda=-30/9=-10/3

Substitute the value of \lambda in the parametric equations:

x=-5-2(-10/3)=-5+20/3=5/3\\y=-5-2(-10/3)=5/3\\z=-3+(-10/3)=-19/3\\

Those values are the coordinates of Q

Q(5/3,5/3,-19/3)

The distance from Po to the plane

d=\left| {\to} \atop {PoQ}} \right|=\sqrt{(\frac{5}{3}-(-5))^2+(\frac{5}{3}-(-5))^2+(\frac{-19}{3}-(-3))^2} \\d=\sqrt{(\frac{5}{3}+5))^2+(\frac{5}{3}+5)^2+(\frac{-19}{3}+3)^2} \\d=\sqrt{(\frac{20}{3})^2+(\frac{20}{3})^2+(\frac{-10}{3})^2}\\d=\sqrt{\frac{400}{9}+\frac{400}{9}+\frac{100}{9}}\\d=\sqrt{\frac{900}{9}}=\sqrt{100}\\d=10u

7 0
3 years ago
How to solve a mixed number fraction
lesantik [10]
You find a common denominator (the bottom one) and then add the numerator (the top) but you don't add the denominator, you just make it the same
8 0
3 years ago
Can someone please help me with this math problem.
mr Goodwill [35]
PART A:
Find the rate of change between 1980 and 1989
d for P₁ = 80 - 60
d for P₁ = 20

d for P₂ = 76 - 82
d for P₂ = -6

The rate of change in P₁ is 20 hundred per year. The rate of change in P₂ is -6 hundred per year.

PART B:
Find the rate of change between 1989 and 1996
d for P₁ = 100 - 80
d for P₁ = 20

d for P₂ = 70 - 76
d for P₂ = -6

The rate of change in P₁ is 20 hundred per year. The rate of change in P₂ is -6 hundred per year.

PART C:
Find the rate of change between 1980 and 1996
d for P₁ = 100 - 60
d for P₁ = 40

d for P₂ = 70 - 82
d for P₂ = -12

The rate of change in P₁ is 40 hundred per year. The rate of change in P₂ is -12 hundred per year.
4 0
3 years ago
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