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quester [9]
3 years ago
13

In order for an object to be a polygon it must have (choose the best answer) *

Mathematics
2 answers:
photoshop1234 [79]3 years ago
6 0
All sides must be a line segment
HACTEHA [7]3 years ago
3 0

Answer:

1. all side must intersect exactly two others sides but only at their endpoints

2. The sides must be non collinear and have a common endpoint

3. All sides must be a line segment

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If f(x) =x/2 -3 and g(x) = 4x^2+ x-4, find( f + g )(x)
Andrei [34K]

f(x)=\dfrac{x}{2}-3=\dfrac{1}{2}x-3\\\\g(x)=4x^2+x-4\\\\(f+g)(x)=f(x)+g(x)\\\\(f+g)(x)=\left(\dfrac{1}{2}x-3\right)+(4x^2+x-4)=4x^2+1\dfrac{1}{2}x-7

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How many real roots does the function f(x)=x^2-3x+4 have
nadya68 [22]
Since b^2 - 4ac = 9-16 < 0

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5 0
3 years ago
The formula to find the period of orbit of a satellite around a planet is T^2=(4pi^2/GM)r^3 where r is the orbit's mean radius,
Natalija [7]
The answer is r= \sqrt[3]{GMT^{2}/4 \pi^{2}}

T^{2} = \frac{4 \pi^{2}}{GM} r^{3}

Move \frac{4 \pi^{2} }{GM} to the other side of the equation:
T^{2} /\frac{4 \pi^{2} }{GM} = r^{3}  \\ &#10;T^{2} *\frac{GM}{4 \pi^{2} } = r^{3}

Rearrange:
r^{3} = T^{2} *\frac{GM}{4 \pi^{2} } \\ &#10;r^{3}= \frac{T^{2} *GM}{4 \pi^{2} }  \\ &#10;r^{3}= \frac{GMT^{2}}{4 \pi^{2} } \\ &#10;r^{3} = GMT^{2}/4 \pi^{2}

Since x^{3}= \sqrt[3]{x}, then
r^{3} = GMT^{2}/4 \pi^{2} \\ &#10;r= \sqrt[3]{GMT^{2}/4 \pi^{2}}
3 0
3 years ago
Read 2 more answers
Please help!!!!
vfiekz [6]
I solved this problem by setting 240/80 equal to x/100 . Once you set them equal to each other, multiply 240 and 100 and divide the product by 80 and you get 300.
3 0
3 years ago
PLEASE HELP ME!!!!!!! I think the answer is exponential but I am not 100% sure!!!!!!!
babunello [35]
It is exponentially each time you are dividing by 4/3
7 0
3 years ago
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