Which of the following shapes are polygons with more than 444 sides?

Mathematics

1 answer:

Leviafan [203]3 years ago

6 0

Hexagon and pentagon

You might be interested in

For f(x) = 4x +1 and g(x) = x2 - 5, find (g/f) (x).

Softa [21]

Answer:

$\left(g/f\right)\left(x\right)=\frac{x}{4}-\frac{1}{16}-\frac{79}{16\left(4x+1\right)}$

Step-by-step explanation:

$f(x)=4x+1$

$g\left(x\right)=x^2\:-\:5$

(g/f)(x) = g(x) / f(x)

$=\:\frac{x^2\:-\:5}{4x+1}\:\:\:\:\:\:$

$=\frac{x}{4}+\frac{-\frac{x}{4}-5}{4x+1}$

$=\frac{x}{4}-\frac{1}{16}+\frac{-\frac{79}{16}}{4x+1}$

$=\frac{x}{4}-\frac{1}{16}-\frac{79}{16\left(4x+1\right)}$

Therefore,

$\left(g/f\right)\left(x\right)=\frac{x}{4}-\frac{1}{16}-\frac{79}{16\left(4x+1\right)}$

8 0

2 years ago

The position equation for a particle is s of t equals the square root of the quantity t cubed plus 1 where s is measured in feet

vladimir1956 [14]

$\bf s(t)=\sqrt{t^3+1}
\\\\\\
\cfrac{ds}{dt}=\cfrac{1}{2}(t^3+1)^{-\frac{1}{2}}\cdot 3t^2\implies \boxed{\cfrac{ds}{dt}=\cfrac{3t^2}{2\sqrt{t^3+1}}}\leftarrow v(t)
\\\\\\
\cfrac{d^2s}{dt^2}=\cfrac{6t(2\sqrt{t^3+1})-3t^2\left( \frac{3t^2}{\sqrt{t^3+1}} \right)}{(2\sqrt{t^3+1})^2}\implies 
\cfrac{d^2s}{dt^2}=\cfrac{ \frac{6t(2\sqrt{t^3+1})-1}{\sqrt{t^3+1}} }{4(t^3+1)}$

$\bf \cfrac{d^2s}{dt^2}=\cfrac{6t[2(t^3+1)]-1}{4(t^3+1)\sqrt{t^3+1}}\implies 
\boxed{\cfrac{d^2s}{dt^2}=\cfrac{12t^4+12t-1}{4t^3+4\sqrt{t^3+1}}}\leftarrow a(t)\\\\
-------------------------------\\\\a(2)=\cfrac{215~ft^2}{44~sec}$

8 0

2 years ago

Read 2 more answers

Write the equation for the area (A) in terms of length (l) of a playground if the width (w) is one half as long as the length (l

Dvinal [7]

Let, the length = l
Width = l/2

Area = l * w = l * l/2 = l²/2

So, Your Final Answer would be A = l²/2

Hope this helps!

5 0

2 years ago

What is the value of this expression when d equals three

kotykmax [81]

Answer:

What is the expresion I am confused?