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pantera1 [17]
2 years ago
9

*I NEED THIS ASAP*

Mathematics
1 answer:
BigorU [14]2 years ago
5 0

Answer:

5 terms

to the fourth degree

leading coeff of 1

3 turning points

end behavior (when x -> inf, y -> inf. When x -> - inf, y -> -inf)

x intercepts are (0,-4) (0,-2) (0,1) (0,3)

Relative min: (-3.193, -25) (2.193, 25)

Relative max: (-0.5, 27.563)

Step-by-step explanation:

The terms can be counted, seperated by the + and - in the equation given.

The highest exponent is your degree.

The number before the highest term is your leading coeff, if there is no number it is 1.

The turning points are where the graph goes from falling to increasing or vice versa.

End behaviour you have to look at what why does when x goes to -inf and inf.

X int are the points at which the graph crosses the x-axis.

The relative min and max are findable if you plug in the graph on desmos or a graphing calculator.

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Convert from rectangular to polar coordinates: note: choose rr and θθ such that rr is nonnegative and 0≤θ<2π0≤θ<2π (a)(9,0
DochEvi [55]

Answer:

Step-by-step explanation:

Convert rectangle (x , y) to polar coordinates ( r , θ)

x=r  \cos \theta, y= r \sin \theta

r=\sqrt{x^2+y^2} , \theta =tan^-^1 (\frac{y}{x} )

a) converts (9, 0) to polar coordinates  ( r , θ)

r=\sqrt{x^2+y^2} \\\\=\sqrt{9^2+0} \\\\=9

\theta= \tan^-^1 (\frac{0}{9} )\\\\=0

b) Convert (18,\frac{18}{\sqrt{3} } ) to polar coordinates ( r, θ)

r = \sqrt{18^2+(\frac{18}{\sqrt{3} })^2 } \\\\=\sqrt{324+108} \\\\=\sqrt{432}

\frac{x}{y} \theta = \tan^-^1(\frac{\frac{18}{\sqrt{3} } }{18} )\\\\= \tan ^-^1(\frac{1}{\sqrt{3} } )\\\\= \frac{\pi}{6}

c)  converts (-5, 5) to polar coordinates  ( r , θ)

r =\sqrt{(-5)^2+(5)^2} \\\\=\sqrt{50} \\\\=5\sqrt{2}

\theta=\tan^-^1(\frac{5}{-5} )\\\\= \tan^-^1(-1)\\\\=\frac{3\pi}{4}

d)  converts (-1, √3) to polar coordinates  ( r , θ)

r=\sqrt{(-1)^2+(\sqrt{3})^2 } \\\\= \sqrt{4} \\\\=2

\theta=\tan^-^1(\frac{\sqrt{3} }{-1} )\\\=\tan^-^1(-\sqrt{3} )\\\\=\frac{2\pi}{3}

= \frac{2\pi}{\sqrt{3} }

6 0
3 years ago
the large Square above has area 9 is divided into 9 smaller squares of equal area what is the length of how strong and bold
Kay [80]
If one square is divided into 9 smaller equal squares, then they have to be arranged in 3 lines of 3, that is 3 smaller equal squares per side of the original big square. That said, the area of the big square is equal to the multiplication of 3 small squares sides times 3 small squares sides, call x the length of the small squares.
So,
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therefore the smaller squares have sides of 1 unit
8 0
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