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Sliva [168]
3 years ago
10

Sum of (4x^4-x2-x)+(x^4+x^3+x^2+x-1)

Mathematics
2 answers:
Flauer [41]3 years ago
6 0
<span>                -9x4 + 5x2 - 26x - 12
—————————————————————
                                3

</span>
Sedaia [141]3 years ago
6 0
<span>(4x^4-x2-x)+(x^4+x^3+x^2+x-1)
=</span><span>4x^4 - x2 - x + x^4 + x^3 + x^2 +  x- 1
=5x^4 </span>+ x^3 - 1

(x^2 and x are canceled out)...
x^2 - x^2 = 0
x -x =0
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Convert from rectangular to polar coordinates: note: choose rr and θθ such that rr is nonnegative and 0≤θ&lt;2π0≤θ&lt;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 height of a triangle is 7 cm longer than its base.
katrin [286]
A) Height -7 = base
B) .5 * height * base = 60
Substituting A into B
B) .5 * height * (height -7) = 60
B)  .5*height^2  -3.5*height = 60
B)  .5*height^2 -3.5*height -60 = 0
Using the quadratic formula:
Height = 15
Subtracting 7 gives us the base length
Base = 8
**************************************************************
Double-Check
.5 * 15 * 8 = 60


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Hence, Jake's running time in decimal is: 3.2 hours

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