I dont know actually but you can Google it
<span>r=2sin0-3cos0</span>
explain<span>using the formula that links Cartesian to Polar coordinates.<span>∙y=r<span>sinθ</span></span><span>∙x=r<span>cosθ</span></span>then : <span>r<span>sinθ</span>=3r<span>cosθ</span>+2</span>and <span>r<span>sinθ</span>−3r<span>cosθ</span>=2</span>hence <span>r<span>(<span>sinθ</span>−3<span>cosθ</span>)</span>=2</span></span>
<span>⇒r=<span><span>2<span><span>sinθ</span>−3<span>cosθ</span></span></span></span></span>
Step-by-step explanation:
x/(x + 2) + 1/x = 1
x² + (x + 2) = x(x + 2)
x² + x + 2 = x² + 2x
x = 2.
Since x = 2 neither makes (x + 2) or x become 0, there are no extraneous solutions.
The answer is "The equation has 1 valid solution and no extraneous solutions." (C)
Answer:
1st one
Step-by-step explanation:
spinner colours: green, blue, red
card colours: purple, green
<em>possible outcomes:</em>
green + purple
green + green
blue + purple
blue + green
red + purple
red + green
(3x−2)(2x^2+5)
=(3x+−2)(2x^2+5)
=(3x)(2x^2)+(3x)(5)+(−2)(2x^2)+(−2)(5)
=6x^3+15x−4x^2−10
=6x^3−4x^2+15x−10
