<em>Answer:</em><em> </em>
<h2>
<em><</em><em>3</em><em> </em><em>and </em><em><</em><em>2</em></h2>
<em>Complementary</em><em> </em><em>angles</em><em> </em><em>are </em><em>those </em><em>angles </em><em>which </em><em>are </em><em>exactly </em><em>9</em><em>0</em><em> </em><em>degree.</em>
<em>Hope </em><em>it</em><em> </em><em>will</em>
<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em>
5,8,13 is not possible because when you add 5 and 8, they equal 13, the longest side/hypotenuse. Basically, when your two shorter side lengths add to be equal to or less than the longest side length, it’s not a triangle
Step-by-step explanation:
For a moment let y = x^2 + 5
Then by the distributive property
(2x^2 + 4x - 3 ) P = . 2x^2p + 4xp - 3p
Again by the distributive
2x^2 p = 2x^2(x^2 - 2x +5) = 2x^4 - 4x^3 + 10x^2
4xp = 4x(x^2 + 5) = 4x^3 - 8X^2 +20x
-3p = -3(x^2 - 2x + 5) = -3x^2 + 6x -15
Taking everything together we get
(2x^2 + 4x -3 )(x^2-2x + 5) = (2x^4 -4x^3 +10X^2) + (4x^3 - 8x^2 + 20) + -3x^2 + 6x -15)(2X^2 + 4x - 3 )(x^2 -2x = 5) = 2x^4 + 26 - 15
Answer:
x^2+y^2=6^2
Step-by-step explanation:
The equation of a circle is (x-h)^2+(y-k)^2=r^2
where h is the x coordinate of the center, k is the y coordinate of the center, and r is the radius of the circle.
We can see the center is on 0,0, and the radius is 6, so our equation is
(x-0)^2+(y-0)^2=6^2
or
x^2+y^2=6^2