Neither P, nor A are on the sketch
I guess P is the upper right corner of the rectangle
and A=(0,1)
P belongs to the line going through (1,0) and B(0,y)
<span>but we don't know the y-coordinate of B </span>
<span>the triangle is right and isosceles, so pythagoras a²+a²=2² ... 2a²=4 ... a²=2 ... a=sqrt2 </span>
now look at the right triangle BOA
<span>his hypotenuse is AB=sqrt2 and the <span>the kathete</span> OA is 1 </span>
so y²+1²=(sqrt2)² ... y²+1=2 ... y²=1.. y=1
so the coordinates of B are (0,1)
the line going through (1,0) and (0,1) is L(x)=-x+1
P belongs to this line, so the coordinates of P are P(x,-x+1) (0<x<1)
b) so if that's P, the height of the rectangle is -x+1 and the width=2x
<span>so its area A(x)=2x*(-x+1)= -2x²+2x
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>
They would have to work 15 hours and 50 minutes
First, you need to find the zero of x+3, which is -3
then you multiply -3×1, which equals -3
the reason why I multiplied -3×1 is because the 1 is the coefficient of x^2 but we just don't see it because we don't need to.
once we get -3 we add that to the next coefficient (-5)
-3+-5=-8
we take the sum (-8) and multiply that to -3 (the zero of x+3)
we get 24, so we add -22+24=2
-22 Is the next coefficient to come and 2 is our remainder
the new equation is: x-8 remainder 2
Solution:
First find the slope
Secondly, write the equation
Answer:
Check:
<em>Hope this was helpful.</em>