EinsteinJr

Your answers for Question 1 seem to be good; here's your brownie:

I found the answer my teacher gave me:

A(2,0); let B be the point of coordinates \((x,\sqrt{x}) \\\text{With } x \geq 0\)

The formula to calculate the distance between to points M(x_M,y_M) and N(x_N,y_N) is \(MN=\sqrt{(x_N-x_M)^2+(y_N-y_M)^2}\)

So \(AB=\sqrt{(x-2)^2+(\sqrt{x}-0)^2} \\AB^2=(x-2)^2+\sqrt{x}^2 \\AB^2=x^2-4x+4+x=x^2-3x+4\)

We end up with a quadratic polynomial (the expression has the form ax²+bx+c, here a=1, b=-3 and c=4). Since a=1>0, the polynomial has a minimum. The formula to calculate the extremum E (minimum or maximum) of a quadratic polynomial ax²+bx+c is \(E=\frac{-b}{2a}\)

So the x coordinate of the minimum M of the polynomial (the x coordinate of B) is \(\frac{3}{2}=1.5\)

And the minimum (y coordinate of B) is \(\sqrt{3/2}\)

So the coordinates of B are \((\frac{3}{2}; \sqrt{3/2})\)

Ah, and I'm silly too, I never found a distance AB of 1.25; here's the result I got:

\(AB=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2} \\AB=\sqrt{(\frac{3}{2}-2)^2+(\sqrt{\frac{3}{2}}-0)^2} \\AB=\sqrt{\frac{1}{4}+\frac{3}{2}}=\sqrt{\frac{7}{4}}\)

Click 2nd then asin

Well done Guest, your answers are exact!

You deserved a 20/20 and a brownie:

1 kg = 1 000 000 mg so 24.68 kg = 24 680 000 mg =2.468×10⁷ mg

k(a+b) = ka+kb

So :

-2(t+8)=-2t-16

and -5(3f+11)=-15f-55

1 mg = 0.001 g so 50 mg = 0.05 g = 5×10^-2 g

3+11=14