could you explain how the 2n appeared please

thank you, but can you please explain the first line and where you got that from, apart from that part it all makes sense

thank you very much

NVM i used 2 for ub in my e equation instead of -2 in all 4 of my attempts..

How did you find those numbers?

i subbed in 2 and 7 into the second differentiated equation but it wasnt those numbers, sorry im just really bad at these

thank you so much for teaching me

ok...

so TE = 116 + (58*9.81*6) = 5921.8

my gpe under 6M: .(58*9.81*6)= 3529.88

5805.8 - 3529.88 = 2275.92

2507.2 = 1/2 58v^2

86.48= v^2

v= 9.3 ms^-1

Thank you very much

thank you

question B is similar to a so i figured id just try it myself

try subbing it in formula b^2-4ac < 0 ( this represents when a quadratic formula has no real solution)

then rearrange it

thank you!

1. find gradient 2--6/7-3 = 8/4 = 2

m*m = -1

m= -0.5

sub the information u know into formula

y = mx + c or y - y1 = m(x -x1)

y = -0.5x + 0.5

ok thank you so much for helping!

ah ok thank you

one more question: does this method apply to all polar graphs such as when you have to solve tan and sin or is there a different method to do so

can you please explain how you find out -pi/3 = 5pi/3