what did u do to get 38
#can anyone verify
celest zenith
cuz i got something different
ok the vol of the sphere
2/3 pi (60)^3
which is 144000 pi
hemi*
ohh
ye
send them i’ll try help
so
f(x)= (x+p)(x+q)
f(x)= g(x)
g(x)=r/x
so we can begin by making (x+p)(x+q)= r/x
expand double brackets to get
x^2 + px + qx+ qp = r/x
then multiply by x
to get
x^3 + (p+q)x^2 + qpx -r
cubic graph so the answer would be A
for this question
these q’s look like good practise for the non calc
i’ll try the other one too
label the sides of triangle BAE it’ll help more if u have paper label AE as 2x and AB as 3root5 and BE as x
so pythagoras to find x
x^2 + 4x^2 = 45
5x^2= 45
x^2 = 9
x = 3
i just did (x^2) + (2x^2) = (3root5)^2
so we know BE is 3
AE = 6
yeah sorry i didn’t specify
yes if u square 3root5
that’s what u get
AD = 5/3 AE as the q says so the sf is 5/3
so to find CD u do 3 x 5/3 which is 5
so CD is 5
the shaded region is a trapezium
so we use a+b/2 x h
(5+3)/2 x 4 = 16
height is 4 bc we do AD - AE which is 10-6
for this labelling the angles would help first
if u do that u would see that (5y-6)+ (y^2+15y+90)= 180 and if u rearrange u get
y^2 - 10 y - 96= 0
ur factors are gonna be
16 and -16 i think
and then u obvs use 16 bc angles can’t be negative
then yk angle BAC and BCD = 180
this is so long 😭
pretty sure a paper wouldn’t consist of many hard questions
usually end of the apper
paper*
but yk for the q do u get what to do next
oh
the questions aren’t too bad though just practise
basically once u find x and y
yeah but just try one opposite pair
and then
wait no
ADC
and LCD
not LCD
CDX
whoops
yeah
good practise i would say
if u need any help jus send it again cuz it is practise for me too 😭😭
what grade u at
pine jolt
same my skl has done one mock and thats it
celest zenith
r u aiming for a 9
u can probs get a 9 if you’ve improved and done loads of revision
silly mistakes always gets me
celest zenith
winning game A bc (Wa) and winning game B be (Wb) and losing game A be (La) and losing game B be (Lb)
Wa x Wb = 9/25
losing = (1-Wa)
Wa x 4(1-Wa)= 9/25
expand
-4Wa^2 + 4Wa = 9/25
4Wa^2 - 4Wa - 9/25 =0
4x^2 - 4x -9/25 =0
multiply by 25
100x^2 - 400 x - 9=0
factorise
i think gets
(10x-9)(10x-1)
x=9/10 or x = 1/10
two possible values for Wa😭
now i think u gotta sub it into
Wb= 4(1-Wa)
Wb= 4 -3.6
Wb= 0.4
Wb= 4(1-0.1)
Wb= 4 - 0.4
Wb=3.6
since probability can’t be over 1 our value for Wa must be 0.9 and Wb must be 0.4
and we r asked abt him winning only one game
which is only Wa and Lb or Wb and La
P (Wa , Lb) 9/10 x 6/10 = 54/100 = 27/50
P (Wb,La) 4/10 x 1/10 = 4/100 = 2/50
27/50 + 2/50 = 29/50
so probability would be 29/50
u can’t bc there’s two winning a and winning b so u have to differentiate them
same as losing a and losing b
this gotta be 6 marks
same idea of doing 1-x