Earlyface

Pls I Need Maths 08169282286

yes it getting late.we nid it fast

Please maths answers

Please we need mathematics answers to help our brothers and sisters...its getting late now

(8a)
Let Ms Maureen's Income = Nx
1/4x = shopping mall
1/3x = at an open market

Hence shopping mall and open market = 1/4x + 1/3x
= 3x + 4x/12 = 7/12x

Hence the remaining amount
= X-7/12x = 12x-7x/12 =5x/12

Then 2/5(5x/12) = mechanic workshop
= 2x/12 = x/6
Amount left = N225,000
Total expenses
= 7/12x + X/6 + 225000
= Nx

7x+2x+2,700,000/12 =Nx
9x + 2,700,000 = 12x
2,700,000 = 12x - 9x
2,700,000/3 = 3x/3
X = N900,000

(ii) Amount spent on open market = 1/3X
= 1/3 × 900,000
= N300,000

(8b)
T3 = a + 2d = 4m - 2n
T9 = a + 8d = 2m - 8n
-6d = 4m - 2m - 2n + 8n
-6d = 2m + 6n
-6d/-6 = 2m+6n/-6
d = -m/3 - n
d = -1/3m - n

Thanks dear pls more answers

(1a)
Given A={2,4,6,8,...}
         B={3,6,9,12,...}
         C={1,2,3,6}
         U= {1,2,3,4,5,6,7,8,9,10}

A' = {1,3,5,7,9}
B' = {1,2,4,5,7,8,10}
C' = {4,5,7,8,9,10}
A'nB'nC' = {5, 7}

(1b)
Cost of each premiere ticket = $18.50
At bulk purchase, cost of each = $80.00/50 = $16.00

Amount saved = $18.50 - $16.00
=$2.50

Thanks baba

Yes u can post if is the correct ones @ Obinna

3
3a.....pls draw the diagram

CBD = CDB(Base angles of an issoceles triangle)
BCD + CBD + CDB = 180°(sum of angles in a triangle)
2CDB + BCD = 180°
2CDB + 108° = 180°
2CDB = 180° - 108° =72°
CDB = 72/2 = 36°
BDE = 90°(angle in a semi-circle)
CDE = CDB + BDE
= 36° + 90°
= 126°

(3b)
(CosX)² - SinX/(SinX)²+ CosX
Using Pythagoras theorem, third side of triangle
y² = 1² + |3²
y² = 1 + 3 = 4
y = square root e = 2
Cos X = 1/2(adj/hyp)
Sin X = root 3/2(opp/hyp)
(CosX)² - SinX/(SinX)² + CosX
= (1/2)² - root3/2 / (root3/2)² + 1/2
= 1/4 - root3/2 / 3/4 + 1/2
= 1 - 2root3/4 / 3+2/4
= 1-2root3/5

: 2ai).

p = (rk/Q -ms)2/3
p3/2 = (rk/Q -ms)2/8 * 3/2
p3/2 = rk/Q = ms
p3/2 + ms = rk/Q
q(p3/2 + ms ) = rk
Q =rk/p3/2 + ms

2aii)
Where :
r = 10, k = 4, s = 0.2
p = 3, m = 15

Q = 10 * 4/3^3/2 + (15 * 0.2)
Q = 40/5.2 + 3
Q = 40/8.2
Q = 4.88
. 🕵: 5a)
Using elimination method
2x: 2p-4q=4
3x: 4p+2q=9

Hence
4p-6q=8
9p+6 =27
Adding equation (i)   to (ii)
Therefore
13p=35

Thus p=35/13

Substituting P=35/13 in equation (ii)

Hence
3p+2q=9........... (ii)
3(35/13) + 2q= 9
=105/13+2q=9

Multiplying throughout by 13
 ~13~ *105/ ~13~ + ~13~*2q =13*9
Thus 105+26q=117

Collecting like terms
26q =117-105
26q=12
Q=12/26 or 6/13
*2bi* 2mn^2x/(3m-n)

But m= 3 n = -2 x= -1

= 2*3*(-2)^2*(-1)/3(3)-(-2)

=  2*3*4*-1/9+2
=   -24/11

=2.     2/11
*2a.)*

P= (rk/Q  - ms)⅔

Take the reciprocal of ⅔ by the power

P⅔= rk/Q - ms

rk/Q= P⅔ + ms

Q = rk/ P⅔+ ms

ii.) When P=3, m=15, s=0.2, k=4 and r=10

Q= rk/ P⅔ + ms
   

Q= 10(4)/(3)⅔ + (15)(0.2)
              =40/5.196+3
              =40/8.196
              =4.88 (1dp)

(9a)
Draw the triangle

(9b)
(i)Using cosine formulae
q² = x² + y² - 2xycosQ
q² = 9² + 5² - 2×9×5cos90°
q² = 81 + 25 - 90 × 0
q² = 106
q = square root 106
q = 10.30 = 10km/h
Distance = 10 × 2 = 20km

(ii)
Using sine formula
y/sin Y = q/sin Q
5/sin Y = 10.30/sin 90°
Sin Y = 5 × sin90°/10.30
Sin Y = 5 × 1/10.30
Sin Y = 0.4854
Y = sin‐¹(0.4854), Y = 29.04

Bearing of cyclist X from y
= 90° + 19.96°
= 109.96° = 110°

(9c)
Speed = 20/4, average speed = 5km/h