Question

pls answer all the question.

5f0d33dd29a88aa0749535f1e91aadd6.pdf

Answer 1

1) In a polynomial the exponents of the variable is always

b) Positive

2) Which of the following is correct ?

a) (a – b)² = a² – 2ab + b²

3) The sum of –7pq and 2pq is

–7pq + 2pq = –5pq

d) –5pq

4) If we subtract –3x²y² from x²y², then we get

–3x²y² from x²y²

x²y² – (–3x²y2)

x²y²+3x²y²

4x²y²

d) 4x²y²

5) Area of rectangle with length 4ab and breadth 6b² is

$\frak {Area = length \times breadth}$

$\frak{Area = 4ab \times 6 {b}^{2} }$

$\frak{Area = 24 {a}^{2} {b}^{2} }$

a) 24a²b²

6) Volume of a rectangular box with Length = 2ab, Breadth = 3ac and height = 2ac.

Volume of cuboid = Length × Breadth × Height

Volume = 2ab × 3ac × 2ac

So Volume = 12a³bc²

a) 12a³bc²

7) Common factor of 17abc, 34ab²,51a² is

To get the common factor we need to factorise 17abc, 34ab²and 51a²

17abc, 34ab², 51a²b

17ab(c, 2b, 3a)

b) 17ab

8) Factorised form r² – 10r + 21

r² –(7+3)r + 21

r² – 7r – 3r + 21 __________ [By middle term factorisation]

r(r–7) – 3(r–7)

(r–7)(r–3)

b) (r–7)(r–3)

II)1) Answer the following :

i) (x+3)(x+7)

x²+ (7+3)x + 7×3 Using identity (x+a)(x+b)

x²+10x+21

ii) (xy + yz)²

(xy)² + 2•xy•yz + (yz)² Using identity (a+b)²

x²y² + 2xy²z + y²z²

2) Using suitable identity, evaluate :

I) (52)²

(50+2)² Using identity (a+b)²

(50)² + 2•50•2 + (2)²

2500 + 200 + 4

2700 + 4

2704

II) (103)²

(100+3)²

(100)² + 2•100•3 + (3)²

10000 + 600 + 9

10600 + 9

10609

III) 47 × 53

(50–3)(50+3) Using identity (a–b)(a+b)

(50)²–(3)²

2500–9

2491

IV) 9.8 × 10.2

(10–0.2)(10+0.2) Using identity (a–b)(a+b)

(10)² – (0.2)²

100–0.04

100.04

3) Factorise :

I) 6ab + 12 bc

6b(a+c)

II) –xy–ay

–y(x+a)

III) ax³ – bx² + cx

x(ax² – bx + c)

Answer 2

Step-by-step explanation:

Hey!

Given=mass of moon=7.36×10^22

mass of earth=6.37×10^24

radius of moon=1.74×10^6

g=GMm/Rm^2

g=6.67×10^-11*7.36×10^22/(1.74×10^6)^2

g=1.6ms-2

we can put the value of g=9.8ms-2

so,gm/ge=1.6/9.8=1/6

hope it helps you