Science, asked by NK3665033, 5 months ago

Q.2 VERY SHORT ANSWER QUESTIONS:
1.____
travel through air.
2. Larger vibrations have____
amplitude.
3._____
is the part of ear visible
outside.
4. You hear a bomb blast 10 seconds after it
happens. Do you agree?
5. Similar to light waves, sound waves also
undergo reflection. True or false?
6. Where are the vocal cords located in a
human body?
7. We can get echo in a room of 10 m length
and width. Do you agree?
8. Frequencies below 20 Hz_____
9. The frequent to and fro movement of an
object about its mean position______.
10. Musical sounds of a specific pitch._____​

Answers

Answered by ardhipremananda
2

Answer:

1 Sound 7 No please follow and thank me

Answered by nibha028128
2

Answer:

In Class 9 Maths Chapter 2- Polynomials

= (x−5)x(x+2)+10(x+2)

= (x−5)(x+2)(x+10)

(iv) 2y3+y2–2y–1

Solution:

Let p(y) = 2y3+y2–2y–1

Factors = 2×(−1)= -2 are ±1 and ±2

By trial method, we find that

p(1) = 0

So, (y-1) is factor of p(y)

Now,

p(y) = 2y3+y2–2y–1

p(1) = 2(1)3+(1)2–2(1)–1

= 2+1−2

= 0

Therefore, (y-1) is the factor of p(y)

Ncert solutions class 9 chapter 2-4

Now, Dividend = Divisor × Quotient + Remainder

(y−1)(2y2+3y+1) = (y−1)(2y2+2y+y+1)

= (y−1)(2y(y+1)+1(y+1))

= (y−1)(2y+1)(y+1)

3b)3–3(4a)2(3b)+3(4a)(3b)2

64a3–27b3–144a2b+108ab2=

(4a)3–(3b)3–3(4a)2(3b)+3(4a)(3b)2

=(4a–3b)3

=(4a–3b)(4a–3b)(4a–3b)

Here, the identity, (x – y)3 = x3 – y3 – 3xy(x – y) is used.

(v) 7p3– (1/216)−(9/2) p2+(1/4)p

Solution:

The expression, 27p3–(1/216)−(9/2) p2+(1/4)p

can be written as (3p)3–(1/6)3–3(3p)2(1/6)+3(3p)(1/6)2

27p3–(1/216)−(9/2) p2+(1/4)p =

(3p)3–(1/6)3–3(3p)2(1/6)+3(3p)(1/6)2

= (3p–16)3

= (3p–16)(3p–16)(3p–16)

9. Verify:

(i) x3+y3 = (x+y)(x2–xy+y2)

(ii) x3–y3 = (x–y)(x2+xy+y2)

Solutions:

(i) x3+y3 = (x+y)(x2–xy+y2)

We know that, (x+y)3 = x3+y3+3xy(x+y)

⇒ x3+y3 = (x+y)3–3xy(x+y)

⇒ x3+y3 = (x+y)[(x+y)2–3xy]

Taking (x+y) common ⇒ x3+y3 = (x+y)[(x2+y2+2xy)–3xy]

⇒ x3+y3 = (x+y)(x2+y2–xy)

(ii) x3–y3 = (x–y)(x2+xy+y2)

We know that,(x–y)3 = x3–y3–3xy(x–y)

⇒ x3−y3 = (x–y)3+3xy(x–y)

⇒ x3−y3 = (x–y)[(x–y)2+3xy]

Taking (x+y) common ⇒ x3−y3 = (x–y)[(x2+y2–2xy)+3xy]

⇒ x3+y3 = (x–y)(x2+y2+xy)

10. Factorize each of the following:

(i) 27y3+125z3

(ii) 64m3–343n3

Solutions:

(i) 27y3+125z3

The expression, 27y3+125z3 can be written as (3y)3+(5z)3

27y3+125z3 = (3y)3+(5z)3

We know that, x3+y3 = (x+y)(x2–xy+y2)

27y3+125z3 = (3y)3+(5z)3

= (3y+5z)[(3y)2–(3y)(5z)+(5z)2]

= (3y+5z)(9y2–15yz+25z2)

(ii) 64m3–343n3

The expression, 64m3–343n3can be written as (4m)3–(7n)3

64m3–343n3 =

(4m)3–(7n)3

We know that, x3–y3 = (x–y)(x2+xy+y2)

64m3–343n3 = (4m)3–(7n)3

= (4m-7n)[(4m)2+(4m)(7n)+(7n)2]

= (4m-7n)(16m2+28mn+49n2)

11. Factorise: 27x3+y3+z3–9xyz

Solution:

The expression27x3+y3+z3–9xyz can be written as (3x)3+y3+z3–3(3x)(y)(z)

27x3+y3+z3–9xyz = (3x)3+y3+z3–3(3x)(y)(z)

We know that, x3+y3+z3–3xyz = (x+y+z)(x2+y2+z2–xy –yz–zx)

27x3+y3+z3–9xyz = (3x)3+y3+z3–3(3x)(y)(z)

= (3x+y+z)[(3x)2+y2+z2–3xy–yz–3xz]

= (3x+y+z)(9x2+y2+z2–3xy–yz–3xz)

12. Verify that:

x3+y3+z3–3xyz = (1/2) (x+y+z)[(x–y)2+(y–z)2+(z–x)2]

Solution:

We know that,

x3+y3+z3−3xyz = (x+y+z)(x2+y2+z2–xy–yz–xz)

⇒ x3+y3+z3–3xyz = (1/2)(x+y+z)[2(x2+y2+z2–xy–yz–xz)]

= (1/2)(x+y+z)(2x2+2y2+2z2–2xy–2yz–2xz)

= (1/2)(x+y+z)[(x2+y2−2xy)+(y2+z2–2yz)+(x2+z2–2xz)]

= (1/2)(x+y+z)[(x–y)2+(y–z)2+(z–x)2]

13. If x+y+z = 0, show that x3+y3+z3 = 3xyz.

Solution:

We know that,

x3+y3+z3-3xyz = (x +y+z)(x2+y2+z2–xy–yz–xz)

Now, according to the question, let (x+y+z) = 0,

then, x3+y3+z3 -3xyz = (0)(x2+y2+z2–xy–yz–xz)

⇒ x3+y3+z3–3xyz = 0

⇒ x3+y3+z3 = 3xyz

Hence Proved

14. Without actually calculating the cubes, find the value of each of the following:

(i) (−12)3+(7)3+(5)3

(ii) (28)3+(−15)3+(−13)3

Solution:

(i) (−12)3+(7)3+(5)3

Let a = −12

b = 7

c = 5

Explanation:

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