ASSIGNMENT
CLASS-IX
MATHEMATICS
Q1) 0.323232 when expressed in the form p/q(p, q are integers) is
a) 8/25 b) 29/90 c) 32/99
d) 32/199
Q2) Which of the following numbers can be represented as non-terminatin
repeating decimals?
a 39/24 b) 3/16 c) 3/11 d) 137/25
Q3) which of the following is irrational?
a) 0.14 b) 0.14161616. c) 0.141614161416. d
0.10140014000014.
Q4) If a+b =10 and ab=21, the value of a 3+b3 is
a) 370 b) 372 c) 375 d) none of these
Q5) If x+a is a factor of x' a'x'+3x-6a, then a is
a) 0 b)-1 c) 2 d) 1
Q6) The factors of x -x y-xy +y are
a) (x+y)(x-xy+y) b) (x+y) (x}+xy+y) c) (x+y)? (x-y) d) (x-y)?(x+y)
Q7) If x' +1/x =110 then x+1/x is
a) 5 b) 10 C) 15
d) none of these
3
2
2
a
Q8) In between any two numbers there are
a) Only one rational number b) two rational numbers c) infinite ra
numbers
d) No rational number
09) 3 V6 + 4 V6 is equal to
a) 676 b) 776 c) 4712 d) 7 / 12
Q10) Can we write 0 in the form of p/q?
a) Yes b) No
c) Cannot be explained d) None of the above
Q11) A binomial of degree 20 in the following is
a) 20x+1 b) x/20+1 c) x +1 d) x +20
Q12) The value of f(x) =5X-4x' +3 when x=-1 is
a) 3 b)-12
c) -6 d) 6
Q13) The zero of the polynomial (8)2 2x+7 is
a) 2/7
h)-217 C 772 d) -712
m
=
please solve this
Answers
Q1) 0.323232 when expressed in the form p/q(p, q are integers) is
a) 8/25 b) 29/90 c) 32/99 d) 32/199
Solution :-
let ,
→ p = 0.323232_____
→ 100p = 32.323232____
subtracting ,
→ 100p - p = 32
→ 99p = 33
→ p = 32/99 (c)
Q2) Which of the following numbers can be represented as non-terminating repeating decimals ?
a) 39/24 b) 3/16 c) 3/11 d) 137/25
Solution :-
checking all options we get,
a) 39/24
→ 13/8
→ Prime factors of denominator = 8 = 2 * 2 * 2
since prime factors of denominator have 2 only , (a) is a terminating decimals .
a) 3/16
→ Prime factors of denominator = 16 = 2 * 2 * 2 * 2
since prime factors of denominator have 2 only , (b) is a terminating decimals .
c) 3/11
→ Prime factors of denominator = 11 = 1 * 11
since prime factors of denominator are other than 2 and 5 , (c) is a non terminating repeating decimals .
d) 137/25
→ Prime factors of denominator = 25 = 5 * 5
since prime factors of denominator have 5 only , (d) is a terminating decimals .
solving next,
→ a + b = 10
→ ab = 21
so,
→ (a + b) = 10
squaring both sides,
→ (a + b)² = 10²
→ a² + b² + 2ab = 100
→ a² + b² + 2 * 21 = 100
→ a² + b² = 100 - 42
→ a² + b² = 58
then,
→ a³ + b³ = (a + b)(a² + b² - ab)
→ a³ + b³ = 10(58 - 21)
→ a³ + b³ = 10 * 37
→ a³ + b³ = 370
Learn more :-
https://brainly.in/question/22246812
. Find all the zeroes of the polynomial x4
– 5x3 + 2x2+10x-8, if two of its zeroes are 4 and 1.
https://brainly.in/question/39026698
Answer:
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