Question

সংক্ষিপ্ত প্রশ্ন
১. 0.6 x 0, $ =কত?
21
এর আবৃত্ত দশমিকে প্রকাশিত মান কত?
র
২. 32
1 = 2 - 3, p C Wএবং কে 4 দ্বারা ভাগ করলে ভাগশেষ কত হবে?
৪. tU
{2, 3, 4, 5, 6, 7,9}, A = {2, 4, 6}এবংB = {3, 5, 7}হলে, A 7 B =কত?
সৃজনশীল প্রশ্ন: ০১
1+3+c
f(x) = x + kx2 - 4x - 12এবং g(a) = -দুইটি ফাংশন।
ক. {x € N: 2 > 3 এবং3 < 30}কে তালিকা পদ্ধতিতে প্রকাশ কর।
খ. f(-2) = 0 হলে দেখাও যে, yk একটি অমূলদ সংখ্যা।
গ, প্রমাণ কর যে, g(2) = g(a 2 )​

Answer 1

Step-by-step explanation:

Given : 0.6 x 0.9

$0.32$

n = 2p-3 divided by 4

n = 2p-3 divided by 4U = {2, 3, 4, 5,6,7,9}, A = {2,4, 6 } B = {3, 5, 7}

n = 2p-3 divided by 4U = {2, 3, 4, 5,6,7,9}, A = {2,4, 6 } B = {3, 5, 7} To Find : Value of 0.6 x 0.9

n = 2p-3 divided by 4U = {2, 3, 4, 5,6,7,9}, A = {2,4, 6 } B = {3, 5, 7} To Find : Value of 0.6 x 0.9Rational number 0.\overline{32}0.

n = 2p-3 divided by 4U = {2, 3, 4, 5,6,7,9}, A = {2,4, 6 } B = {3, 5, 7} To Find : Value of 0.6 x 0.9Rational number 0.\overline{32}0. 32

n = 2p-3 divided by 4U = {2, 3, 4, 5,6,7,9}, A = {2,4, 6 } B = {3, 5, 7} To Find : Value of 0.6 x 0.9Rational number 0.\overline{32}0. 32

n = 2p-3 divided by 4U = {2, 3, 4, 5,6,7,9}, A = {2,4, 6 } B = {3, 5, 7} To Find : Value of 0.6 x 0.9Rational number 0.\overline{32}0. 32 Remainder

n = 2p-3 divided by 4U = {2, 3, 4, 5,6,7,9}, A = {2,4, 6 } B = {3, 5, 7} To Find : Value of 0.6 x 0.9Rational number 0.\overline{32}0. 32 RemainderA' ∩ B

n = 2p-3 divided by 4U = {2, 3, 4, 5,6,7,9}, A = {2,4, 6 } B = {3, 5, 7} To Find : Value of 0.6 x 0.9Rational number 0.\overline{32}0. 32 RemainderA' ∩ BSolution:

n = 2p-3 divided by 4U = {2, 3, 4, 5,6,7,9}, A = {2,4, 6 } B = {3, 5, 7} To Find : Value of 0.6 x 0.9Rational number 0.\overline{32}0. 32 RemainderA' ∩ BSolution:0.6 x 0.9 = 0.54

n = 2p-3 divided by 4U = {2, 3, 4, 5,6,7,9}, A = {2,4, 6 } B = {3, 5, 7} To Find : Value of 0.6 x 0.9Rational number 0.\overline{32}0. 32 RemainderA' ∩ BSolution:0.6 x 0.9 = 0.54N = 0.32

=> 100N = 32

=> 100N = 32

=> 100N = 32 => 99N = 32

=> 100N = 32 => 99N = 32=> N = 32/99

=> 100N = 32 => 99N = 32=> N = 32/990.32

= 32/99

= 32/99n = 2p-3 divided by 4

= 32/99n = 2p-3 divided by 4p = 2k

= 32/99n = 2p-3 divided by 4p = 2k=> n = 2(2k) - 3

= 32/99n = 2p-3 divided by 4p = 2k=> n = 2(2k) - 3=> n = 4k - 3

= 32/99n = 2p-3 divided by 4p = 2k=> n = 2(2k) - 3=> n = 4k - 3=> n = 4k - 4 + 1

= 32/99n = 2p-3 divided by 4p = 2k=> n = 2(2k) - 3=> n = 4k - 3=> n = 4k - 4 + 1=> n = 4(k - 1) + 1 remainder = 1

= 32/99n = 2p-3 divided by 4p = 2k=> n = 2(2k) - 3=> n = 4k - 3=> n = 4k - 4 + 1=> n = 4(k - 1) + 1 remainder = 1p = 2k + 1

= 32/99n = 2p-3 divided by 4p = 2k=> n = 2(2k) - 3=> n = 4k - 3=> n = 4k - 4 + 1=> n = 4(k - 1) + 1 remainder = 1p = 2k + 1=> n = 2(2k+1) - 3

= 32/99n = 2p-3 divided by 4p = 2k=> n = 2(2k) - 3=> n = 4k - 3=> n = 4k - 4 + 1=> n = 4(k - 1) + 1 remainder = 1p = 2k + 1=> n = 2(2k+1) - 3=> n = 4k +2 - 3

= 32/99n = 2p-3 divided by 4p = 2k=> n = 2(2k) - 3=> n = 4k - 3=> n = 4k - 4 + 1=> n = 4(k - 1) + 1 remainder = 1p = 2k + 1=> n = 2(2k+1) - 3=> n = 4k +2 - 3=> n = 4k - 1

= 32/99n = 2p-3 divided by 4p = 2k=> n = 2(2k) - 3=> n = 4k - 3=> n = 4k - 4 + 1=> n = 4(k - 1) + 1 remainder = 1p = 2k + 1=> n = 2(2k+1) - 3=> n = 4k +2 - 3=> n = 4k - 1=> n = 4k - 4 + 3

= 32/99n = 2p-3 divided by 4p = 2k=> n = 2(2k) - 3=> n = 4k - 3=> n = 4k - 4 + 1=> n = 4(k - 1) + 1 remainder = 1p = 2k + 1=> n = 2(2k+1) - 3=> n = 4k +2 - 3=> n = 4k - 1=> n = 4k - 4 + 3=> n = 4(k - 1) + 3 remainder = 3

= 32/99n = 2p-3 divided by 4p = 2k=> n = 2(2k) - 3=> n = 4k - 3=> n = 4k - 4 + 1=> n = 4(k - 1) + 1 remainder = 1p = 2k + 1=> n = 2(2k+1) - 3=> n = 4k +2 - 3=> n = 4k - 1=> n = 4k - 4 + 3=> n = 4(k - 1) + 3 remainder = 3Remainder = 1 or 3

= 32/99n = 2p-3 divided by 4p = 2k=> n = 2(2k) - 3=> n = 4k - 3=> n = 4k - 4 + 1=> n = 4(k - 1) + 1 remainder = 1p = 2k + 1=> n = 2(2k+1) - 3=> n = 4k +2 - 3=> n = 4k - 1=> n = 4k - 4 + 3=> n = 4(k - 1) + 3 remainder = 3Remainder = 1 or 3U = {2, 3, 4, 5,6,7,9},

= 32/99n = 2p-3 divided by 4p = 2k=> n = 2(2k) - 3=> n = 4k - 3=> n = 4k - 4 + 1=> n = 4(k - 1) + 1 remainder = 1p = 2k + 1=> n = 2(2k+1) - 3=> n = 4k +2 - 3=> n = 4k - 1=> n = 4k - 4 + 3=> n = 4(k - 1) + 3 remainder = 3Remainder = 1 or 3U = {2, 3, 4, 5,6,7,9},A = {2,4, 6 } => A' { 3 , 5 , 7 , 9 }

= 32/99n = 2p-3 divided by 4p = 2k=> n = 2(2k) - 3=> n = 4k - 3=> n = 4k - 4 + 1=> n = 4(k - 1) + 1 remainder = 1p = 2k + 1=> n = 2(2k+1) - 3=> n = 4k +2 - 3=> n = 4k - 1=> n = 4k - 4 + 3=> n = 4(k - 1) + 3 remainder = 3Remainder = 1 or 3U = {2, 3, 4, 5,6,7,9},A = {2,4, 6 } => A' { 3 , 5 , 7 , 9 }B = {3, 5, 7}

= 32/99n = 2p-3 divided by 4p = 2k=> n = 2(2k) - 3=> n = 4k - 3=> n = 4k - 4 + 1=> n = 4(k - 1) + 1 remainder = 1p = 2k + 1=> n = 2(2k+1) - 3=> n = 4k +2 - 3=> n = 4k - 1=> n = 4k - 4 + 3=> n = 4(k - 1) + 3 remainder = 3Remainder = 1 or 3U = {2, 3, 4, 5,6,7,9},A = {2,4, 6 } => A' { 3 , 5 , 7 , 9 }B = {3, 5, 7} A' ∩ B = {3, 5, 7}