Solution:- ধৰো, m = 5
n = 3
\(m^2 – n^2\) = \(5^2 – 3^2\) = 25 – 9 = 16 যুগ্ম
Ans:- option (a)
Solution:- Option d
Solution :- দিয়া আছে, \(y_1 = 20\) , \(x_1 = 5\)
\(y_2 = y \) , \(x_2 = 50\)
\(x_1× y_1\) = \(x_1× y_1\)
100 = 50y
y = 2
Option (C)
Solution:-
আমি জানো যে, α + β = b
αβ = k
3α + 2β = 20
ধৰো, α = 8 β = -2
3(8) + 2(-2) = 24 -4 =20
So, αβ = k
8 × -2 =k
-16 = k
Ans:- Option (A)
Solution:-
আমি জানো যে, α + β = – \(\frac{6}{a}\)
αβ = \(\frac{c}{a}\)
A/Q
α + β = αβ
– \(\frac{6}{a}\) = \(\frac{c}{a}\)
C = -6
সমীকৰণটোৰ পৰা পাও যে, b = 6
= b + c
= 6 – 6 = 0
Ans:- Option (b)
Solution:-
3ax + 4y = -2
3(-3)a + 4(4) = -2
= -9a + 16 = -2
= -9a = -18
= a = 2
and, 2x + by = 14
2 (-3) + b (4) = 14
= – 6 + 4b =14
= 4b = 20
= b = 5
Ans:- Option (A)
Solution:-
\(a_1 = a, b_1 = 2 c_1 = – 9\)
\(a_2 = 3, b_2 = b c_2 = – 18\)
\(\frac{a_1}{a_2}\ = \frac{b_1}{b_2}\)
\(\frac{a}{3}\ = \frac{2}{b}\)
ab =6
Ans:- option (d)
Ans:- Option (b)
Solution:- Option (a)
Solution:- Option (b)
= \( \sqrt3\)× tan 60 – 3
= \( \sqrt3\) ×\( \sqrt3\) – 3 = 3 – 3 = 0
Solution:- Option (A)
কোনো এডাল লেখৰ শূণ্য হবলৈ হলে সেই বিন্দুটোৰ মাজেৰে লেখডাল যাব লাগিব
Solution:-
Given , \(a_n = 7n – 4\)
\(a_1 = 3)
\(a_2 = 10)
\(a_3 = 17)
d = 10 – 3 = 7
Ans:- Option (a)
Ans:- Option (b)
Solution:-
we know that,
sin 30 = 1/2
cos 60 = 1/2
A + B = 30 + 60 = 90
Ans:- Option (b)
Solution:-
we know that
So, Option (d)
Solution:-
A/q
2πrh = 264
πrh = 132
π\(r^2\)h = 924
132r = 924
r = 7
So, 2πrh = 264
= 2 × \(\frac{22}{7}\)×7 × h =264
= 44h = 264
= h = 6
ব্যাস = 14
ব্যাস : উচ্চতা
14 : 6
7 : 3
Ans:- Option (b)
Solution:- ধৰো, ব্যাসাৰ্ধ = r
চুঙাটোৰ উচ্চতা = 5
চুঙাৰ আয়তন = শংকুৰ আয়তন
π\(r^2\)h = 1/3 π\(r^2\)h
5 = 1/3 h
h = 15
So, Option (b)