Class 11th

2x + y = 4
2x + 6y = 14
} y=2, x=3
B (1, 2)
Let C (k, 4–2k)
Now AB² = AC²
=> 5k² – 24k + 19 = 0
α = (6+1+10/5)/3 = 18/5
Now 15 (α+β)
15 (17/5) = 51

According to Newton's law of cooling
(T? -T? )/t = K [ (T? +T? )/2 - T? ]
For 1? cup of coffee,
(90-80)/t = K [ (90+80)/2 - 20]
For 2? cup of coffee,
(80-60)/t' = K [ (80+60)/2 - 20]
Divide (1) by (2)
t'/t = 65/50 ⇒ t' = (13/5)t

E ∝ F? A? T?
[M¹L²T? ²] ∝ [M¹L¹T? ²]? [LT? ²]? [T]?
a = 1
a + b = 2 ⇒ b = 1
-2a - 2b + c = -2
⇒ c = 2
a = 1 b = 1 c = 2
E ∝ [F] [A] [T²]

cos? ¹ (y/2) = log? (x/5)? , |y| < 2
Differentiating on both side
(1/√ (1- (y/2)²) * (-y'/2) = 5 * (5/x) * (1/5)
(-xy')/ (2√ (1- (y²/4) = 5
Square on both side
(x²y'²)/4 = 25 * (4-y²)/4
Diff on both side
xy' + y'x² + 25y = 0

f (x) = x? – 4x + 1 = 0
f' (x) = 4x³ – 4
= 4 (x–1) (x²+1+x)
=> Two solution

dy/dx = (ax-by+a)/ (bx+cy+a)
=> bxdy + cydy + ady = axdx – bydx + adx
cy²/2 + ay – ax²/2 – ax + bxy = k
ax² + ay² + 2ax – 2ay = k
=> x² + y² + 2x – 2y = λ
Short distance of (11,6)
= √12²+5² – 5
= 13 – 5
= 8

x = Σ a? = 1/ (1-a); y = Σ b? = 1/ (1-b); z = Σ c? = 1/ (1-c)

Now,
a, b, c → AP
1-a, 1-b, 1-c → AP
1/ (1-a), 1/ (1-b), 1/ (1-c) → HP
x, y, z → HP

z? = iz²
Let z = x + iy
x – iy = I (x² – y² + 2xiy)

Case-I
x = 0
–y² = –y
y = 0, 1

Case - II
y = – 1/2
=> x² – 1/4 = 1/2 => x = ±√3/2

Area of polygon
= 1/2 | (0, 1, 1), (√3/2, -1/2, 1), (-√3/2, -1/2, 1) |
= 1/2 | -√3/2 - √3/2 | = 3√3/4

Let height of bridge $=\mathrm{h}$

Displacement of ball, $S=-h$

$\begin{array}{rr}& S=ut+\frac{1}{2}{\mathrm{a}\mathrm{t}}^2\\ & -\mathrm{h}=4*4+\frac{1}{2}(-10)(4{)}^2\\ & \Rightarrow \mathrm{h}=64\mathrm{m}\end{array}$

Least count = 1 mm / 100 = 0.01 mm
Diameter = main scale reading + circular scale reading
Diameter = 0 + 52 * 0.01 mm
= 0.52 mm = 0.052 cm