Question

SOLVE.................↑↑

Answer 1

$\huge\tt\red{\underline{Answer}}$

$8 x\cos(4 {x}^{2} ) \\ \cos(4 {x}^{2} ) \\ - \cos(4 {x}^{2} ) \\ - 8x \cos(4 {x}^{2} )$

Solution-11:

$y_{1} = \sin4 {x}^{2} \\ \frac{d y_{1} }{dx} = \cos4 {x}^{2} (8x) = 8x \cos4 {x}^{2}$

Answer 2

Answer:

(sin o + cos o ) ( sec o+ cosec o)

=(sin o+ cos o) (1/ cos o + 1/ sin o)

= (sin o + cos o)(sin o+ cos o/sin o. cos o)

=(sin o+ cos o)^2/ sin o . cos o

=1+2sin o cos o/ sin o.cos o

=1/sin o. cos o+2 sin o cos o/ sin o cos o

=1/sin o . 1/cos o+2

=2+cosec o . sin o

√2\√5-√2

6√5-6√2\5-2

6√5-6√2\3

3[2√5-2√2]\1

2√5-2√2.Here, x = – 2 is the root of the equation

3x^2 + 7x + p = 0

⇒ 3(– 2)2 + 7(– 2) + p = 0

⇒ p = 2

Root of the equation x^2 + 4kx + k^2 – k + 2 = 0  are equal.

⇒ b^2 – 4ac = 0

⇒ 16k^2 – 4(k^2 – k + 2) = 0

⇒ 3k^2 + k – 2 = 0

⇒ (3k – 2)(k + 1) = 0

⇒ k =  2  ÷ 3  , − 1

Hey Dear,

◆ Answer -

F = mv^2/r

◆ Explaination -

First, we'll write down known dimensions of given quantities.

[F] = [L1M1T-2]

[v] = [L1T-1]

[m] = [M]

[r] = [L]

Let x, y & z are numbers such that F = k.v^x.m^y.r^z

In dimensional form, this can be written as -

[F] = [v]^x.[m]^y. [r]^z

[L1M1T-2] = [L1T-1]^x.[M]^y.[L]^z

[L1M1T-2] = [L^(x+z).M^(y).T^(-x)

Comparing indexes on both sides -

x + z = 1

y = 1

-x = -2

Solving these equations,

x = 2

y = 1

z = -1

Hence,

F = k.ν^x.m^y.r^z

F = k × v^2 × m^1 × r^-1

F = k.v^2.m/r

F = mv^2/r ...(k=1)

Thus, centripetal force of the particle in UCM is mv^2/r.