Question

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Answer 1

$6.(a)1.2$

This is the answer for Qû

Answer 2

Q 5.

• Radius of the circle = 12 cm
• Increasing rate of the radius, dr/dt = 0.01 cm/s

The rate with which the area increases, dA/dt = ?

Change in radius :-

$\sf{\dfrac{dr}{dt}=0.01 }$

We know,

Area, A = πr²

$\sf{\dfrac{dA}{dt}=\dfrac{d}{dr}(\pi r^2) }\\\\\sf{\implies {\dfrac{dA}{dt}=\dfrac{d}{dr}(r^2) }$

$\sf{\implies {\dfrac{dA}{dt}=\pi.2r}$

$\sf{\implies {\dfrac{dA}{dt}=2\pi r}$

$\sf{\implies {\dfrac{dA}{dt}=2\pi \times 12}$

$\sf{\implies {\dfrac{dA}{dt}=24\pi}$

So, the area increases with a rate of 24 sq. cm/s.

The answer is C) 24 sq. cm/s.

________________

Q 6.

(1.0002)³⁰⁰⁰ = (1 + 0.0002)³⁰⁰⁰

It is in the form (1 + x)ᵃ.

The value of x is very less than 1. (x <<< 1)

Therefore we get,

= 1 + ax

= 1 + 3000 × 0.0002

= 1 + (3000 × 2/10000)

= 1 + 3 × 0.2

= 1 + 0.6

= 1.6

Therefore, the approximate value is 1·6.

The answer is C) 1·6.