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4 years ago

Are the product of 3/4×2/9 greater than 2/9 plz explain

2 answers:

7 0

Hey there!

Let's first multiply 3/4 and 2/9. When we multiply fractions, we multiply the numerators and denominators, so we have:

3/4 times 2/9 = 6/36 = 1/6

Now, we have to compare 1/6 and 2/9. In order to do this, we can multiply them both by a version of one to get equivalent fractions with the same denominator. This is because anything times one is itself, but we can keep it as itself, just with a different denominator, as I said. The least common multiple of the denominators are 18, so we multiply:

1/6(3/3) --> (3/3 equals one) = 3/18
2/9(2/2) = 4/18

Now that we have the same denominator, we can compare the numerators to see that 2/9 is actually greater.

Hope this helps!

adoni [48]4 years ago

4 0

Less because if you multiply any number by a fraction (not mixed or improper) than it will always be less because multiplying it by 1 will keep it the same anything less than one will make the number smaller the answer of 3/4 x 2/9 is 1/6 btw

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garri49 [273]

Answer:

<h3>METHOD I:</h3>

(by using the first principle of differentiation)

We have the "Limit definition of Derivatives":

$\boxed{\mathsf{f'(x)= \lim_{h \to 0} \{\frac{f(x+h)-f(x)}{h} \} ....(i)}}$

Here, f(x) = sec x, f(x+h) = sec (x+h)

Substituting these in eqn. (i)

$\implies \mathsf{f'(x)= \lim_{h \to 0} \{\frac{sec(x+h)-sec(x)}{h} \} }$

sec x can be written as 1/ cos(x)

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{1}{h} \{\frac{1}{cos(x+h)} -\frac{1}{cos(x)} \} }$

Taking LCM

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{1}{h} \{\frac{cos(x)-cos(x+h)}{cos(x)cos(x+h)} \} }$

By Cosines sum to product formula, i.e.,

$\boxed{\mathsf{cos\:A-cos\:B=-2sin(\frac{A+B}{2} )sin(\frac{A-B}{2} )}}$

=> cos(x) - cos(x+h) = -2sin{(x+x+h)/2}sin{(x-x-h)/2}

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{2sin(x+\frac{h}{2} )}{cos(x+h)cos(x)}\:.\: \lim_{h \to 0} \frac{sin(\frac{h}{2} )}{h} }$

I shifted a 2 from the first limit to the second limit, since the limits ar ein multiplication this transmission doesn't affect the result

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x+\frac{h}{2} )}{cos(x+h)cos(x)}\:.\: \lim_{h \to 0} \frac{2sin(\frac{h}{2} )}{h} }$

2/ h can also be written as 1/(h/ 2)

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x+\frac{h}{2} )}{cos(x+h)cos(x)}\:.\: \lim_{h \to 0} \frac{1\times sin(\frac{h}{2} )}{\frac{h}{2} } }$

We have limₓ→₀ (sin x) / x = 1.

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x+\frac{h}{2} )}{cos(x+h)cos(x)}\:.\: 1 }$

h→0 means h/ 2→0

Substituting 0 for h and h/ 2

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x+0)}{cos(x+0)cos(x)} }$

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x)}{cos(x)cos(x)} }$

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x)}{cos(x)}\times \frac{1}{cos x} }$

sin x/ cos x is tan x whereas 1/ cos (x) is sec (x)

$\implies \mathsf{f'(x)= tan(x)\times sec(x) }$

Hence, we got

$\underline{\mathsf{\overline{\frac{d}{dx} (sec(x))=sec(x)tan(x)}}}$

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

<h3>METHOD II:</h3>

(by using other standard derivatives)

$\boxed{ \mathsf{ \frac{d}{dx} ( \sec \: x) = \sec x \tan x }}$

sec x can also be written as (cos x)⁻¹

We have a standard derivative for variables in x raised to an exponent:

$\boxed{ \mathsf{ \frac{d}{dx}(x)^{n} = n(x)^{n - 1} }}$

Therefore,

$\mathsf{ \frac{d}{dx}( \cos x)^{ - 1} = - 1( \cos \: x) ^{( - 1 - 1} } \\ \implies \mathsf{\ - 1( \cos \: x) ^{- 2 }}$

Any base with negative exponent is equal to its reciprocal with same positive exponent

$\implies \: \mathsf{ - \frac{1}{ (\cos x) {}^{2} } }$

The process of differentiating doesn't just end here. It follows chain mechanism, I.e.,

while calculating the derivative of a function that itself contains a function, the derivatives of all the inner functions are multiplied to that of the exterior to get to the final result.