Compare 3/5 and 6/9.

Mathematics

2 answers:

mamaluj [8]3 years ago

7 0

3/5 and 6/9 simplified is 2/3.
so 3/5 and 2/3
make them equal so you get : 9/15 and 10/15
6/9 is greater that 3/5 :)

tiny-mole [99]3 years ago

5 0

To compare you should make the down part equal first
27/45 < 30/45

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Can a sequence be both arithmetic and geometric?

artcher [175]

An aritmetic sequence is like this
$a_n=a_1+d(n-1)$ where a1=first term and d=common difference

geometric is $a_n=a_1(r)^{n-1}$ where a1=first term and r=common ratio

can it be both aritmetic and geometric
hmm, that means that the starting terms should be the same

therfor we need to solve $d(n-1)=(r)^{n-1}$
what values of d and r make all natural numbers of n true?
are there values that make all natural numbers for n true?

when n=1, then d(1-1)=0 and r^(1-1)=1, so already they are not equal

the answer is no, a sequence cannot be both aritmetic and geometric

4 0

3 years ago

Find dy/dx of y = <img src="https://tex.z-dn.net/?f=log_%7B2%7D" id="TexFormula1" title="log_{2}" alt="log_{2}" align="absmiddle

NikAS [45]

Step-by-step explanation:

$y = log_2 x^2 \\ \therefore y = \frac{log x^2}{log2}\\ (By\:change\:of\:base\:law) \\ \\ \therefore y = \frac{2log x}{log2}\\\\ differentiating \: both \: side \: w \: r \: t \: x. \\ \\ \frac{dy}{dx} = \frac{2}{log2} \bigg(\frac{d}{dx} log x \bigg) \\ \\ \therefore \frac{dy}{dx} = \frac{2}{log2} \bigg( \frac{1} {x} \times \frac{d}{dx} x \bigg) \\ \\ \therefore \frac{dy}{dx} = \frac{2}{log2} \bigg(\frac{1}{x} \times 1\bigg) \\ \\ \huge \red{ \boxed{\therefore \frac{dy}{dx} = \frac{2}{xlog2}}}$