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lianna [129]
2 years ago
13

Compute x/y if

Mathematics
2 answers:
Citrus2011 [14]2 years ago
6 0

Let first consider the equations one by one and will be solving one by one ;

{:\implies \quad \sf x+\dfrac{1}{y}=4}

Multiplying both sides by y will lead ;

{:\implies \quad \sf xy+1=4y}

{:\implies \quad \boxed{\sf xy=4y-1\quad \cdots \cdots(i)}}

Now, consider the second equation which is ;

{:\implies \quad \sf y+\dfrac{1}{x}=\dfrac14}

Multiplying both sides by x will yield

{:\implies \quad \sf xy+1=\dfrac{x}{4}}

{:\implies \quad \sf xy=\dfrac{x}{4}-1}

{:\implies \quad \boxed{\sf xy=\dfrac{x-4}{4}\quad \cdots \cdots(ii)}}

As LHS of both equations (i) and (ii) are same, so equating both will yield;

{:\implies \quad \sf 4y-1=\dfrac{x-4}{4}}

Multiplying both sides by 4 will yield

{:\implies \quad \sf 16y-4=x-4}

{:\implies \quad \sf 16y=x}

Dividing both sides by y will yield :

{:\implies \quad \boxed{\bf{\dfrac{x}{y}=16}}}

Hence, the required answer is 16

Ivenika [448]2 years ago
4 0

Answer:

Y = 1/4x-1

X = 4y-1

Step-by-step explanation:

layout the equations :

x+1/y = 4        y+1/x = 1/4

solve the equations, remember to switch sides switch signs solving the equations will give us the value of the letter

x+1/y = 4

x+1 = 4y

x=4y-1

***

y+1/x = 1/4

y+1  = 1/4x

y = 1/4x-1

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Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe
ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for n\geq 78

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: \sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4

For n=1: \sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4

From this point we will assume that n\geq 2

As we can see, \sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4 and (4n)^4=256n^4. Then,

\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4

Now, we will use the formula for the sum of the first 4th powers:

\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}

Therefore:

\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0

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465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1

Observe that, because n \geq 2 and is an integer,

n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5

In concusion, the statement is true if and only if n is a non negative integer such that n\leq 77

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute  (4n)^4- \sum^{n}_{i=0} (2i)^4 for 77 and 78 you will obtain:

(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064

(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992

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