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Zinaida [17]
3 years ago
12

There are exactly four positive integers $n$ such that \[\frac{(n + 1)^2}{n + 23}\] is an integer. Compute the largest such $n$.

Mathematics
1 answer:
ololo11 [35]3 years ago
5 0
Let's present the given equation first. Deciphering the given code, I think the equation is (n+1)²/n+23. Then, we want to find the maximum value of n. Suppose the complete equation is:

f(n) = (n+1)²/n+23

To find the maximum,let's apply the concepts in calculus. The maxima can be determined by setting the first derivative to zero. Therefore, we use the chain rule to differentiate the fraction. For a fraction u/v, the derivative is equal to (vdu-udv)/v². 

f'(n) = [(n+23)(2)(n+1)-(n+1)²(1)]/(n+23)² = 0
[(n+23)(2n+2) - (n+1)²]/(n+23)² = 0
(2n²+2n+46n+46-n²-2n-1)/(n+23)²=0
n²+46n+45=0
n = -1, -45

There are two roots for the quadratic equation. Comparing the two, the larger one is -1. Thus, the maximum value of n is -1.
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sigma (difference)=√(sigma1/n1 + sigma2/n2), Plug:

sigma(d)= √(49/100 + 36/50)

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6 0
4 years ago
Read 2 more answers
7. Sapna travelled 48 km to meet her parents. She travelled the first 18 km by car and the remaining distance by train. Find the
Anton [14]

The fraction of the journey Sapna traveled by car and train is 3/8 and 5/8 respectively.

We know that fraction is an expression in mathematics used to denote the division of two whole numbers.

Given that the total distance = 48 km.

The distance Sapna traveled by car = 18 km.

So, the fraction is = distance traveled by car/total distance = 18/48 = 3/8

The distance Sapna traveled by train = 48 - 18 km = 30 km.

So, the fraction is = distance traveled by train/total distance = 30/48 = 5/8

Therefore the fraction of the journey Sapna traveled by car and train is 3/8 and 5/8 respectively.

Know more about fractions here -

brainly.com/question/10354322

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6 0
2 years ago
IF I HAVE A LAPTOP COMPUTER THAT MEASURES 30.8×6.6×39.8 CENTIMETERS...WHAT SIZE CASE DOES IT REQUIRE
soldi70 [24.7K]

Answer:

8090.54

Step-by-step explanation:

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4 0
3 years ago
Please help me 33/2 + 3y/5 = 7y/10 + 15
8090 [49]

We are given with an equation in <em>variable y</em> and we need to solve for <em>y</em> . So , now let's start !!!

We are given with ;

{:\implies \quad \sf \dfrac{33}{2}+\dfrac{3y}{5}=\dfrac{7y}{10}+15}

Take LCM on both sides :

{:\implies \quad \sf \dfrac{165+6y}{10}=\dfrac{7y+150}{10}}

<em>Multiplying</em> both sides by <em>10</em> ;

{:\implies \quad \sf \cancel{10}\times \dfrac{165+6y}{\cancel{10}}=\cancel{10}\times \dfrac{7y+150}{\cancel{10}}}

{:\implies \quad \sf 165+6y=7y+150}

Can be <em>further written</em> as ;

{:\implies \quad \sf 7y+150=165+6y}

Transposing <em>6y </em>to<em> LHS</em> and <em>150</em> to<em> RHS </em>

{:\implies \quad \sf 7y-6y=165-150}

{:\implies \quad \bf \therefore \quad \underline{\underline{y=15}}}

8 0
2 years ago
I need help with this
miv72 [106K]
Your answer would be 3
3 0
3 years ago
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