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Helga [31]
3 years ago
7

Simultaneous equations x^2+3xy=10 x=2y

Mathematics
1 answer:
satela [25.4K]3 years ago
8 0

9514 1404 393

Answer:

  (x, y) = (-2, -1) or (2, 1)

Step-by-step explanation:

Substitute for x in the first equation:

  (2y)^2 +3(2y)y = 10

  10y^2 = 10

  y^2 = 1

  y = ±1

  x = 2y = ±2

Solutions are (x, y) = (-2, -1) or (2, 1).

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What is the square root of -225
NemiM [27]

Answer:

15i

Step-by-step explanation:

since 15 x 15 is 225, the imaginary number comes into play here. i x i = -1, so 15i x 15i = -225

8 0
3 years ago
Read 2 more answers
A retail company estimates that if it spends x thousands of dollars on advertising during the year, it will realize a profit of
Gwar [14]

Answer:

Step-by-step explanation:

If the profit realized by the company is modelled by the equation

P (x) = −0.5x² + 120x + 2000, marginal profit occurs at dP/dx = 0

dP/dx = -x+120

P'(x) = -x+120

Company's marginal profit at the $100,000 advertising level will be expressed as;

P '(100) = -100+120

P'(100) = 20

Marginal profit at the $100,000 advertising level is $20,000

Company's marginal profit at the $140,000 advertising level will be expressed as;

P '(140) = -140+120

P'(140) = -20

Marginal profit at the $140,000 advertising level is $-20,000

<u>Based on the marginal profit at both advertising level, I will recommend the advertising expenditure when profit between $0 and $119 is made. At any marginal profit from $120 and above, it is not advisable for the company to advertise because they will fall into a negative marginal profit which is invariably a loss.</u>

7 0
3 years ago
Please calculate this limit <br>please help me​
Tasya [4]

Answer:

We want to find:

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n}

Here we can use Stirling's approximation, which says that for large values of n, we get:

n! = \sqrt{2*\pi*n} *(\frac{n}{e} )^n

Because here we are taking the limit when n tends to infinity, we can use this approximation.

Then we get.

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n} = \lim_{n \to \infty} \frac{\sqrt[n]{\sqrt{2*\pi*n} *(\frac{n}{e} )^n} }{n} =  \lim_{n \to \infty} \frac{n}{e*n} *\sqrt[2*n]{2*\pi*n}

Now we can just simplify this, so we get:

\lim_{n \to \infty} \frac{1}{e} *\sqrt[2*n]{2*\pi*n} \\

And we can rewrite it as:

\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n}

The important part here is the exponent, as n tends to infinite, the exponent tends to zero.

Thus:

\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n} = \frac{1}{e}*1 = \frac{1}{e}

7 0
3 years ago
How many bars would the team need to purchase from
Anna007 [38]

The numbers of bars that the team have to get from each company in order for the total cost to be equal is 40.

<h3>What is the purchase about?</h3>

Since n = numbers of bars purchase.

y= total cost in dollars

Company A = 0.75n

Company B = 10 + 0.50n

Therefore, to find the numbers of bars that the team would need to purchase from each company in order for the total cost to be equal =

0.75n = 10 + 0.50n

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n = 40

Learn more about purchase from

brainly.com/question/5168855

#SPJ1

5 0
1 year ago
One kilogram of cheese costs £6.60 how much would 200g cost?
Helga [31]

Answer:

0.132

Step-by-step explanation:

if 1kg of chees cost 6.60 100g cost 0.0660 So if 0.0660 multiplied by 2        

= 0.132

7 0
2 years ago
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