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Hatshy [7]
3 years ago
7

What is

v%202" id="TexFormula1" title="6.73 - 2 \times 5to \: the \:zero \: power \div 2" alt="6.73 - 2 \times 5to \: the \:zero \: power \div 2" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
attashe74 [19]3 years ago
5 0
\bf 6.73-2\cdot 5^0\div 2\impliedby \mathbb{PEMDAS}
\\\\\\
6.73-\underline{2\cdot 1}\div 2\implies 6.73-\underline{2\div 2}\implies 6.73-1\implies 5.73
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Find two positive integers such that the sum of the first number and four times the second number is 1000, and the product of th
vladimir2022 [97]

Answer:

The two numbers are:

x = 500

y = 125

Step-by-step explanation:

We want to find two numbers x and y, such that:

x + 4*y = 1000

f(x, y) = x*y is maximum.

From the first equation, we can isolate one of the variables to get

x = 1000 - 4y

now we can replace it in f(x, y):

x*y = (1000 - 4*y)*y  = 1000*y - 4*y^2

So now we want to maximize the function:

f(y) =- 4*y^2 + 1000*y

where y must be an integer.

Notice that this is a quadratic equation with a negative leading coefficient (so the arms of the graph will open downwards), thus, the maximum will be at the vertex.

Remember that for a general quadratic equation:

y = a*x^2 + bx + c

the x-value of the vertex is:

x = -b/(2*a)

so, in the case of:

f(y) =- 4*y^2 + 1000*y

the y-value of the vertex will be:

y = -1000/(2*-4) = 1000/8 = 125

So we found the value of y.

now we can use the equation:

x = 1000 - 4*y

x = 1000 - 4*125 = 1000 - 500 = 500

x = 500

Then the two numbers are:

x =500

y = 125

6 0
3 years ago
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