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kherson [118]
2 years ago
12

Please solve with explanation. High points

Mathematics
1 answer:
Bezzdna [24]2 years ago
8 0

Answer:

Step-by-step explanation:

um

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Simplify (2x+y^2) (2y-4x-3y^2)
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explaining

 

Algebra

 

Simplify (2x^2)(4x^3y^2)

(2x2)(4x3y2)(2x2)(4x3y2)

Multiply x2x2 by x3x3 by adding the exponents.

Tap for more steps...

2x5(4y2)2x5(4y2)

Rewrite using the commutative property of multiplication.

2⋅4(x5y2)2⋅4(x5y2)

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8x5y2

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3 years ago
Consider the following polynomial:
guajiro [1.7K]
Linear is 4
Quadratic is 2
Cubic is 3
6 0
3 years ago
72in is how much yards
Gelneren [198K]

Answer:

2 yards is the answer

Step-by-step explanation:

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3 0
3 years ago
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A travel agent currently has 80 people signed up for a tour. The price of a ticket is $5000 per person. The agency has chartered
Nikolay [14]
So hmm let's take a peek at the cost first

so, they chartered the plane for 150 folks with a fixed cost of 250,000
now, incidental fees are 300 per person, if we use the quantity "x", for how many folks, then if "x" persons are booked, then incidental fees are 300x

so, more than likely an insurance agency is charging them 300x for coverage

anyway, thus the cost C(x) = 250,000 + 300x

now, the Revenue R(x), is simple is jut price * quantity

well, the price, thus far we know is 5000 for 80 folks, but it can be lowered by 30 to get one more person, thus increasing profits

so... let's see what the price say y(x) is  \bf \begin{array}{ccllll}
quantity(x)&price(y)\\
-----&-----\\
80&5000\\
81&4970\\
82&4940\\
83&4910
\end{array}\\\\
-----------------------------\\\\

\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%   (a,b)
&({{ 80}}\quad ,&{{ 5000}})\quad 
%   (c,d)
&({{ 83}}\quad ,&{{ 4910}})
\end{array}
\\\quad \\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{-90}{3}\implies -30
\\ \quad \\\\
% point-slope intercept
y-{{ 5000}}={{ -30}}(x-{{ 80}})\implies y=-30x+2400+5000\\
\left.\qquad   \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y=-30x+7400

so.. now we know y(x) = -30x+7400

now, Revenue is just price * quantity
the price y(x) is -30x+7400, the quantity is "x"

that simply means R(x) = -30x²+7400x


now, for the profit P(x)

the profit is simple, that is just incoming revenue minus costs, whatever is left, is profit
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P(x) = (7400x - 30x²) - (250,000+300x)

P(x) = -30x² + 7100x - 250,000

now, where does it get maximized? namely, where's the maximum for P(x)?

well \bf \cfrac{dp}{dx}=-60x+7100

and as you can see, if you zero out the derivative, there's only 1 critical point, run a first-derivative test on it, to see if its a maximum
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