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beks73 [17]
3 years ago
10

PLS HELP ME!!! I AM GIVING ALL MY POINTS OUT( look I don't have alot of points don't bully me(⌐■_■))

Mathematics
1 answer:
hodyreva [135]3 years ago
4 0

Answer:

9x

Step-by-step explanation:

3x3=9 +x=9x hope this helps

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Maggie's brother is 6 years younger than four times her age. The sum of their ages is 39. How old is Maggie?
poizon [28]

We can set Maggie's age to x

If her brother is six years younger than four times her age, than her brother must be equal to: 4x-6

Since the sum of their ages is 39, we can set up an equation.

(Sum of Maggie's age) + (sum of Maggie's brother's age) = 39

x + (4x -6) = 39

5x -6 = 39

5x = 45

x = 9 , Maggie's age is 9

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5 0
3 years ago
This is how dora subtracts 73-26. 73-3=70. 70-?=50. ?-?=? fill in the blanks to solve
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Answer:

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7 0
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6 0
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Read 2 more answers
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
so P(x) = R(x) - C(x)

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
7 0
3 years ago
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