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den301095 [7]
3 years ago
9

Stewart has a home-based business putting on children’s parties. He charges $60 to design the party and then $6.00 per child. Wr

ite a function rule that relates the total cost of the party to the number of children n.
f(n) = 6 + 60n

f(n) = 6n – 55

f(n) = 60 + 6n

f(n) = 6 – 55n
Mathematics
2 answers:
Aleksandr [31]3 years ago
7 0

Answer:

<h2>f(n) = 60 + 6n</h2>

Step-by-step explanation:

numbers of students = n

party decoration charge = 60

per child cost = 6

so, therefore

f(n) = 60 + 6n

<h2 /><h2>MARK ME AS BRAINLIST</h2><h3>PLZ FOLLOW ME</h3>
Varvara68 [4.7K]3 years ago
4 0

Answer: A. f(n) 60 + 6n

60 is constant

6 per child is variable

Let c = number of children

total cost = 60 +6c

Step-by-step explanation: i got it from somebody else's brainly

If there are n children, and each additional child who joins will add another $6 to the cost, then the cost due to the number of children will be 6n. We also have the fixed cost of $60, no matter how many children join. Therefore, we add up these costs to get Cost = f(n) = 60 + 6n.

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How do you solve this?​
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\text{L.H.S}\\\\=\begin{vmatrix} 1 & bc& b+c\\ 1 & ca& c+a\\ 1 &ab&a+b\end{vmatrix}\\\\\\=\begin{vmatrix} 1 & bc& b+c-(a+b+c)\\ 1 & ca& c+a-(a+b+c)\\ 1 &ab&a+b-(a+b+c)\end{vmatrix}~~~~~~~~~~~~~~~;c_3\rightarrow c_3 -(a+b+c)c_1\\\\\\=\begin{vmatrix} 1 & bc& -a\\ 1 & ca& -b\\ 1 &ab&-c\end{vmatrix}\\\\\\=\dfrac 1{abc}\begin{vmatrix} a & abc& -a^2\\ b & abc& -b^2\\ c &abc&-c^2\end{vmatrix}\\\\\\=-\dfrac {abc}{abc}\begin{vmatrix} a & 1& a^2\\ b & 1& b^2\\ c &1&c^2\end{vmatrix}\\

=-1\begin{vmatrix} a & 1& a^2\\ b & 1& b^2\\ c &1&c^2\end{vmatrix}\\\\\\=-1 \times -1 \begin{vmatrix} 1 & a& a^2\\ 1 & b& b^2\\ 1 &c&c^2\end{vmatrix}\\\\\\=\begin{vmatrix} 1 & a& a^2\\ 1 & b& b^2\\ 1 &c&c^2\end{vmatrix}\\\\\\=\text{R.H.S}\\\\\text{Showed.}

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