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3 years ago

13

suppose it takes 48 chickens fingers to feed Mr . Youngs 4th grade class of 20 students how many chicken fingers would be needed

for 30 students

1 answer:

Mrrafil [7]3 years ago

6 0

30 is 1.5 times greater than 20 so 48*1.5 = 72
it would take 72 chicken fingers to feed 30 students

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Can someone please help convert these so they have the same denominator??

ratelena [41]

The LCM of the denominators is 120. In other words, the denominators can multiply to 120.

3 * 40 = 120

40 * 2 = 80

Tigers: P = 80/120

24 * 5 = 120

24 * 4 = 96

Redbirds: P = 96/120

8 * 15 = 120

3 * 15 = 45

Bulldogs P = 45/120

60 * 2 = 120

60 * 1 = 60

Titans P = 60/120

Answer

Tigers: P = 80/120

Redbirds: P = 96/120

Bulldogs: P = 45/120

Titans: P = 60/120

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2 years ago

Translate: Sixteen decreased by fifteen times a number

Serhud [2]

That would be 16 - 15n

8 0

3 years ago

Read 2 more answers

A triangle has two sides of lengths 5 and 12 what value could be the third side check all that apply

seropon [69]

Answer:

9 and 11

Step-by-step explanation:

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2 years ago

What the recursive formula for the sequence

klemol [59]

$\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ a_n=2-5(n-1)\implies a_n=\stackrel{\stackrel{a_1}{\downarrow }}{2}+(n-1)(\stackrel{\stackrel{d}{\downarrow }}{-5})$

so, we know the first term is 2, whilst the common difference is -5, therefore, that means, to get the next term, we subtract 5, or we "add -5" to the current term.

$\bf \begin{cases} a_1=2\\ a_n=a_{n-1}-5 \end{cases}$

just a quick note on notation:

$\bf \stackrel{\stackrel{\textit{current term}}{\downarrow }}{a_n}\qquad \qquad \stackrel{\stackrel{\textit{the term before it}}{\downarrow }}{a_{n-1}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{current term}}{a_5}\qquad \quad \stackrel{\textit{term before it}}{a_{5-1}\implies a_4}~\hspace{5em}\stackrel{\textit{current term}}{a_{12}}\qquad \quad \stackrel{\textit{term before it}}{a_{12-1}\implies a_{11}}$

8 0

3 years ago

The 6 consecutive integers is 393 what is the third number in this sequence

MAXImum [283]

X+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)=393
6x+15=393
6x=393-15
6x=378
x=63
third one is 63+2=65

3 0

2 years ago