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Radda [10]
3 years ago
6

Bonnie will buy up to 200 paper cups for a party. she found a store that sells paper cups in packages of 25. the function c(p)=2

5p represents the number of cups Bonnie would have if she bought p packages of cups from the store
what are the theoretical domain and range of function c
what are the practical domain and range of function c
Mathematics
1 answer:
taurus [48]3 years ago
5 0
The answer is that she bought 300 packages of  paper cups for a party
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A restaurant sells burritos for $8.00 each. To buy supplies, the restaurant pays $1.50 for the meat and $1.00 for the other ingr
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isnt it just 2.50$ ?

Step-by-step explanation:

1$ + 1.5$ = 2.5$ ?

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I have a question that says " jada has read 3/5 of a book. She has read 75 pages so far. How many pages are in the whole book?"
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Answer and Step-by-step explanation:

To solve, we would multiply by the reciprocal of the portion Jada has read, and multiply that number to 75.

The reciprocal means the the value, but flipped, so \frac{3}{5} will become \frac{5}{3}.

Now we do 75 times \frac{5}{3}/

\frac{75 * 5}{3} \\\\\\\\\frac{375}{3} \\\\\\\\125

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3 years ago
A teacher places n seats to form the back row of a classroom layout. Each successive row contains two fewer seats than the prece
Alex_Xolod [135]

Answer:

The number of seat when n is odd S_n=\frac{n^2+2n+1}{4}

The number of seat when n is even S_n=\frac{n^2+2n}{4}

Step-by-step explanation:

Given that, each successive row contains two fewer seats than the preceding row.

Formula:

The sum n terms of an A.P series is

S_n=\frac{n}{2}[2a+(n-1)d]

    =\frac{n}{2}[a+l]

a = first term of the series.

d= common difference.

n= number of term

l= last term

n^{th} term of a A.P series is

T_n=a+(n-1)d

n is odd:

n,n-2,n-4,........,5,3,1

Or we can write 1,3,5,.....,n-4,n-2,n

Here a= 1 and d = second term- first term = 3-1=2

Let t^{th} of the series is n.

T_n=a+(n-1)d

Here T_n=n, n=t, a=1 and d=2

n=1+(t-1)2

⇒(t-1)2=n-1

\Rightarrow t-1=\frac{n-1}{2}

\Rightarrow t = \frac{n-1}{2}+1

\Rightarrow t = \frac{n-1+2}{2}

\Rightarrow t = \frac{n+1}{2}

Last term l= n,, the number of term =\frac{ n+1}2, First term = 1

Total number of seat

S_n=\frac{\frac{n+1}{2}}{2}[1+n}]

    =\frac{{n+1}}{4}[1+n}]

     =\frac{(1+n)^2}{4}

    =\frac{n^2+2n+1}{4}

n is even:

n,n-2,n-4,.......,4,2

Or we can write

2,4,.......,n-4,n-2,n

Here a= 2 and d = second term- first term = 4-2=2

Let t^{th} of the series is n.

T_n=a+(n-1)d

Here T_n=n, n=t, a=2 and d=2

n=2+(t-1)2

⇒(t-1)2=n-2

\Rightarrow t-1=\frac{n-2}{2}

\Rightarrow t = \frac{n-2}{2}+1

\Rightarrow t = \frac{n-2+2}{2}

\Rightarrow t = \frac{n}{2}

Last term l= n, the number of term =\frac n2, First term = 2

Total number of seat

S_n=\frac{\frac{n}{2}}{2}[2+n}]

    =\frac{{n}}{4}[2+n}]

     =\frac{n(2+n)}{4}

    =\frac{n^2+2n}{4}  

4 0
3 years ago
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