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ololo11 [35]
3 years ago
9

A new phone sells for $250 and a used one sells for $75. How many of each are sold if the profit for the day is $625?

Mathematics
2 answers:
Mnenie [13.5K]3 years ago
7 0

Answer:

1 new and 5 used

Step-by-step explanation:

If we subtract multiples of 250 from 625 and divide by 75, we can determine the number of phones.  

We know that 250(2) = 500; if we subtract this from 625, this leaves

625-500 = 125 for the old phones.  Dividing this by 75, we see that it is not a whole number; this means there cannot have been two of the new $250 phones sold.

625-250 = 375.  Dividing this by 75, we get 5; this means 1 new phone and 5 old were sold.

dusya [7]3 years ago
5 0
5 used phones and one new phone 75 x 5 + 250 = 625
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3 years ago
Find the number of terms, n, in the arithmetic series whose first term is 13, the common difference is 7, and the sum is 2613.
siniylev [52]

Answer:

A

Step-by-step explanation:

Recall that the sum of an arithmetic series is given by:

\displaystyle S = \frac{n}{2}\left(a + x_n\right)

Where <em>n</em> is the number of terms, <em>a</em> is the first term, and <em>x</em>_<em>n</em> is the last term.

We know that the initial term <em>a</em> is 13, the common difference is 7, and the total sum is 2613. Since we want to find the number of terms, we want to find <em>n</em>.

First, find the last term. Recall that the direct formula for an arithmetic sequence is given by:

x_n=a+d(n-1)

Since the initial term is 13 and the common difference is 7:

x_n=13+7(n-1)

Substitute:

\displaystyle S = \frac{n}{2}\left(a + (13+7(n-1)\right)

We are given that the initial term is 13 and the sum is 2613. Substitute:

\displaystyle (2613)=\frac{n}{2}((13)+(13+7(n-1)))

Solve for <em>n</em>. Multiply both sides by two and combine like terms:

5226 = n(26+7(n-1))

Distribute:

5226 = n (26+7n-7)

Simplify:

5226 = 7n^2+19n

Isolate the equation:

7n^2+19n-5226=0

We can use the quadratic formula:

\displaystyle x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

In this case, <em>a</em> = 7, <em>b</em> = 19, and <em>c</em> = -5226. Substitute:

\displaystyle x  =\frac{-(19)\pm\sqrt{(19)^2-4(7)(-5226)}}{2(7)}

Evaluate:

\displaystyle x = \frac{-19\pm\sqrt{146689}}{14} = \frac{-19\pm 383}{14}

Evaluate for each case:

\displaystyle x _ 1 = \frac{-19+383}{14} = 26\text{ or } x _ 2 = \frac{-19-383}{14}=-\frac{201}{7}

We can ignore the second solution since it is negative and non-natural.

Therefore, there are 26 terms in the arithmetic series.

Our answer is A.

6 0
3 years ago
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