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qaws [65]
3 years ago
11

The admission fee at a county fair is $2 for children and $4 dollars for adults. Suppose that on the last day, 1600 people enter

the fair and $5000 is collected. Choose the two equations that can be solved as a system of equations to determine how many children and how many adults attended the fair.
1.) a-c=1600
2.) 4a+2c=5000
3.) 4a-2c=5000
4.) a+c=1600
Mathematics
2 answers:
tigry1 [53]3 years ago
7 0

<em>Answer:</em>

<em>2500 children and 625 adults could have attended on this day.</em>

<em>Step-by-step explanation:</em>

5000 divided by 2 is 2500, so 5000 - 2500 = 2500

2500 divided by 4 is 625.

--Emilie Xx

Darya [45]3 years ago
7 0

Answer:

The answer is 1 and 3

Step-by-step explanation:

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Morgarella [4.7K]

Answer:

26 + 10i is your answer.

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Law Incorporation [45]

Given the center (x_0,y_0) and the radius r of a circle, its equation is

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In your case, the center is (-4,0), and the radius is 2. Plug these values into the generic formula and you'll get the equation.

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3 years ago
A total of $12,000 is invested at an annual interest rate of 9%. Find the balance
sineoko [7]

Answer:

$18,726.11

Step-by-step explanation:

Lets use the compound interest formula provided to solve this:

A=P(1+\frac{r}{n} )^{nt}

<em>P = initial balance</em>

<em>r = interest rate (decimal)</em>

<em>n = number of times compounded annually</em>

<em>t = time</em>

<em />

First lets change 9% into a decimal:

9% -> \frac{9}{100} -> 0.09

Since the interest is compounded quarterly, we will use 4 for n. Lets plug in the values now:

A=12,000(1+\frac{0.09}{4})^{4(5)}

A=18,726.11

<u>The balance after 5 years is $18,726.11</u>

6 0
3 years ago
Lesson 6.2 math 7th grade
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What about lesson 6.2 math 7th grade??

5 0
3 years ago
Read 2 more answers
The sum of the first three terms in a GP is 38.Their product is 1728.Find the values of the three terms.
dusya [7]

Answer:

The value of the three terms is 8 and 18

Step-by-step explanation:

Let "a" be the first term and "r" be the common ratio.

Then from the condition, we have these two equations

   a + ar + ar^2  =   38,      (1)

   a*(ar*)*(ar^2) = 1728.      (2)

From equation (2),  a^3*r^3 = 1728,  or  (ar)^3 = 1728,   which implies

   ar = root%283%2C1728%29 = 12;          (3)    

hence,  

   r  = 12%2Fa.                   (4)

Now, in equation (1) replace the term  ar  by 12, based on (3).  You will get

   a + 12 + ar^2 =  38,   which implies

   a + ar^2 = 26.              (5)

Next, substitute  r = 12%2Fa  into equation (5), replacing "r" there.  You will get

   a + a%2A%28144%2Fa%5E2%29 = 26,   or

   a + 144%2Fa = 26.

Multiply by "a" both sides and simplify

   a^2 - 26a + 144 = 0,

   %28a-13%29%5E2 - 169 + 144 = 0

   %28a-13%29%5E2 = 25

   a - 13 = +/- sqrt%2825%29 = +/- 5.

Thus two solutions for "a" are  a = 13 + 5 = 18  or  a = 13 - 5 = 8.

If  a =  8, then from (4)  r = 12%2F8 = 3%2F2.

If  a = 18, then from (4)  r = 12%2F18 = 2%2F3.

   

In the first case, if a = 8,  then the three terms are  8, 8%2A%283%2F2%29 = 12  and  8%2A%283%2F2%29%5E2 = 18.

   In this case, the sum of terms is  8 + 12 + 18 = 38, so this solution does work.

In the second case, if a = 18,  then the three terms are  18, 18%2A%282%2F3%29 = 12  and  18%2A%282%2F3%29%5E2 = 8.

   In this case, the sum of terms is  18 + 12 + 8 = 38, so this solution does work, too.

ANSWER.  The problem has two solution:  

        a)  first term is 18;  the common difference is 2%2F3  and the progression is  18, 12, 8.

        b)  first term is  8;  the common difference is 3%2F2  and the progression is   8, 12, 18.

4 0
2 years ago
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